In that case you can use Euler formula exp(2kPI) where k€{...,-1,0,1,...} - iterate through k from 0 to 4 and proof, that rest of ks will be the same due to periodical nature
Solutions of x^n-1=0 : Consider the unit circle and the point with maximum x value. This is a solution, trivially. Now divide 360 degrees by n. Move along the circle counter clockwise and select points evyer (360/n) degrees. These are all solutions. To convert to complex numbers , z = (cos theta + i \sin theta) where theta = 360k/n for 0
But he literally finds the sine and cosine of those angles in surd form. For example, the 4 key values are: sin(36⁰ ± 18⁰) = (√5 ± 1)/4 sin(54⁰ ± 18⁰) = √(10 ± 2*√5)/4 The solutions use these values in ± form.
(Typos forgiven.) Exactly, and even taking time to explain this a little more clearly this solution require much less time than the half hour devoted to the video. Besides which this is easily generalized for xⁿ = y and still takes essentially constant time to accomplish. God help me if I should have to spend a half an hour on a single entrance exam problem.
If you are running a kebab "restaurant" or a gas station, then yes: you do not need to have anything in common with any university 😀 to be 100% honest when I saw this video graphics I knew the answer within 3s, but I would never propose such a solution! Chapeau bas to everyone who knows how to solve it without looking into any math books!
@@ashtheflyingjew9520 it is not useless - we wouldn't be writing to each other if not this kind of people who are able to solve complex mathematical problems. Someone will take a joy from running a gas station other from solving an absurd mathematical problems :) Both are needed :)
@@thedarthrage you won't be making money from studying in cambridge you will spend the rest of your life paying student debt, so yeah college is useless. On the other hand everyone needs food to survive so running a kebab restaurant is more useful and profitable.
This comment section be like: 1. “Isn’t it just 1?” 2. “This video is stupid” 3. Actual solutions, that are a bit harder to understand for people who just stumble onto this video.
If you really intend to get in to Cambridge, you probably ought to know that exp(2kπi) = 1 where k ∈ ℤ. Then x^5 = exp(2kπi), giving the five solutions as x = exp(2kπi/5) where k = { 0 ... 4 }. The trivial solution is when k=0, giving x = 1. The principal root is therefore exp(2πi/5) = cos(2π/5) + i.sin(2π/5), and the other three are cos(4π/5) + i.sin(4π/5), cos(6π/5) + i.sin(6π/5) and cos(8π/5) + i.sin(8π/5). The time you save (about 23 minutes, apparently) will allow you to do several more trivial questions.
The solution on the video has a nice algebraic approach which is also useful and applicable to other algebra problems. Going for the complex exponential solution requires a whole background on the complex plane in order to render the final solution simple. Both solutions are useful in different moments of math instruction.
@@luisrocha26 The "nice" algebraic approach works in this case because we can solve quartic equations. Have a think about how you might extend the technique to the six roots of x^6 = 1 or the seven roots of x^7 = 1, etc. You may conclude that the complex exponential is a more general solution. I find the particular solution in the video interesting because it allows us to introduce the complex method and then check the results against a purely algebraic method. But beyond that, I'd be hard pressed to recommend that students should consider it. YMMV.
@@RexxSchneider I'm not talking about being able to generalize the solution nor demonstrate more complex things; it's useful as as didactic exercise on algebra
But he found the sin and cos of those angles in surd form. You're stuck with answers such as cos(72⁰) + i*sin(72⁰) and have to figure out how to go further.
@@oahuhawaii2141 But 99% of the time, nobody needs to go further. The trig form is an exact form, just like the surd form. We find that cos(2π/5) is generally just as useful as (√5 - 1)/4. To use it we're almost always going to use a numerical approximation, and either form will give the same approximation, about 0.309. I concede that if there is a problem where the result has to be multiplied by (√5+1), you can get a simplification, but those cases must be vanishingly rare. And finally, the question never specified that an answer in surd form was required. I assure you that when I took my Cambridge Scholarship exams in 1968, I never wasted time trying to answer questions I wasn't asked.
Wisdom is learned knowledge. Wisdom is knowing there are 5 solutions and finding them, because it's your best shot at getting into the university. Wisdom is knowing that, although I might not need them today, there are plenty of fields (engineering, wave mechanics, quantum mechanics, computing, stats etc) that use complex numbers, so these solutions do carry value.
@michaellee7313 lmfao no wisdom is the practical application of knowledge thats really it. You can know a million random facts and have minimal wisdom
@@michaellee7313 As the problem does not specify that x is a complex number, I would consider it wiser to just write "Assuming that x is a real number, because the problem did not specify otherwise." and solving the problem by writing "x=1". Did you solve the problem the way it was meant to be solved? No. Did you put in the extra effort of solving it the hard way? No. Did you show everyone that you are smart enough to recognize that it technically wasn't required to solve the problem in the complex plane? Yes.
@David-pz4gy we're told that this is a university admissions exam. If you don't solve for all solutions, they will assume that you can't find the complex solutions and that will never go in your favour. Recognising that the reason that you ate solving this problem is to show off your ability and therefore solving for all complex solutions is ALWAYS going to be your best bet. Anyone who has studied maths to this sort of level would understand that you aren't getting a college place just because you can say x=1 by inspection Edit typo
x⁵ = 1 so 5 roots. Angle between roots will be 360º / 5 = 72º. So, answers are: (1, 0º), (1, 72º), (1, 144º), (1, 216º), (1, 288º). EDIT: "Cheap way out?" Agreed. Why make things more difficult than they need to be? EDIT2: "How did u guess they'd have the same distance?" They always do! Look up: "Multiplying Complex Numbers in Polar Form" You multiply the magnitudes but add the angles. So, for a fifth root you need a set of angles that when multiplied by 5 will always end up with the same answer. So: 5 * 0º = 0º 5 * 72º = 360º or 0º 5 * 144º = 720º or 0º 5 * 216º = 1030º or 0º 5 * 288º = 1440º or 0º EDIT3: Anyone who does not think the answers are evenly spaced round the circle should look carefully at 24:22. Look carefully at the red dots and ignore the other lines (axes). They are evenly spaced.
@@autumn948 Irrelevant. They are still correct and root is still a root, no matter the representation you use. Radial coordinates can exist in ANY form of radial system be it degrees radians or the gradian system.
The thing I don't understand is why do you keep explaining basic math (such as "-x+5 = 5-x") and spend several minutes to rewrite a bunch of obvious things when you're explaining a college level problem? It's like explaining how to build a rocket and then you spend half an hour talking about how the screwdriver works.
Is really funny for you to say that, because I don't know why this appeared in my feed, I was literally just watching a Minecraft video😂. All of a sudden I was here, watched half video and didn't understand nothing jajaj (I'm in Uni, I'm cooked).
@@suryanshvarshney111 "high school" level and "college" level are really subjective as it really isn't one type of math is higher than the other. I've taken Linear Algebra and wasn't required to take one statistics class. So saying this is grade 8th 12th or higher isn't a strong argument.
@@Marinealver i call bs, there clearly are levels to mathematics. You wouldn't go straight to learning Stochastic processes when you don't know shit about basic probability and statistics. And no shit you don't need stats for LA 🤦🏻♂️ two topics can lie on same level. So, no high-school level and college level isn't subjective. It literally implies that these topics were taught in high-school so you should know it🤦🏻♂️ imagine coping so hard that you try to say college level and high school level is subjective 😭🙏🏻
Instead of doing this long way you can just rewrite it as x^5 = 1, which means x represents the fifth roots of unity. By expressing 1 in polar form as e^(2πi k) for k = 0, 1, 2, 3, 4, we can take the fifth roots to get x = e^(2πi k / 5). This gives the five solutions: x_k = cos(2πk / 5) + i sin(2πk / 5) for k = 0, 1, 2, 3, 4, representing points evenly spaced on the unit circle.
Anyone who knows complex numbers well can just write down all 5 solutions. But k only ranges from 0 to 4, any other integer duplicates one of those answers. (That is, if m = n (mod 5) then e^(2m𝜋i/5) = e^(2n𝜋i/5).)
Yep, so solutions would be 1; 1e^j2π/5; 1e^j-2π/5; 1e^j4π/5; 1e^j-4π/5 with k = -2, -1, 0, 1, 2. Didn't even watched the video just jumped from front picture to comment section.
The idea is to provide the exact solution without recourse to the trigonometric forms. This also gives you a slick way to find sin(2Pi/5) and cos(2Pi/5)
@@paulbritt588 The trigonometric form is just as exact and usually more useful. The main utility of complex numbers is to describe rotation, and since this is also on the unit cicrle, it's even a subgroup of U(1) (under multiplication). In fact, this subgroup represents the same abstract group as Z5.
@@haakoflo I am aware of the complex solution and the ease with which it solves the problem. I think that the displayed algebraic solution is shown, as it would appeal more to a high school student who may not have seen the trigonometric method.
Cambridge exam questions are typically not difficult but there are too many of them in the paper to pass if you spend a lot of time on each. This is a good example, solve laboriously as in the video in 25 minutes and you may get the right answer but no time left to answer enough questions to pass. So the exam is actually testing your depth of understandingand ability use it to see the fast paths to solution. ( which is the euler relation shown by others here. 20 seconds, boom, done.)
Perhaps you are smarter than I, but I would have to disagree. Some of those questions really are just truly hard. This level of question isnt usually present.
Everyone saying this method takes too long. Sure on an exam you'd use a different method, but can we appreciate that he showed how to get exact values for trig functions analytically? I think that's neat
I appreciate you working through all the steps so methodically. It kept me from getting lost. Also, thanks for calling out all the formulas and identities you used by their proper names - it helps me look them up becasue I need to go reference them again.
