The solution given isn’t complete. We started by asking whether the solution was real, and we’re left with another number raised to an imaginary power so we don’t know. The answer given is actually in Re^i(theta) format, with R = e^(-pi/4) (we don’t need the multiples of 2pi) and theta = ln(2) / 2. To get to the a + bi format, use a calculator to find R cos theta and R sin theta. Then we’re done. (The answer is complex. )
This made me wonder about the general situation of (a + bi)^ci (a, b, c /= 0). Going through the same steps, we get R = e^[-c arctan (b/a)] and theta = c [ln (a^2 + b^2)] / 2. We can see two conditions where the solution will have a real value. First, when a^2 + b^2 = 1, the natural log of their sum = 0 and the solution will be real. A simple approach is to set a = b = SQRT(2) / 2, and c = anything. The other option for a real solution is when c [ln (a^2 + b^2)] / 2 = 2 pi k where k is 1, 2, 3, ... . There appear to be infinite solutions! As a check on the problem in the video, we see that a = b = c = 1, which gives [(ln 2) / 2] /= 2 pi k. That is why the solution to the problem in the video is complex. I enjoyed where this took me. Thanks for posting it.
@@SyberMath why don't you include the i factor when taking the magnitude of a co.pkex number again? Since you are multiplying by i..because thst is the imaginary part and you jjst take it out?
@@leif1075 I used the values of R and theta I derived and calculated a and b from there. I didn’t take the last step of showing it in a + bi form, since I thought that was understood.
Convert 1+i to sqrt( 2 )*exp( i•pi/4 ) or even better: exp( ln( sqrt( 2 ) ) ) * exp( i•pi/4 ) and then take power of i and get: exp( -pi/4 ) • exp( i•ln( sqrt( 2 ) ) ). For even more fun, remember that the conversion of 1+i is multi-valued. It makes a factor of exp( 2k•pi ) where k is an integer in the result.
Since you are so much in to complex numbers I would like to ask a general question. I get the idea of the complex number system initiated as a result of square rooting a negative number and how it helped making the engineering calculations so much easier. However if there are no real solutions to an equation does it imply there are complex solutions? Would it be possible to have no real or complex solutions to some ridiculous equations like 1^x=2 or even Sin x =2? I have seen people solving those but there is no method to check the answer besides reverse calculation which means nothing. I don’t see any such equation being generated by an engineering application.
I wasn't so happy with the multivalue solution and multiplicity is not always available for complex exponents, but the point, where we get multiple values, is the complex log when we transform (1 + i) to e^(log(1 + i))... (1 + i) = e^(log(1 + i)) = e^(ln(|1 + i|) + i (π/4 + 2 k π)). ln() is the real log, thus legal! ln(|1 + i|) = ln(√2) = ½ ln(2).
I got e^(i * (ln(root2)) - pi/4 + 2*n*pi). Steps: (1 + i) = root(2i) = root(2) * e^(pi*i/4) Raised to the i: root(2)^i * e^(pi*i/4)^i or (e^ln(root2))^i * (e^(pi*i/4))^i Therefore power rule e^(i * ln(root2)) * e^(i * pi * i/4) Simplify e^(i * ln(root 2)) * e^(-pi/4) Power rules e^(i * ln(root 2) - pi/4) Since 2*n*pi for any n e Z = some multiple of 360, we must add it on so answer is: exp(i * ln(root(2)) - (pi/4) + 2*n*pi) Right?
I would compute log(1+i) = ln(sqrt(2))+ i*(pi/4+2*pi*n) for some integer n i*log(1+i) = -(pi/4+2*pi*n) + i*ln(sqrt(2)) (1+i)^i=exp(i*log(1+i))= exp(-pi/4+2*pi*n) exp(i*ln(sqrt(2)))
@@gdtargetvn2418 yeah but k is in exp(-pi/4 +2*k*pi) so it seems to me as well that there's an infinite set of possible solutions (because the number of rotations k in (i+1) can be any integer regardless... If you know the answer do tell, but don't act like this because we know k is an integer
Any complex number has an infinite number of _representations_ in polar coordinates, since any number multiplied by 1 retains the same value. Note that 1 = e^2nπi. Let z = r.e^iθ. Then z * 1 = r.e^iθ * e^2nπi = r.e^i(θ + 2nπ). I think the _value_ of z is the same, although that value may be expressed in an infinite number of ways by picking different values of n.
If my maths is correct, it should have infinitely many solutions! These would have the form: (e^(-2*pi*n))(0.4288+i*0.1549). Note that the "n" is just a positive or negative integer (or 0 if you want the principle value!). The negative integers signify a rotation clockwise within the complex plane, and the positive integers signify a positive rotation (with discrete steps of 2pi)! Hope this was helpful!! Edit: This can also take the form (e^(2*pi*n))(0.4288+i*0.1549) in which the 2pi factor has lost the minus sign, this would just mean that the rotations provided by the integer "n" would travel in the opposite direction!