I've posted something like this before. You want to go to your neighbor's house, which is 100 feet to your west. You could walk 100' and arrive at your destination. OR you could travel East the entire diameter of the earth minus that 100'. ALL TOO OFTEN, the solutions shown take the long way to get there.
What does the "traveling" metaphor" means in relationship to math? You have a very naïve understanding concerning what solving a math problem means. Your 100' solution will give only a 1/5 of a correct score.
@@harrisorourke6926 obviously, he could have walked in expanding circles around his house that follow the fibonacci sequence until there is an interception with the backwards traced fibonacci sequence radiating from the neighbour which he then walks backwards. so naive
Except when you are in a polar Region, a bit south of the northpole where he w way eastwards May be the same considered he ist exactly on the opposite Side 180 degrees. Sadly o do Not know how far He must be south of the northpole. IT reminds me of a Joke when a carholic priett ate meat and someone reminded him it was friday, when eating meat IS forbidden for catholics. The priest walke some distance and remarked "Here IT IS still thursday!"
But he found the sine and cosine of angles n*2*π/5 in surd form. You're stuck with answers such as cos(72⁰) + i*sin(72⁰) and have to figure out how to go further.
yes @@oahuhawaii2141 I am not an advanced student. I was fascinated and enlightened by the clever substitution that revealed a 4th degree equation as two "nested" quadratics. Even more satisfying were the algebraic expressions in the respective real and imaginary parts for the sines of the various angles n*18° or n(pi/10). Clearly, algebraic expressions for the trig functions are encoded in the complex roots of specific polynomials generally. More elegant ways to calculate them probably exist, such as whatever algorithms likely programmed into calculators, but this blows my mind. I'm old, so my high school math texts contained appended trig and log tables, both rendered to 4 significant digits probably. I may have imagined that some monastary dwellers had produced these in the 16th century using prohibitively laborious methods. I pulled a shit ton of inspiration and knowledge from this vid.
@@oahuhawaii2141From most cases I've seen is that if you have a angle that does not have a exact value like 72° you don't have to spend extra time rewriting it.
@@Apple-vm5gc: I've looked at a few textbooks, and sin(72⁰) isn't computed in surd form. Even past CRC tables didn't have sin(n*18⁰). Only 0⁰, 15⁰, 30⁰, 45⁰, 60⁰, 75⁰, and 90⁰. I did all multiples of 3⁰.
If you know polar coordinates, the solution would be more simple and beautiful. By knowing the only real solution (1), the complexes would be calculated just dividing 360 degrees by 5. So, the other four solutions are 1
@@boringsupernova4836 it is. In north america it's common to drop the s from maths, and in the UK they say maths ... one of those "no wrong answer but some people want to sound pretentious" and feign superiority. :D I do it sometimes too.
Depending on the time allotted for this, I would agree that the trig and complex exponential version "answers" for the complex roots might not get you far at Cambridge - as a mere demonstration of rote learning. Another approach is to notice that the roots are all unimodular - that is: taking x=a+i*b, (a & b Real), a^2+b^2=1. Substituting this in the quartic equation for x, we know both the Real and Imaginary parts must vanish. The imaginary part is i*b*(4*a^3+3*a^2+a*(2-4*b^2)-b^2+1)=0. As b is nonzero for roots of interest (we know from the monotonicity of x^5 there aren't other real roots), we can ignore the overall factor of b, so substituting in b^2=1 - a^2 elsewhere, and extracting a common factor of a (joining the b since a is nonzero too) leads to a quadratic for a : 4*a^2+2*a-1=0, roots of which are -1/4 +Sqrt[5]/4, and -1/4 - Sqrt[5]/4, and the corresponding Imaginary parts can be straightforwardly calculated (from Sqrt[1 - a^2]). Whether this is "cleverer" than the "completing the square" approach presented might be a matter of taste - a sharper eye there versus using more knowledge of the complex roots? Another connection with the presented solution is that for a unimodular complex number x, x +1/x is twice the real part, which is why the solutions above for a are half the presenter's "t" roots, indeed could use this instead of solving the second quadratic where the complex numbers first arise in the presentation. Of course for completeness one might wonder about the Real part of the x-quartic: after substituting for b^2 one gets a quartic for a - with four real roots, the pair above and two more (+1/Sqrt[2[, -1/sqrt[2]) that don't make the Imaginary part vanish - must confess that took the edge off a little.
Note that if we set x =2π/5, we can use the fact that cos(4π/5) = cos(2π - 4π/5) = cos(6π/5) to get the relationship cos(2x) = cos(3x). We can then use the Pythagorean identities. Hence cos(2x) = 2cos^2(x) - 1 is equal to cos(3x) = 4cos^3(x) - 3cos(x). If we set c = cos(x), we now have 2c^2 - 1 = 4c^3 - 3c. Therefore 4c^3 - 2c^2 - 3c + 1 = 0. A factor is c=1 giving: (c - 1)(4c^2 + 2c -1) = 0. The c=1 is when x=0, which also satisfies cos(2x) = cos(3x), so we just need to solve the quadratic, giving c = (-2 ± √20)/8 = (-1 ± √5)/4. Since x is in the first quadrant, we can see that cos(2π/5) = (-1 +√5)/4) which is also the value for cos(8π/5). We can see that cos(4π/5) = cos(6π/5) = (-1 - √5)/2. Similarly for sin(2kπ/5).
Decades ago, my 9th grade teacher gave a bonus trigonometry problem to find sin(θ) in surd form, for θ = 18⁰ . I solved it at my desk for sin(θ), cos(θ), tan(θ), cot(θ), sec(θ), csc(θ) . Nobody else in class figured out how to do it. At home, I did the same for θ = n*18⁰ , n ∈ ℤ . The next day, I showed my teacher the table, and said I'll work on θ = n*3⁰ during the weekend. I used lots of scratch paper for those calculations. When I showed her my completed table, she asked me to join the math team. I did and she gave me a CRC math book. The first thing I did was look at its trig section, which I saw only had θ = n*15⁰ . I submitted a copy of my table to the publisher. I never got a response.
But he found the sine and cosine of angles n*2*π/5 in surd form. You're stuck with answers such as cos(72⁰) + i*sin(72⁰) and have to figure out how to go further.
@@oahuhawaii2141 you can use trigonometric identities to find em , also considering a regular polygon with five sides(pentagon) whose circumcircle has radius one with its centre at origin will help .
Very good you got one of the five solutions to the problem in 15 seconds. What took you so long? However apparently you did not notice that the problem presents an exponent of 5 which leads to a solution with 5 answers.
@@robert.eduard Not imaginary, complex. Complex numbers are pretty essential to a lot of practical applications, so being able to find complex roots to an equation is significant. "Imaginary" is an outdated term for complex numbers.
So it's basically finding 5th root of 1, but for complex numbers which have the general form: r*exp(i*x) It's clear that r has to be 1 (a non-negative real number that has absolute value 1 when raised to the 5th, that is just 1). So you're really just looking for a number of the form: e^(ix) But when you raise it to 5: e^(i*(5x)) That means x must be an angle that multipled by 5 gives you 2pi. So x = 2pi/5 (call it the base case X1) But whenever you sum a whole turn, you get to the same spot, so all the answers would be: X1= 2pi/5 X2= 2pi/5 + 2pi/5 = 4pi/5 X3= 4pi/5 + 2pi/5 = 6pi/5 X4= 6pi/5 + 2pi/5 = 8pi/5 X5= 8pi/5 + 2pi/5 = 2pi "=" 0 (X5 is the real root, that's when you stop because x is confined to [0,2pi), doing this again will give you the first root ) This is how i would solve it. The roots would be X1= exp(i * 2pi/5) X2= exp(i * 4pi/5) X3= exp(i * 6pi/5) X4= exp(i * 8pi/5) X5= 1 Verifying this is the same as the video is left as an exercise...
This is exactly what makes the difference between students in these exams, if you can come up with a solution that can help you solve the problem much more quickly than the intuitive solution as shown in this video, you can then answer much more questions.
But he found the sine and cosine of angles n*2*π/5 in surd form. You're stuck with answers such as cos(72⁰) + i*sin(72⁰) and have to figure out how to go further.
x1 = 1 ; x2, x3 = cos ( 2 * PI / 5 ) + / - i * sin ( 2 * PI / 5 ); x4, x5 = cos ( 4 * PI / 5 ) + / - i * sin (4 * PI / 5 ) - tthe primary values. + all 2 PI rotations of these roots. Where 2 * PI / 5 ~ 72 degrees, 5 * 72 degrees = 360 degrees.
But he found the sine and cosine of angles n*2*π/5 in surd form. You're stuck with answers such as cos(72⁰) + i*sin(72⁰) and have to figure out how to go further.
What? x = 1^(1/5) = exp(2ni*pi/5), n = 0; 1; 2; 3; 4 because other integers give the same numbers. So x_1 = 1; x_2to5 = exp(2ni*pi/5) = cos(n*72°) + i*sin(n*72°), n=1; 2; 3; 4
There's 5 complex solutions because of the fundamental theorem of algebra. The real solution is obviously x=1. The complex solutions are all the rotations of n/5 of the unit circle in the complex plane. So e^(2n/5πi) with n equal to 1, 2, 3, 4 and 5. Notice that when n=5 this reduces to 1. Funnily enough we could let n be any integer and it would reduce to the same answer as n mod 5.
Simple. Draw a unit circle centered on the origin of the complex plane. Mark 5 points on the circle at 72° intervals starting on the real axis at x=1. What do I think about the answer? All that is needed is to know how to evaluate sine and cosine of 72°, 144°, 216°, and 288°.
He takes 24 minutes to solve a problem that can be solved in 30 seconds if you do it the proper way using de Moivre theorem's, which every pre University student should be familiar with
Specialist Maths in the Victorian (Australia) curriculum includes teaching students how to do operations with complex numbers in both Cartesian and Polar form. De Moivre's is bread and butter here.
I personally prefer the visual approach to this problem. First transform to x^5 = 1. There is a way to look at complex numbers as the vectors from 0 0 to (complex part, imaginary part) Then because multiplying something in the complex plane means multiplying the "lenght" of the number, and adding the "angles" of the number to gether First i know the len of all solution vectors is 1 Then i i know that the angle of all solution vectors is 0° -> 5 × n = 0° => the 5 solutions have the angle 0/5×360 1/5×360 2/5×360 3/5×360 4/5×360 And by knowing the lenght and angel of all solution vectors i can calculate theire real part with cos, and imaginary part with sinus
I guess in other countries, not only Croatia where i am from, highschoolers are tought about the de Moivre formula (or atleast Eulers formula) , my God! Then it is trivial.
@@philippelepage1609 This was not mentioned at the beginning. Therefore the fastest and most elegant way might be preferred. Also, using polar coordinates you can also solve similar equations with larger exponents easily, like x^10-1=0.
Why just don't you use complex numbers and solve fast. Entrance examiner will give you zero for this step they want you to solve it using complex number which is way faster.
True. This should take the average highschool student who knows about complex numbers less than 2mins to solve, using polar representation of complex numbers.
Me, too. I've seen this guy before, giving ridiculously complex solutions to some very simple problems. Any first day algebra student would say, "What minus 1 is zero? Okay, 1-1 is zero. So x^5 = 1. Well, 1 to any power is 1, so the answer is 1. Duh."
@@Astrobrant2 You are doing naïve math. Actually you are not doing math. The solution requires determining the 5 solutions to the equation. If you submitted this on a homework assignment you might get partial credit and definitely a note from the instructor that says, "See me about this".
@@Shri For someone with any math background, this is still extremely trivial when using polar coordinates. Took me about 0.5 seconds to see the pentagon represented by this equation. Or to unfold it as the Z5 cyclic group.
@@Anu_here_148Yes, but there are actually 4 more roots, that are all complex numbers. If you rotate the complex number ("point") "1 + 0*i " on the unit circle 72 degrees or 2 * PI / 5 radians , you will end up in the complex number "cos ( 2 * PI / 5 ) + i * sin ( 2 PI / 5 )", where i is the imaginary unit. This is pr. definition e^( i *2 * PI/5 ), since the "length" or modulus rather is 1 ( we are on the unit circle in the complex plane ;-) ). Raising this complex number to the fifth power then gives us ( rotating it to the argument ( angle ) that is 5 times larger ): z^5 = ( e^( i *2*PI/ 5 ))^5 = e^( i * 2*PI /5 * 5 (!!!!!) ) = e^( i * 2PI ) = cos(2*PI) + i *sin( 2*PI) = 1 + 0*i = 1 And we are back to 1 ;-) This also applies to the symmetric complex roots at - 72 degrees, and +/- 144 degrees on the unit circle in the complex plane - and to all full 360 degrees ( or 2*PI ) rotations of these 4 primary roots.
@@Bjowolf2, yeah and aren't we allowed to consider that the real reason for more than one root x = 1 is , that the equation is a fifth grade one . so there should be five roots as a full solution .
@@keescanalfp5143 Yes, precisely - a n'th order polynomium always has n roots ( complex or real, including multiple roots ), and if the coefficients of the polynomium are all real (as in this case!), then any complex roots will always appear as complex conjugated pairs ( symmetrically around the real axis in the complex plane ) 😊
This is a perfect example of why I hated math in school so much. The teachers were never interested in actually helping their students learn math. They were mostly interested in showing everyone how smart they were by intentionally overcomplicating everything and speaking in a language that they knew none of us understood so that we would bow down in awe of their massive brains. It wasn't until I got to university that I finally met a humble math teacher who was genuinely interesting in helping me understand.
But he found the sine and cosine of angles n*2*π/5 in surd form. You're stuck with answers such as cos(72⁰) + i*sin(72⁰) and have to figure out how to go further.
The function has only one real root x = 1. For complex roots there was some theorem (I'm to old to remember it :D ) I think the guys in Cambridge would want the person to know the theorem and use it straight away instead of doing algebra. The test should check whether the person knows something more than algebra.
They could use abstract algebra to solve this, and say that the solution is the numbers 0,1,2,3 and 4 in the cyclic Z5 group, just represented on a homomorphic representation of the same abstract group on the complex unit circle :) That MIGHT have scored full points, especially if the homomorphism was accompanied by an illustration.
As someone who went to MIT and is a contrarian both solution are valid. Actually any solution you can properly present is valid for solving the problem and can't be downgraded. This is generally true for all university exams and test makers generally make so you don't have enough able time to try algebraic solutions instead of theorems with easier solutions. In this case the theorem you would use is the "De Moivres". I would do algebraic solutions anyway because I am really fast at mental math and the teachers would always complain =D I was kind of able to get away because people still need enough time to try different theorems until see witch one solves the problem with an easy solution but you can just brute force. It always work.
1, i^4/5, i^8/5, i^12/5, i^16/5 Nailed it. Joking aside, there's 3 factors that determine how I should answer: - What major am I applying for? - What is the question? - What situations will my answer be applied to?
Yes! It's just 1! Why don't just x=1??? But if you want, there is a better solution. :) If 𝑥5 = 1, we are looking for values of x that satisfy the equation: x 5 =1. This equation has five complex roots, known as the fifth roots of unity. These roots are located on the complex unit circle, meaning they lie on a circle with a radius of 1 centered at the origin, and are evenly spaced in angle. We can find these roots using Euler’s formula: x k =e 2πik/5 for k=0,1,2,3,4, where 𝑖 is the imaginary unit, and 𝑘 represents the root index. Thus, 𝑥 can take any one of these five values. One of these is the real number 1, while the other four are complex numbers evenly distributed around the unit circle in the complex plane.
Z^5 = cos theta +isin theta, using de moivres, z = cos theta/5 + isin theta/5 theta = 2*k*pi substitute k = 0,1,2,-1 into z to get the final four solutions aswell as z = 1
So much easier to solve with complex polar coordinates and Euler's equation. x^5=e^(5*i*t)=1. We need to find a "t" such that our angle in the complex plane is 2*pi*(an integer) so that our value of x^5 wraps around to being on the +real axis. 0 works obviously (x=1 solution), 2pi/5 works (quadrant 1 solution in the complex plane), 4pi/5 (quadrant 2), 6pi/5 (quadrant 3), 8pi/5 (quadrant 4), then 10pi/5 brings us back to 2pi (which is equivalent to the t=0 solution). x = r*e^(i*t), with r=1, and e^(i*t) = cos(t) + i*sin(t). Each of our values of t above give us our different solutions (including t=0 which leads to x=1*(1+0i)).
To simply put it for those who do not fully understand the math they’re referring to is in this video is for exploring variants, not the solution. they are not asking for the answers to the question they are asking for the variants to the answer of the questions.
On the complex plane, draw a circle of radius 1. The points in the circle that make angles to the horizontal of 72°, 144°, 216°, 288° and 0° will be your solutions.
Complex number theory is in college for us. I assume we go over things that you probably don't in high school and vice versa. I know that some high schools go over complex number theory but not mathematical induction and structural induction (to my surprise).
Thank you for watching. A great question today x^5 - 1 = 0 (Finding all roots) Have a great day and take care! Wish you all the best in you life and career❤❤❤
It's great that you found sin(θ) and cos(θ) in surd form for θ = k*2*π/5 = k*72⁰ , k ∈ ℤ . Everyone knows the values when k = 0 , but few know them for k = ±1, ±2 . Most are stuck with answers such as cos(72⁰) ± i*sin(72⁰) and have to figure out how to go further. The 4 complex solutions use the following key values in ± form: sin(36⁰ ± 18⁰) = (√5 ± 1)/4 sin(54⁰ ± 18⁰) = √(10 ± 2*√5)/4
The easiest solution is dividing the unit circle into five equal pie pieces. Each piece has a 72 degree tip. So the first solution is x=1. The second is x=cos 72 + i sin72 and so on.