@@birb1947 That's not correct. The way you wrote it, your first term is a real number that scales your second term, which is a complex number. Each 'n' results in a different complex number that is longer but points in the same direction in the complex plane. But there is only one answer: 0.4288+0.15487i. In polar notation a complex number has multiple representations as r*e^(i*2*pi*n). So in rectangular form your answer should look like r*cos(2*pi*n)+r*i*sin(2*pi*n)
@@MushookieMan Oh! I think I somewhat understand what you mean? Although I'm a little confused? My first step was to convert the inside of (1+i)^i to exponential form re^it (t = theta) where it takes on the value (sqrt(2)^i) * (e^(it+2pi*n))^i (this is where I got the "n" term) this can then simplify to become (sqrt(2)^i) * (e^-t)*(e^-2pi*n) in which case the exponential containing the "n" term becomes real. The rest is just converting the remaining complex numbers to polar form which we already did in the previous comments. I'd appreciate it if you could show me the step I messed up! The working out is a bit oversimplified as I didn't want to bombard you with pages of typed maths shorthand haha! Although I can go more rigorous if you need me to! Thank you!
isn't it easier to call our number z, tranform everything ith exponential notation, take the log on both sides (at this point you should have an equation looking like logz=(ipi/4)exp(ipi/2)log(sqrt2)). Converting ipi/4 into its exponential notation you get (pi/4)exp(ipi/2). Combine the exponentials in the equation and ouu get logz=pi/4(exp(ipi))log(sqrt2)=-log(2)pi/8 which implies that z=exp(-log(2)pi/8).
The rule (a^b)^c = a^bc … IS NOT VALID FOR IMAGINARY NUMBERS. If it were, then it is easy to show that -1 = 1: Simply replace 1 by e^(2pi*i) in the right part of the equation 1 = sqrt(1).
TMHO there are 2 questions in 1: (1) let the function f(Z)=Z^i, Z in C, evaluate : f(1+i)= ? (2) solve for Z in C the following equation : Z - (1+i)^i = 0 The video gives the answer for the second formulation, but for the first, we must modify it because the complex Log function (given Z=R e^(i Theta)) has the following restrictions in C: - R is non zero (complex log is defined for C*) - 0
It's not the modulus, it's the argument, so the angle, since it's an angle, all measures that are in the form - pi/4 (mod 2pi), are the same angle, that's why cos and sin are 2pi-periodic functions
@@thecrazzxz3383 In the answer in the video the argument is given as "ln(√2)" (with the modulus, that also includes "e" being on the right side of the first "e" instead of the left.) I think maybe the answer just isn't completely simplified and technically needs more work.
@@JefiKnight Oh sorry i was completely missing it, yes there should be only one solution ! Actually, the whole video doesn't make any sense, nothing defines raising a number to a complex exponent, it doesn't make any sense !
The calculation is cool. But can the answer be written as a+bi? The answer doesn't look satisfying. I have no idea how to comprehend this number at all...
I haven't watched the video yet. First, convert the complex number on the base into polar form, where r²=x²+y² and tanθ=y/x. x=1 and y=1, so r²=1²+1²=2 and r=√2. tanθ=y/x=1/1, so θ=π/4+2πn, where n is an arbitrary integer. That means that 1+i=√2·e^{πi/4+2πin}. Raising that to the power of i multiples the exponent by i, giving us √2·e^{-π/4-2πn}. (1+i)^i has infinite real solutions, following a geometric progression. Its principal root is √2·e^{-π/4}.
What tells you that raising to the power of i multiplies the exponent by i? Can you prove it? By definition, all we know is that e^ix = cosx + isinx for all real numbers x...
e^(iθ)=cosθ+i sinθ can be proven from a much simpler claim, that d(e^x)=e^x dx. If d(e^x)=e^x dx, then d[e^(ix)]=e^(ix) i dx. d(cos x + i sin x)=-sin x + i cos x = i² sin x + i cos x = i (cos x + i sin x). As well, e^0=1, and cos 0 + i sin 0 =1. With these two facts established, we now know that e^(ix)=cos x + i sin x. We can also show, starting from d(e^x) = e^x dx, that (e^a)(e^b)=e^(a+b); and from there, it's trivial to show that (e^a)^b=e^(ab). If b=I, then (e^a)^i=e^(ai). Despite i's reputation for being exotic, it's actually a really well-behaved number
I feel it is enlightening to expose the answer in rectangular form so that the real and imaginary parts are clearly visible. 2kπ - π/4 ____ ____ e . ( cos √ln 2 + %i . sin √ln 2 ) Applying the half-angle identities is perhaps overkill but pretty nevertheless: 2kπ - π/4 e ____________ ___________ ------------------ . (√1 + cos ln 2 + %i . √1 - cos ln 2 ) √2
You can't just multiple powers of i time i since they are complex numbers. That rule doesn't apply to complex numbers, even though you got lucky the answer came out correct. That rule only applies to real numbers. Please re-think is over and make the necessary correction to the steps. You are on the right track.