ok but why not just use the polar form and Eulers identity x^5 - 1 = 0 x^5 = 1 x^5 = e^i(0+2kπ ) x = (e^i(0+2kπ))^(1/5) x = e^i(2kπ/5) k=0 ; x = 1 k=1 ; x = e^i(2π/5) k=2 ; x = e^i(4π/5) k=-1 ; x = e^i(-2π/5) k=-2 ; x = e^i(-4π/5)
a trick to this question is: draw a circle (radius=1) in the argand-gauss plane and inscribe a pentagon in it, each one of the vertices will be one of the solutions
You can understand it by considering the phase of the complex numbers with the exponential representation. Solving this equation is equivalent to find the phases for which when it is multiplied by 5 you get the phase equal to 2pi (rad) because the phase of the complex number 1 is 0 or 2pi. So first trivial solution is x=1. Second solution is the complex number which phase equal to 2pi/5 because 2pi/5×5=2pi. Third solution is the complex number which phase is equal to 2×2pi/5 because 2×2pi/5×5=2×2pi => same complex number =1. And then the same explanation you get x=3×2pi/5 and x=4×2pi/5 which are the 5 only possible solutions because then if x=5×2pi/5 and x=6×2pi/5 it is already the first and second solutions. CQFD 😊 (without to solve any complex equations I found the complex solutions 🤭)
1. x = 1 2. x = e ^ (i * (2pi)/5) 3. x = e ^ (i * (4pi)/5) 4. x = e ^ (i * (6pi)/5) 5. x = e ^ (i * (8pi)/5) These can also be expressed in terms of cosine and sine: 1. x = 1 2. x = cos((2pi)/5) + i * sin((2pi)/5) 3. x = cos((4pi)/5) + i * sin((4pi)/5) 4. x = cos((6pi)/5) + i * sin((6pi)/5) 5. x = cos((8pi)/5) + i * sin((8pi)/5)
X⁵-1=0 X⁵=1 X=⁵√1 X=1 as ever root and potentiation of 1 is 1. If i write X⁵-2= 0 there's no results as 2 don't have a ⁵√ sp there's no result, bu if i chose a number with a ⁵√ such as 32 it will as 2*2*2*2*2 is 32, so 2⁵=32 So x⁵-32=0 X⁵=32 X=⁵√32 and so on If you change the x for x³ you just implies that if the results is also 0 so the subtracted number must be ³√ such as 8 or 27 and so on
if u divide in 5 (bcz the x to the 5) the 1-unit circle centered in 0, you have the 5 solutions. 360/5 = each 72 degrees. x1 = cos(0°) + isen(0°) x2 = cos(72°) + isen(72°) x3 = cos(144°) + isen(144°) x4 = cos(216°) + isen(216°) x5 = cos(288°) + isen(288°)
We can do simple algebraic manipulation: Add 1 to both sides and we end up with: x^5 = 1. What multiplied by itself 5 times is equal to 1? Since the power itself is ODD it cannot be -1 since -1 * -1 * -1 * -1 * -1 = -1 AND -1 Does Not Equal 1. It must be 1 since 1^5 = 1. This is the only answer within the Real Numbers. If we move to the Complex plane and we consider modulus arithmetic and the possibility of All Roots, then we can use a few various techniques either through vector notation, simple rotations around the unit circle, etc. and from there we can obtain the other 4 roots of unity that only exist in the complex plane. For example: we know that sqrt(-1) is the imaginary unit i. We know that i^2 = -1. Since i^2 = -1 then i^4 = 1 since -1 * -1 = 1. Thus, i^4 is also a root. In other words, 1^5 and i^4 evaluate to the same thing as they are equivalent. The rest is up to you to figure out.
For this specific problem there is an easier way by decomposing the result in an angle and radius in the complex plane. Then computing that taking an exponent is taking an exponent of the radius an multiplying the angle (r exp(i theta))^n is r^n exp(i theta n). The radius is then 1 and the resulting angle need to be 0 modulo 2pi. This give us 2 pi k/n. With k from 0 to n-1. Set n to 5 for this specific problem.
The main thing this made me think is that this is how any given event can be balanced throughout all dimensions even in a manifold which is more complete than 4 dimensional spacetime, but that is because I am generally thinking about the more complete dimensional manifold of actual spacetime and quantum dimensions outside of of conventional understanding. The circular graphical representation of all solutions made me think about the potential that imaginary solutions might actually give me the mathematical clue I'm looking for to describe my 5 dimensional spacetime with its extra cosmological scale spatial dimension and the full manifold including the additional quantum scale dimensions, also plural or singular because I haven't seen enough yet to be fully confident of the full manifold spectrum and framework. I know they are there, but I don't know their full shape yet.
The roots must be of unit length, so either positive angle which lands on one when raised to the fith power, or 0 angle which is the real root, - 1/5th 2pi =4/5th 2pi in terms of angle so only 5 roots all unit lenght complex numbers of the form cos(n 2pi/5)+ i*sin(n 2pi/5) n; 0, +-1,+-2. Thats doable in the head.
Also if you know complex multiplication, you know the same works when the constant is not unit length, you just need to look at the fifth root of the lenght of the constant, to get the initial magnitude of the complex and real solutions. And add angles and multiply lenght to find the roots, not too complicated.
I was about to write a comment about you spending 20+ minutes on a simple solution, however I didn't know that complex solutions with imaginary numbers have real applications in several fields, so I will like to thank you for the video. It was a great review of my times at school
Prove that x4+x3+... +1>0. Split into x4+x2 which is 0 or grater and the rest are always positive (proof with determinant etc). So 0+positive= positive q. E. D.
![The most beautiful equation in math, explained visually [Euler’s Formula]](http://i.ytimg.com/vi/f8CXG7dS-D0/mqdefault.jpg)
1
Oh you also want the complex solutions.
Ya it's simple
X⁵ - 1 = 0
X⁵ = 1
⁵√x⁵ = ⁵√1
x = 1
@@Zomsteve theres more to it than that
@@Zomsteve that's a solution.
What it isn't is ALL the solutions.
I mean, you have to find x, so you want all the solutions. If a equation have fifth grade that means that have 5 solutions.
In that case you can use Euler formula exp(2kPI) where k€{...,-1,0,1,...} - iterate through k from 0 to 4 and proof, that rest of ks will be the same due to periodical nature
Why don't just x=1
Because an equation of the Nth degree has N solutions.
Its too easy for the Mathematics, they do not like Pragmatism. 😉
Because complex numbers exists
@@itzsoweezee9980😂😂
Because this is that 'new algebra'. Nothing is objective.
You see the answer here is; Strudel.
Don't ask.
In ten minutes, the answer will be cupcake.
Solutions of x^n-1=0 : Consider the unit circle and the point with maximum x value. This is a solution, trivially. Now divide 360 degrees by n. Move along the circle counter clockwise and select points evyer (360/n) degrees. These are all solutions. To convert to complex numbers , z = (cos theta + i \sin theta) where theta = 360k/n for 0
de moirve theorem
But he literally finds the sine and cosine of those angles in surd form. For example, the 4 key values are:
sin(36⁰ ± 18⁰) = (√5 ± 1)/4
sin(54⁰ ± 18⁰) = √(10 ± 2*√5)/4
The solutions use these values in ± form.
(Typos forgiven.) Exactly, and even taking time to explain this a little more clearly this solution require much less time than the half hour devoted to the video. Besides which this is easily generalized for xⁿ = y and still takes essentially constant time to accomplish. God help me if I should have to spend a half an hour on a single entrance exam problem.
@@DaneBrookea
Indeed... The longest n-root ever seen. If I had to review this exam I would consider it as failed. What a waste
The answer is either:
1. Don't go to Cambridge.
2. Never hire someone who went to Cambridge.
If you are running a kebab "restaurant" or a gas station, then yes: you do not need to have anything in common with any university 😀 to be 100% honest when I saw this video graphics I knew the answer within 3s, but I would never propose such a solution! Chapeau bas to everyone who knows how to solve it without looking into any math books!
@@thedarthragewhat the hell does chapeau bas mean talk like a normal person smartass
@@thedarthragewho cares it's useless anyways, also I'm willing to bet that a kebab restaurant owner makes more money than a maths professor.
@@ashtheflyingjew9520 it is not useless - we wouldn't be writing to each other if not this kind of people who are able to solve complex mathematical problems. Someone will take a joy from running a gas station other from solving an absurd mathematical problems :) Both are needed :)
@@thedarthrage you won't be making money from studying in cambridge you will spend the rest of your life paying student debt, so yeah college is useless. On the other hand everyone needs food to survive so running a kebab restaurant is more useful and profitable.
This comment section be like:
1. “Isn’t it just 1?”
2. “This video is stupid”
3. Actual solutions, that are a bit harder to understand for people who just stumble onto this video.
He wants the real and complex solutions
Didn’t say that
@@peterexner5979 who asked?
How is it called the identity?
Do think that the creator could say a word or two as a introduction
If you really intend to get in to Cambridge, you probably ought to know that exp(2kπi) = 1 where k ∈ ℤ.
Then x^5 = exp(2kπi), giving the five solutions as x = exp(2kπi/5) where k = { 0 ... 4 }. The trivial solution is when k=0, giving x = 1.
The principal root is therefore exp(2πi/5) = cos(2π/5) + i.sin(2π/5), and the other three are cos(4π/5) + i.sin(4π/5), cos(6π/5) + i.sin(6π/5) and cos(8π/5) + i.sin(8π/5).
The time you save (about 23 minutes, apparently) will allow you to do several more trivial questions.
The solution on the video has a nice algebraic approach which is also useful and applicable to other algebra problems. Going for the complex exponential solution requires a whole background on the complex plane in order to render the final solution simple. Both solutions are useful in different moments of math instruction.
@@luisrocha26 The "nice" algebraic approach works in this case because we can solve quartic equations. Have a think about how you might extend the technique to the six roots of x^6 = 1 or the seven roots of x^7 = 1, etc. You may conclude that the complex exponential is a more general solution.
I find the particular solution in the video interesting because it allows us to introduce the complex method and then check the results against a purely algebraic method. But beyond that, I'd be hard pressed to recommend that students should consider it. YMMV.
@@RexxSchneider I'm not talking about being able to generalize the solution nor demonstrate more complex things; it's useful as as didactic exercise on algebra
But he found the sin and cos of those angles in surd form. You're stuck with answers such as cos(72⁰) + i*sin(72⁰) and have to figure out how to go further.
@@oahuhawaii2141 But 99% of the time, nobody needs to go further. The trig form is an exact form, just like the surd form.
We find that cos(2π/5) is generally just as useful as (√5 - 1)/4. To use it we're almost always going to use a numerical approximation, and either form will give the same approximation, about 0.309.
I concede that if there is a problem where the result has to be multiplied by (√5+1), you can get a simplification, but those cases must be vanishingly rare.
And finally, the question never specified that an answer in surd form was required. I assure you that when I took my Cambridge Scholarship exams in 1968, I never wasted time trying to answer questions I wasn't asked.
Intelligence is knowing the answer is -1/4-(5**1/2)/4-i*(5/8-(5**1/2)/8)**1/2
Wisdom is recognizing the answer is 1
Wisdom is learned knowledge. Wisdom is knowing there are 5 solutions and finding them, because it's your best shot at getting into the university. Wisdom is knowing that, although I might not need them today, there are plenty of fields (engineering, wave mechanics, quantum mechanics, computing, stats etc) that use complex numbers, so these solutions do carry value.
@@michaellee73131 , 1 - part 2 , 1 again , 1 - reboot and 1 limited edition.
That's 5.
@michaellee7313 lmfao no wisdom is the practical application of knowledge thats really it. You can know a million random facts and have minimal wisdom
@@michaellee7313 As the problem does not specify that x is a complex number, I would consider it wiser to just write "Assuming that x is a real number, because the problem did not specify otherwise." and solving the problem by writing "x=1".
Did you solve the problem the way it was meant to be solved? No.
Did you put in the extra effort of solving it the hard way? No.
Did you show everyone that you are smart enough to recognize that it technically wasn't required to solve the problem in the complex plane? Yes.
@David-pz4gy we're told that this is a university admissions exam. If you don't solve for all solutions, they will assume that you can't find the complex solutions and that will never go in your favour. Recognising that the reason that you ate solving this problem is to show off your ability and therefore solving for all complex solutions is ALWAYS going to be your best bet. Anyone who has studied maths to this sort of level would understand that you aren't getting a college place just because you can say x=1 by inspection
Edit typo
x⁵ = 1 so 5 roots.
Angle between roots will be 360º / 5 = 72º.
So, answers are: (1, 0º), (1, 72º), (1, 144º), (1, 216º), (1, 288º).
EDIT: "Cheap way out?" Agreed. Why make things more difficult than they need to be?
EDIT2: "How did u guess they'd have the same distance?" They always do! Look up: "Multiplying Complex Numbers in Polar Form" You multiply the magnitudes but add the angles.
So, for a fifth root you need a set of angles that when multiplied by 5 will always end up with the same answer. So:
5 * 0º = 0º
5 * 72º = 360º or 0º
5 * 144º = 720º or 0º
5 * 216º = 1030º or 0º
5 * 288º = 1440º or 0º
EDIT3: Anyone who does not think the answers are evenly spaced round the circle should look carefully at 24:22. Look carefully at the red dots and ignore the other lines (axes). They are evenly spaced.
You took the cheap way out and didn't calculate anything, these are radial coordinates and you didn't even convert to radians
@@autumn948I believe this is a genius method.
How did u guess they'd have the same distance?
@@erachmah all roots of unity are evenly spaced radially around the origin, and one of them is always 1
@@autumn948 Irrelevant. They are still correct and root is still a root, no matter the representation you use. Radial coordinates can exist in ANY form of radial system be it degrees radians or the gradian system.
The thing I don't understand is why do you keep explaining basic math (such as "-x+5 = 5-x") and spend several minutes to rewrite a bunch of obvious things when you're explaining a college level problem?
It's like explaining how to build a rocket and then you spend half an hour talking about how the screwdriver works.
Is really funny for you to say that, because I don't know why this appeared in my feed, I was literally just watching a Minecraft video😂. All of a sudden I was here, watched half video and didn't understand nothing jajaj (I'm in Uni, I'm cooked).
@@gabrielcalvo2623 bro how tf you don't understand this? this is literally high-school level mathematics
@@suryanshvarshney111 "high school" level and "college" level are really subjective as it really isn't one type of math is higher than the other. I've taken Linear Algebra and wasn't required to take one statistics class.
So saying this is grade 8th 12th or higher isn't a strong argument.
@@Marinealver i call bs, there clearly are levels to mathematics. You wouldn't go straight to learning Stochastic processes when you don't know shit about basic probability and statistics. And no shit you don't need stats for LA 🤦🏻♂️ two topics can lie on same level. So, no high-school level and college level isn't subjective. It literally implies that these topics were taught in high-school so you should know it🤦🏻♂️ imagine coping so hard that you try to say college level and high school level is subjective 😭🙏🏻
Not even college level they teach us these things in +1 in india
Instead of doing this long way you can just rewrite it as x^5 = 1, which means x represents the fifth roots of unity. By expressing 1 in polar form as e^(2πi k) for k = 0, 1, 2, 3, 4, we can take the fifth roots to get x = e^(2πi k / 5). This gives the five solutions: x_k = cos(2πk / 5) + i sin(2πk / 5) for k = 0, 1, 2, 3, 4, representing points evenly spaced on the unit circle.
Yes, this is the most elegant and shortest solution path, assuming there is limited time during the entrance exam.
Solutions (beyond x=1) are complex numbers :
e^(2k.Pi.i/5) with k being any integer.
Simple geometry in the complex plane.
True
how did you find
Anyone who knows complex numbers well can just write down all 5 solutions. But k only ranges from 0 to 4, any other integer duplicates one of those answers. (That is, if m = n (mod 5) then e^(2m𝜋i/5) = e^(2n𝜋i/5).)
@@agustincaiThese are the points that divide the circle into n equal parts
Yep, so solutions would be 1; 1e^j2π/5; 1e^j-2π/5; 1e^j4π/5; 1e^j-4π/5 with k = -2, -1, 0, 1, 2. Didn't even watched the video just jumped from front picture to comment section.
Polar plot. Radius =1. Rotate the vector 2pi/5. Project each stopping point with an X and Yi coordinate. Four complex roots and X=1. Done
The idea is to provide the exact solution without recourse to the trigonometric forms. This also gives you a slick way to find sin(2Pi/5) and cos(2Pi/5)
@@paulbritt588 The trigonometric form is just as exact and usually more useful. The main utility of complex numbers is to describe rotation, and since this is also on the unit cicrle, it's even a subgroup of U(1) (under multiplication). In fact, this subgroup represents the same abstract group as Z5.
@@haakoflo I am aware of the complex solution and the ease with which it solves the problem. I think that the displayed algebraic solution is shown, as it would appeal more to a high school student who may not have seen the trigonometric method.
@@paulbritt588depends on the education system. In Australia we are taught roots of unity
Cambridge exam questions are typically not difficult but there are too many of them in the paper to pass if you spend a lot of time on each. This is a good example, solve laboriously as in the video in 25 minutes and you may get the right answer but no time left to answer enough questions to pass. So the exam is actually testing your depth of understandingand ability use it to see the fast paths to solution. ( which is the euler relation shown by others here. 20 seconds, boom, done.)
Perhaps you are smarter than I, but I would have to disagree. Some of those questions really are just truly hard. This level of question isnt usually present.
British school is amazing. I didnt have euler equations in school oO..
@@erachmah euler shows up in stats a lot
@@NewxToXThisxBiz in Brazil they only teach it in college.. :( and I didnt attend to a STEM field related course
@@erachmah i took coursera classes for free to learn. I also follow Jensenmath on youtube to learn. good luck!
Everyone saying this method takes too long. Sure on an exam you'd use a different method, but can we appreciate that he showed how to get exact values for trig functions analytically? I think that's neat
Exactly, but people just love to bitch and whine. Man is laying down a work of art and mainly gets shit for it, it's unfair.
I appreciate you working through all the steps so methodically. It kept me from getting lost. Also, thanks for calling out all the formulas and identities you used by their proper names - it helps me look them up becasue I need to go reference them again.
I've posted something like this before. You want to go to your neighbor's house, which is 100 feet to your west. You could walk 100' and arrive at your destination. OR you could travel East the entire diameter of the earth minus that 100'. ALL TOO OFTEN, the solutions shown take the long way to get there.
What does the "traveling" metaphor" means in relationship to math? You have a very naïve understanding concerning what solving a math problem means. Your 100' solution will give only a 1/5 of a correct score.
@@harrisorourke6926 obviously, he could have walked in expanding circles around his house that follow the fibonacci sequence until there is an interception with the backwards traced fibonacci sequence radiating from the neighbour which he then walks backwards. so naive
It says people take the long way (this video) instead of the short cut (de Moivre's theorem in this case).
Except when you are in a polar Region, a bit south of the northpole where he w way eastwards May be the same considered he ist exactly on the opposite Side 180 degrees. Sadly o do Not know how far He must be south of the northpole. IT reminds me of a Joke when a carholic priett ate meat and someone reminded him it was friday, when eating meat IS forbidden for catholics. The priest walke some distance and remarked "Here IT IS still thursday!"
@@harrisorourke6926travell8ng means covering a distance.so IT IS math
Many students cannot follow this long process. Simply apply De Moivres complex roots theorem. Much shorter and precise. Thanks.
But he found the sine and cosine of angles n*2*π/5 in surd form. You're stuck with answers such as cos(72⁰) + i*sin(72⁰) and have to figure out how to go further.
yes @@oahuhawaii2141
I am not an advanced student. I was fascinated and enlightened by the clever substitution that revealed a 4th degree equation as two "nested" quadratics.
Even more satisfying were the algebraic expressions in the respective real and imaginary parts for the sines of the various angles n*18° or n(pi/10). Clearly, algebraic expressions for the trig functions are encoded in the complex roots of specific polynomials generally. More elegant ways to calculate them probably exist, such as whatever algorithms likely programmed into calculators, but this blows my mind.
I'm old, so my high school math texts contained appended trig and log tables, both rendered to 4 significant digits probably. I may have imagined that some monastary dwellers had produced these in the 16th century using prohibitively laborious methods.
I pulled a shit ton of inspiration and knowledge from this vid.
@@oahuhawaii2141 finding the value of sin 72 using trignomatic identities is taught in highschool
@@oahuhawaii2141From most cases I've seen is that if you have a angle that does not have a exact value like 72° you don't have to spend extra time rewriting it.
@@Apple-vm5gc: I've looked at a few textbooks, and sin(72⁰) isn't computed in surd form. Even past CRC tables didn't have sin(n*18⁰). Only 0⁰, 15⁰, 30⁰, 45⁰, 60⁰, 75⁰, and 90⁰. I did all multiples of 3⁰.
If you know polar coordinates, the solution would be more simple and beautiful. By knowing the only real solution (1), the complexes would be calculated just dividing 360 degrees by 5. So, the other four solutions are 1
Complex is an example of how we humans like to make a problem that doesn't exist and then try to solve it
Of course, as an entrance into Cambridge, you would have to realise this is Mathematics, not “Math”.
Or maths.
What's the difference? I figured math is just shorthand for mathematics
@@boringsupernova4836it is, i figure they're just an obtuse brit
@@boringsupernova4836 it is. In north america it's common to drop the s from maths, and in the UK they say maths ... one of those "no wrong answer but some people want to sound pretentious" and feign superiority. :D I do it sometimes too.
I think theyre just joking about cambridge being pretentious lmoa.
Depending on the time allotted for this, I would agree that the trig and complex exponential version "answers" for the complex roots might not get you far at Cambridge - as a mere demonstration of rote learning. Another approach is to notice that the roots are all unimodular - that is: taking x=a+i*b, (a & b Real), a^2+b^2=1. Substituting this in the quartic equation for x, we know both the Real and Imaginary parts must vanish. The imaginary part is i*b*(4*a^3+3*a^2+a*(2-4*b^2)-b^2+1)=0. As b is nonzero for roots of interest (we know from the monotonicity of x^5 there aren't other real roots), we can ignore the overall factor of b, so substituting in b^2=1 - a^2 elsewhere, and extracting a common factor of a (joining the b since a is nonzero too) leads to a quadratic for a : 4*a^2+2*a-1=0, roots of which are -1/4 +Sqrt[5]/4, and -1/4 - Sqrt[5]/4, and the corresponding Imaginary parts can be straightforwardly calculated (from Sqrt[1 - a^2]). Whether this is "cleverer" than the "completing the square" approach presented might be a matter of taste - a sharper eye there versus using more knowledge of the complex roots? Another connection with the presented solution is that for a unimodular complex number x, x +1/x is twice the real part, which is why the solutions above for a are half the presenter's "t" roots, indeed could use this instead of solving the second quadratic where the complex numbers first arise in the presentation. Of course for completeness one might wonder about the Real part of the x-quartic: after substituting for b^2 one gets a quartic for a - with four real roots, the pair above and two more (+1/Sqrt[2[, -1/sqrt[2]) that don't make the Imaginary part vanish - must confess that took the edge off a little.
Note that if we set x =2π/5, we can use the fact that cos(4π/5) = cos(2π - 4π/5) = cos(6π/5) to get the relationship cos(2x) = cos(3x). We can then use the Pythagorean identities.
Hence cos(2x) = 2cos^2(x) - 1 is equal to cos(3x) = 4cos^3(x) - 3cos(x). If we set c = cos(x), we now have 2c^2 - 1 = 4c^3 - 3c. Therefore 4c^3 - 2c^2 - 3c + 1 = 0. A factor is c=1 giving:
(c - 1)(4c^2 + 2c -1) = 0. The c=1 is when x=0, which also satisfies cos(2x) = cos(3x), so we just need to solve the quadratic, giving c = (-2 ± √20)/8 = (-1 ± √5)/4. Since x is in the first quadrant, we can see that cos(2π/5) = (-1 +√5)/4) which is also the value for cos(8π/5). We can see that cos(4π/5) = cos(6π/5) = (-1 - √5)/2. Similarly for sin(2kπ/5).
Decades ago, my 9th grade teacher gave a bonus trigonometry problem to find sin(θ) in surd form, for θ = 18⁰ . I solved it at my desk for sin(θ), cos(θ), tan(θ), cot(θ), sec(θ), csc(θ) . Nobody else in class figured out how to do it. At home, I did the same for θ = n*18⁰ , n ∈ ℤ . The next day, I showed my teacher the table, and said I'll work on θ = n*3⁰ during the weekend. I used lots of scratch paper for those calculations. When I showed her my completed table, she asked me to join the math team. I did and she gave me a CRC math book. The first thing I did was look at its trig section, which I saw only had θ = n*15⁰ . I submitted a copy of my table to the publisher. I never got a response.
X=cos (2n*pi/5)+i*sin (2n*pi/5), with n=0, 1, 2, 3, and 4 for the principal values.
But he found the sine and cosine of angles n*2*π/5 in surd form. You're stuck with answers such as cos(72⁰) + i*sin(72⁰) and have to figure out how to go further.
@@oahuhawaii2141 you can use trigonometric identities to find em , also considering a regular polygon with five sides(pentagon) whose circumcircle has radius one with its centre at origin will help .
@@22tirthsavaliya36: Well, yes, but he didn't do it.
If you can solve it in 24 minutes, why should you spend 15 seconds?
Very good you got one of the five solutions to the problem in 15 seconds. What took you so long? However apparently you did not notice that the problem presents an exponent of 5 which leads to a solution with 5 answers.
@@harrisorourke6926sry mate, he meant all 5 solutions in 15 seconds, which is pretty easy to do considering 1 equals 1.e^j(0+2kπ)
@@harrisorourke6926 imaginary solutions, hence "i".
@@robert.eduard Not imaginary, complex. Complex numbers are pretty essential to a lot of practical applications, so being able to find complex roots to an equation is significant. "Imaginary" is an outdated term for complex numbers.
@@ToasterPizzaFun “i” is still called an imaginary unit, but if it has real life applications, then it’s great. I didn’t know that, I’ll look into it.
So it's basically finding 5th root of 1, but for complex numbers which have the general form:
r*exp(i*x)
It's clear that r has to be 1 (a non-negative real number that has absolute value 1 when raised to the 5th, that is just 1).
So you're really just looking for a number of the form:
e^(ix)
But when you raise it to 5:
e^(i*(5x))
That means x must be an angle that multipled by 5 gives you 2pi.
So x = 2pi/5 (call it the base case X1)
But whenever you sum a whole turn, you get to the same spot, so all the answers would be:
X1= 2pi/5
X2= 2pi/5 + 2pi/5 = 4pi/5
X3= 4pi/5 + 2pi/5 = 6pi/5
X4= 6pi/5 + 2pi/5 = 8pi/5
X5= 8pi/5 + 2pi/5 = 2pi "=" 0
(X5 is the real root, that's when you stop because x is confined to [0,2pi), doing this again will give you the first root )
This is how i would solve it. The roots would be
X1= exp(i * 2pi/5)
X2= exp(i * 4pi/5)
X3= exp(i * 6pi/5)
X4= exp(i * 8pi/5)
X5= 1
Verifying this is the same as the video is left as an exercise...
X1=1
X2=cos(72°) + i*sin(72°)
X3=-cos(36°) + i*sin(36°)
X4=-cos(36°) - i*sin(36°)
X5=cos(72°) - i*sin(72°)
Right, but what do those cos and sin evaluate to?
x^5=1=exp(i2πn)
hence
x=exp(i2πn/5) and using exp(iθ)=cis(θ)=cosθ+isinθ, for n=0..4
we have 5 solutions as required
This is exactly what makes the difference between students in these exams, if you can come up with a solution that can help you solve the problem much more quickly than the intuitive solution as shown in this video, you can then answer much more questions.
But he found the sine and cosine of angles n*2*π/5 in surd form. You're stuck with answers such as cos(72⁰) + i*sin(72⁰) and have to figure out how to go further.
@@oahuhawaii2141 Why are you copy-pasting the same comment over and over again?
@@dieterbohm9700: Because these commenters are raising the same point others have made.
x1 = 1 ; x2, x3 = cos ( 2 * PI / 5 ) + / - i * sin ( 2 * PI / 5 ); x4, x5 = cos ( 4 * PI / 5 ) + / - i * sin (4 * PI / 5 ) - tthe primary values. + all 2 PI rotations of these roots.
Where 2 * PI / 5 ~ 72 degrees, 5 * 72 degrees = 360 degrees.
But he found the sine and cosine of angles n*2*π/5 in surd form. You're stuck with answers such as cos(72⁰) + i*sin(72⁰) and have to figure out how to go further.
@@oahuhawaii2141 True, but I didn't bother calculating them or looking them up - I just wanted to illustrate the concept. 😂
What? x = 1^(1/5) = exp(2ni*pi/5), n = 0; 1; 2; 3; 4 because other integers give the same numbers. So x_1 = 1; x_2to5 = exp(2ni*pi/5) = cos(n*72°) + i*sin(n*72°), n=1; 2; 3; 4
I've tried the same, but then, I remembered that I don't know the trigonometric functions of 72º
@@JoseFernandes-js7ep ... No big deal, u could (better still) leave ur 'x - values' in exponential form.
You can do
cos[(2π/5)*x] + i sin[(2π/5)*x]
Put 0,1,2,3,4 as x and you'll get all 5 roots
There's 5 complex solutions because of the fundamental theorem of algebra. The real solution is obviously x=1. The complex solutions are all the rotations of n/5 of the unit circle in the complex plane. So e^(2n/5πi) with n equal to 1, 2, 3, 4 and 5. Notice that when n=5 this reduces to 1. Funnily enough we could let n be any integer and it would reduce to the same answer as n mod 5.
Simple. Draw a unit circle centered on the origin of the complex plane. Mark 5 points on the circle at 72° intervals starting on the real axis at x=1.
What do I think about the answer? All that is needed is to know how to evaluate sine and cosine of 72°, 144°, 216°, and 288°.
He does the evaluation in the video. It isn't simple.
this video is actually good i have no idea why people feel the need to be so upset in the comments.
He takes 24 minutes to solve a problem that can be solved in 30 seconds if you do it the proper way using de Moivre theorem's, which every pre University student should be familiar with
@@julianbrown7976in which highschool do people learn about it? And im also pretty sure non stem uni students do not need it
Specialist Maths in the Victorian (Australia) curriculum includes teaching students how to do operations with complex numbers in both Cartesian and Polar form. De Moivre's is bread and butter here.
"Fast and quick video", the video length: 24:47!
I personally prefer the visual approach to this problem.
First transform to x^5 = 1.
There is a way to look at complex numbers as the vectors from 0 0 to (complex part, imaginary part)
Then because multiplying something in the complex plane means multiplying the "lenght" of the number, and adding the "angles" of the number to gether
First i know the len of all solution vectors is 1
Then i i know that the angle of all solution vectors is 0°
-> 5 × n = 0°
=> the 5 solutions have the angle
0/5×360
1/5×360
2/5×360
3/5×360
4/5×360
And by knowing the lenght and angel of all solution vectors i can calculate theire real part with cos, and imaginary part with sinus
Never underestimate the ability of a "mathematician" to "simplify" an elementary problem into incomprehensible gibberish.
how you call it a "simplifying" if that's they to find ALL roots, not just one.
Why is "mathmatician" in quotation marks
I guess in other countries, not only Croatia where i am from, highschoolers are tought about the de Moivre formula (or atleast Eulers formula) , my God! Then it is trivial.
Clearly the goal is not to use de Moivre, but it should be stressed in the title indeed!
@@philippelepage1609 This was not mentioned at the beginning. Therefore the fastest and most elegant way might be preferred. Also, using polar coordinates you can also solve similar equations with larger exponents easily, like x^10-1=0.
Why just don't you use complex numbers and solve fast. Entrance examiner will give you zero for this step they want you to solve it using complex number which is way faster.
True. This should take the average highschool student who knows about complex numbers less than 2mins to solve, using polar representation of complex numbers.
Fascinating to see all the complex solutions for x
I mean it in the best possible way. This video helped me go to sleep at 3:30. And I love math
In my head in under 5 seconds.
Me, too. I've seen this guy before, giving ridiculously complex solutions to some very simple problems. Any first day algebra student would say, "What minus 1 is zero? Okay, 1-1 is zero. So x^5 = 1. Well, 1 to any power is 1, so the answer is 1. Duh."
@@Astrobrant2 You are doing naïve math. Actually you are not doing math. The solution requires determining the 5 solutions to the equation. If you submitted this on a homework assignment you might get partial credit and definitely a note from the instructor that says, "See me about this".
@@Astrobrant2 the question is to find ALL ROOTS which includes finding complex numbers. It is not your naive math problem
@@Shri For someone with any math background, this is still extremely trivial when using polar coordinates. Took me about 0.5 seconds to see the pentagon represented by this equation. Or to unfold it as the Z5 cyclic group.
My first thought was: Am I so stupid and forgot the basics of complex numbers? Why 20 minutes, this is not a trivial problem?
Finding all real rational solutions 😀
Finding ALL solutions 🗿💀
Isnt the x just 1?
Exactly!!! That's what I'm thinking
@@Anu_here_148Yes, but there are actually 4 more roots, that are all complex numbers.
If you rotate the complex number ("point") "1 + 0*i " on the unit circle 72 degrees or 2 * PI / 5 radians , you will end up in the complex number "cos ( 2 * PI / 5 ) + i * sin ( 2 PI / 5 )", where i is the imaginary unit.
This is pr. definition e^( i *2 * PI/5 ), since the "length" or modulus rather is 1 ( we are on the unit circle in the complex plane ;-) ).
Raising this complex number to the fifth power then gives us ( rotating it to the argument ( angle ) that is 5 times larger ):
z^5 = ( e^( i *2*PI/ 5 ))^5 = e^( i * 2*PI /5 * 5 (!!!!!) ) = e^( i * 2PI ) =
cos(2*PI) + i *sin( 2*PI) = 1 + 0*i = 1
And we are back to 1 ;-)
This also applies to the symmetric complex roots at - 72 degrees, and +/- 144 degrees on the unit circle in the complex plane - and to all full 360 degrees ( or 2*PI ) rotations of these 4 primary roots.
@@Bjowolf2,
yeah and aren't we allowed to consider that the real reason for more than one root
x = 1
is , that the equation is a fifth grade one .
so there should be five roots as a full solution .
@@keescanalfp5143 Yes, precisely - a n'th order polynomium always has n roots ( complex or real, including multiple roots ), and if the coefficients of the polynomium are all real (as in this case!), then any complex roots will always appear as complex conjugated pairs ( symmetrically around the real axis in the complex plane ) 😊
@@Bjowolf2 Thank you 👍✨
Excellent - great explanation of something that is more “complex” than it seems at first glance 👍
x⁵ - 1 = 0
x⁵ = 1
x⁵ = 1⁵
x = 1 (as powers are same bases are equal )
Done ✅
This is a perfect example of why I hated math in school so much. The teachers were never interested in actually helping their students learn math. They were mostly interested in showing everyone how smart they were by intentionally overcomplicating everything and speaking in a language that they knew none of us understood so that we would bow down in awe of their massive brains. It wasn't until I got to university that I finally met a humble math teacher who was genuinely interesting in helping me understand.
These kind of problems are trivial in polar coordinates.
Exact, j'ai réalisé en moins de 60 secondes.
But he found the sine and cosine of angles n*2*π/5 in surd form. You're stuck with answers such as cos(72⁰) + i*sin(72⁰) and have to figure out how to go further.
The function has only one real root x = 1. For complex roots there was some theorem (I'm to old to remember it :D )
I think the guys in Cambridge would want the person to know the theorem and use it straight away instead of doing algebra.
The test should check whether the person knows something more than algebra.
Examinators usually look for a mix of knowledge and competences, so no, not necessarily
They could use abstract algebra to solve this, and say that the solution is the numbers 0,1,2,3 and 4 in the cyclic Z5 group, just represented on a homomorphic representation of the same abstract group on the complex unit circle :) That MIGHT have scored full points, especially if the homomorphism was accompanied by an illustration.
As someone who went to MIT and is a contrarian both solution are valid. Actually any solution you can properly present is valid for solving the problem and can't be downgraded. This is generally true for all university exams and test makers generally make so you don't have enough able time to try algebraic solutions instead of theorems with easier solutions. In this case the theorem you would use is the "De Moivres". I would do algebraic solutions anyway because I am really fast at mental math and the teachers would always complain =D I was kind of able to get away because people still need enough time to try different theorems until see witch one solves the problem with an easy solution but you can just brute force. It always work.
1, i^4/5, i^8/5, i^12/5, i^16/5
Nailed it.
Joking aside, there's 3 factors that determine how I should answer:
- What major am I applying for?
- What is the question?
- What situations will my answer be applied to?
Yes!
It's just 1!
Why don't just x=1???
But if you want, there is a better solution. :)
If 𝑥5 = 1, we are looking for values of x that satisfy the equation: x 5 =1.
This equation has five complex roots, known as the fifth roots of unity. These roots are located on the complex unit circle, meaning they lie on a circle with a radius of 1 centered at the origin, and are evenly spaced in angle. We can find these roots using Euler’s formula:
x k =e 2πik/5 for k=0,1,2,3,4, where 𝑖 is the imaginary unit, and 𝑘 represents the root index.
Thus, 𝑥 can take any one of these five values. One of these is the real number 1, while the other four are complex numbers evenly distributed around the unit circle in the complex plane.
X equals 1. I am aged 72
CHeater!
You missed the other 4 solutions, which happen to be complex numbers.
@@oahuhawaii2141 I am an orthopaedic surgeon, I deal with reality, not the fuzzy world of complex numbers
Well, I guess I'm not going to Cambridge.
you are too intelligent?
I was a janitor at Cambridge and wrote the answer on a whiteboard in the hallway. Who says you can't get into Cambridge?
@@wes9627few years after got lost in mars
This channel wasting my time more than my procrastination
Z^5 = cos theta +isin theta, using de moivres, z = cos theta/5 + isin theta/5 theta = 2*k*pi substitute k = 0,1,2,-1 into z to get the final four solutions aswell as z = 1
So much easier to solve with complex polar coordinates and Euler's equation. x^5=e^(5*i*t)=1. We need to find a "t" such that our angle in the complex plane is 2*pi*(an integer) so that our value of x^5 wraps around to being on the +real axis. 0 works obviously (x=1 solution), 2pi/5 works (quadrant 1 solution in the complex plane), 4pi/5 (quadrant 2), 6pi/5 (quadrant 3), 8pi/5 (quadrant 4), then 10pi/5 brings us back to 2pi (which is equivalent to the t=0 solution).
x = r*e^(i*t), with r=1, and e^(i*t) = cos(t) + i*sin(t). Each of our values of t above give us our different solutions (including t=0 which leads to x=1*(1+0i)).
To simply put it for those who do not fully understand the math they’re referring to is in this video is for exploring variants, not the solution. they are not asking for the answers to the question they are asking for the variants to the answer of the questions.
You don't have to do all of this. I saw it immediately.
Did he get the right answer? If yes, why do you bother?
You saw all 5 roots immediately? Seriously?
x to the fifth = 1 if x = 1. 1 - 1 = 0
@Martha-jl6eu Did you watch, to learn how to derive the four complex roots algebraically?
@@Martha-jl6eu There are 4 complex roots. Everyone knows the real root is 1.
12:55 boringly complicated
On the complex plane, draw a circle of radius 1. The points in the circle that make angles to the horizontal of 72°, 144°, 216°, 288° and 0° will be your solutions.
Before watching this video, ı spent nearly an hour to find complex solutions. It's such a good and hard question. Thanks for sharing it with us ❤
25minutes,eek!
For everyone saying "the answer is just x=1", did you ever go to highschool?? Jeez...
hiskool iz for loozers
sorry mate
my highschool taught me the answer is simply x=1, maybe it was college you were intending?
Complex number theory is in college for us. I assume we go over things that you probably don't in high school and vice versa. I know that some high schools go over complex number theory but not mathematical induction and structural induction (to my surprise).
Thank you for watching. A great question today x^5 - 1 = 0 (Finding all roots) Have a great day and take care! Wish you all the best in you life and career❤❤❤
It's great that you found sin(θ) and cos(θ) in surd form for θ = k*2*π/5 = k*72⁰ , k ∈ ℤ . Everyone knows the values when k = 0 , but few know them for k = ±1, ±2 . Most are stuck with answers such as cos(72⁰) ± i*sin(72⁰) and have to figure out how to go further. The 4 complex solutions use the following key values in ± form:
sin(36⁰ ± 18⁰) = (√5 ± 1)/4
sin(54⁰ ± 18⁰) = √(10 ± 2*√5)/4
The easiest solution is dividing the unit circle into five equal pie pieces. Each piece has a 72 degree tip. So the first solution is x=1. The second is x=cos 72 + i sin72 and so on.
ok but why not just use the polar form and Eulers identity
x^5 - 1 = 0
x^5 = 1
x^5 = e^i(0+2kπ )
x = (e^i(0+2kπ))^(1/5)
x = e^i(2kπ/5)
k=0 ; x = 1
k=1 ; x = e^i(2π/5)
k=2 ; x = e^i(4π/5)
k=-1 ; x = e^i(-2π/5)
k=-2 ; x = e^i(-4π/5)
Because that's incomplete. You need to actually evaluate those answers
X=1,1,1,1 and 1.
That was easy
And wrong
1:04 1st class student watching this video
a trick to this question is: draw a circle (radius=1) in the argand-gauss plane and inscribe a pentagon in it, each one of the vertices will be one of the solutions
You can understand it by considering the phase of the complex numbers with the exponential representation. Solving this equation is equivalent to find the phases for which when it is multiplied by 5 you get the phase equal to 2pi (rad) because the phase of the complex number 1 is 0 or 2pi. So first trivial solution is x=1. Second solution is the complex number which phase equal to 2pi/5 because 2pi/5×5=2pi. Third solution is the complex number which phase is equal to 2×2pi/5 because 2×2pi/5×5=2×2pi => same complex number =1.
And then the same explanation you get x=3×2pi/5 and x=4×2pi/5 which are the 5 only possible solutions because then if x=5×2pi/5 and x=6×2pi/5 it is already the first and second solutions. CQFD 😊 (without to solve any complex equations I found the complex solutions 🤭)
I'll never fathom how some people find this useful.
Or entertaining.
Or comprehensible.
solutions:
x=1
x=cos(2pi/5)+isin(2pi/5)
x=cos(4pi/5)+isin(2pi/5)
x=cos(6pi/5)+isin(6pi/5)
x=cos(8pi/5)+isin(8pi/5)
X = e^(2nPi/5) for n=0,1,2,3,4
X^5 = e^(2nPi) for those n
Which is 1 for all n
Job done…
Very beautiful solution, thanx for video!
1. x = 1
2. x = e ^ (i * (2pi)/5)
3. x = e ^ (i * (4pi)/5)
4. x = e ^ (i * (6pi)/5)
5. x = e ^ (i * (8pi)/5)
These can also be expressed in terms of cosine and sine:
1. x = 1
2. x = cos((2pi)/5) + i * sin((2pi)/5)
3. x = cos((4pi)/5) + i * sin((4pi)/5)
4. x = cos((6pi)/5) + i * sin((6pi)/5)
5. x = cos((8pi)/5) + i * sin((8pi)/5)
X⁵-1=0
X⁵=1
X=⁵√1
X=1 as ever root and potentiation of 1 is 1.
If i write
X⁵-2= 0 there's no results as 2 don't have a ⁵√ sp there's no result, bu if i chose a number with a ⁵√ such as 32 it will as
2*2*2*2*2 is 32, so 2⁵=32
So x⁵-32=0
X⁵=32
X=⁵√32 and so on
If you change the x for x³ you just implies that if the results is also 0 so the subtracted number must be ³√ such as 8 or 27 and so on
Please keep making these videos despite any negative comments. Really brings back memories of my past.
if u divide in 5 (bcz the x to the 5)
the 1-unit circle centered in 0, you have the 5 solutions. 360/5 = each 72 degrees.
x1 = cos(0°) + isen(0°)
x2 = cos(72°) + isen(72°)
x3 = cos(144°) + isen(144°)
x4 = cos(216°) + isen(216°)
x5 = cos(288°) + isen(288°)
x=e^(2ki/5),k=0,1,2,3,4
Exp(i 2/5 Pi n) for n from 0 to 4
Very interesting, even though I'm not good at math, with the basic algebra I understood mostly of the exercise. Good challenge
The solution is amazing. For sure it should be aworded with some prestigious math prize.
e^(2i×pi×n/5) where n = 1,2,3,4,5
*"x^n - 1 = 0" solution:*
*x = 1,w,w²,w³,w⁴,..., w^(n-1)*
*where w = e^i(2π/n)*
Looking forward to solving (x^5,1)-1=0. I'd always wanted to see a 0,1 math solution.
We can do simple algebraic manipulation:
Add 1 to both sides and we end up with:
x^5 = 1.
What multiplied by itself 5 times is equal to 1?
Since the power itself is ODD it cannot be -1 since -1 * -1 * -1 * -1 * -1 = -1 AND -1 Does Not Equal 1.
It must be 1 since 1^5 = 1.
This is the only answer within the Real Numbers.
If we move to the Complex plane and we consider modulus arithmetic and the possibility of All Roots, then we can use a few various techniques either through vector notation, simple rotations around the unit circle, etc. and from there we can obtain the other 4 roots of unity that only exist in the complex plane.
For example: we know that sqrt(-1) is the imaginary unit i. We know that i^2 = -1. Since i^2 = -1 then i^4 = 1 since -1 * -1 = 1. Thus, i^4 is also a root. In other words, 1^5 and i^4 evaluate to the same thing as they are equivalent. The rest is up to you to figure out.
• Or you could just use De Moivre's theorem. 😊
• By turning the equation to this:
X⁵-1=0
X⁵=1
X⁵= Cos(0) + i Sin(0)
• Then use the theorem. 🙂
13:20 точно, еще существуют комплексные числа...
Как же хорошо было не помнить о их существовании
Why not just
x⁵-1=0
x⁵=1
x=1
An explicit list of all solutions: X = 1, X = e^(2πi⁄5), X = e^(4πi⁄5), X = e^(6πi⁄5), X = e^(8πi⁄5)
Right, but you haven't evaluated them. The video shows how to evaluate them
The nth roots of unity. Cos n(theta) + i Sin n(theta); n 1, 2…5. theta = 360/5. Compute this and it’ll give you all 5 roots. 1 and 4 imaginary roots
More simple question looks the harder it is ! (Atleast at equation problems)
EASYY :
x⁵=1 > x⁵=e^(i2pi) > x1=e^(i⅖pi) x2=e^(i⅘pi) x3=e^(-i⅘pi) x4=e^(-i⅖pi) x5=e^(i2pi)=1
For this specific problem there is an easier way by decomposing the result in an angle and radius in the complex plane. Then computing that taking an exponent is taking an exponent of the radius an multiplying the angle (r exp(i theta))^n is r^n exp(i theta n). The radius is then 1 and the resulting angle need to be 0 modulo 2pi. This give us 2 pi k/n. With k from 0 to n-1. Set n to 5 for this specific problem.
The main thing this made me think is that this is how any given event can be balanced throughout all dimensions even in a manifold which is more complete than 4 dimensional spacetime, but that is because I am generally thinking about the more complete dimensional manifold of actual spacetime and quantum dimensions outside of of conventional understanding. The circular graphical representation of all solutions made me think about the potential that imaginary solutions might actually give me the mathematical clue I'm looking for to describe my 5 dimensional spacetime with its extra cosmological scale spatial dimension and the full manifold including the additional quantum scale dimensions, also plural or singular because I haven't seen enough yet to be fully confident of the full manifold spectrum and framework. I know they are there, but I don't know their full shape yet.
This makes me appreciate the polar form of complex numbers so much
The roots must be of unit length, so either positive angle which lands on one when raised to the fith power, or 0 angle which is the real root, - 1/5th 2pi =4/5th 2pi in terms of angle so only 5 roots all unit lenght complex numbers of the form cos(n 2pi/5)+ i*sin(n 2pi/5)
n; 0, +-1,+-2. Thats doable in the head.
Also if you know complex multiplication, you know the same works when the constant is not unit length, you just need to look at the fifth root of the lenght of the constant, to get the initial magnitude of the complex and real solutions. And add angles and multiply lenght to find the roots, not too complicated.
That portion is doable. You haven't evaluated the root, though
I would love to see a geometrical answer to this question using the complex plane! Could you do a video on that way of obtaining the solution?
I was about to write a comment about you spending 20+ minutes on a simple solution, however I didn't know that complex solutions with imaginary numbers have real applications in several fields, so I will like to thank you for the video. It was a great review of my times at school
x⁵=1
x=1
fantastic 👏🏻👏🏻👏🏻👏🏻
thank you for such a clear explanation)
Prove that x4+x3+... +1>0.
Split into x4+x2 which is 0 or grater and the rest are always positive (proof with determinant etc). So 0+positive= positive q. E. D.