You're right, complex powers are not defined !! And he gets something really weird, because the for the modulus of complex number he gets at the end, so e^-pi/4+2kpi, there's an infinite number of possibilities for (1+i)^i, which doesn't make sense, raising to complex powers doesn't neither ! But when you say "You are on the right track.", what do you mean ?
You will find that the n should not actually be in your answer, however. (1+i)ⁱ is a number, and has only one value. If you want an infinite number of solutions, solve the equation x⁻ⁱ = (1+i). This is similar to the concept that √49 = 7 has only that one value, whereas solving the equation x² = 49 has two solutions, ±7.
No, having n in the final answer is correct. In polar form there are an infinite number of ways to write a single complex number. 1+i = e^i*pi/4, but also, 1+i = e^i*9pi/4, and e^-i*7pi/4, and so on
Also the last step is still missing with transformation of the result into complex number (but that's probably something obvious for all those who got to that point) exp(2kπ - π/4) * (cos(ln(2)/2) + i*sin(ln(2)/2))
Why is x^i this e^ln(x)i, you mean x^i = [e^ln(x)] ^i, but what tells you that [e^ln(x)] ^i is actually, e^(ln(x)i)? Multiplication of powers when you raise a complex to a power is technically only defined for a complex raised to a natural number, or at most a rationnal number, but nothing is defined for a complex to a complex?
Зачем вообще копаться в комплексных числах? Мы ведь точно знаем, что их не существует в реальности. Причём, в отличие от отрицательных, их себе даже образно не представить.
When I wrote it I meant the video ended too soon, before the calculation was completed. But I was wrong, so my comment was stupid. I expected a final result in the form of r . e^iθ and didn't pay enough attention, probably because it's written as e^iθ . r, let's say I was tired ...
@@firelow no, u don’t understand the use of the constant k. k represents some arbitrary number restricted to a specific domain to show all possible numerical solutions. In this case, he specified k to be any integer, which gives the correct answer. Changing any other constants to k like u mentioned at your discretion equivalent is like changing the answer of 1+1 to 3, it makes the whole answer incorrect (unless u change the domain of k and explicitly state it)
This video is incorrect. There is a single principal branch of the complex power function, so the answer should be a single complex number, not an infinite number of them.
Before solving the problem, one should specify what is understood by (1+i)^i. For example, if we are interested in the principal branch, then n=0. As presented, this is a semi-literate nonsense.
The solution given isn’t complete. We started by asking whether the solution was real, and we’re left with another number raised to an imaginary power so we don’t know. The answer given is actually in Re^i(theta) format, with R = e^(-pi/4) (we don’t need the multiples of 2pi) and theta = ln(2) / 2. To get to the a + bi format, use a calculator to find R cos theta and R sin theta. Then we’re done. (The answer is complex. )
This made me wonder about the general situation of (a + bi)^ci (a, b, c /= 0). Going through the same steps, we get R = e^[-c arctan (b/a)] and theta = c [ln (a^2 + b^2)] / 2. We can see two conditions where the solution will have a real value.
First, when a^2 + b^2 = 1, the natural log of their sum = 0 and the solution will be real. A simple approach is to set a = b = SQRT(2) / 2, and c = anything.
The other option for a real solution is when c [ln (a^2 + b^2)] / 2 = 2 pi k where k is 1, 2, 3, ... . There appear to be infinite solutions!
As a check on the problem in the video, we see that a = b = c = 1, which gives [(ln 2) / 2] /= 2 pi k. That is why the solution to the problem in the video is complex.
I enjoyed where this took me. Thanks for posting it.
@@Paul-222 Np. Thank you for the observation! This is cool! 🤩
@@SyberMath why don't you include the i factor when taking the magnitude of a co.pkex number again? Since you are multiplying by i..because thst is the imaginary part and you jjst take it out?
@@Paul-222 I don't think it'd clear what you mean..why would we use a calculator and how? And how does that get rid of the i exponent..
@@leif1075 I used the values of R and theta I derived and calculated a and b from there. I didn’t take the last step of showing it in a + bi form, since I thought that was understood.
correction: It might be nice to include a graph showing the string of solutions spaced e^2pi apart at the angle theta = (ln(2)/2).
We can as well add the two powers of e in the last step and it results in
e ^ (2kπ - πl4 + i.ln(√2) )
Right!
It might be nice to include a graph showing the string of solutions spaced pi apart at the angle theta = (ln(2)/2).
Convert 1+i to sqrt( 2 )*exp( i•pi/4 ) or even better: exp( ln( sqrt( 2 ) ) ) * exp( i•pi/4 ) and then take power of i and get: exp( -pi/4 ) • exp( i•ln( sqrt( 2 ) ) ).
For even more fun, remember that the conversion of 1+i is multi-valued. It makes a factor of exp( 2k•pi ) where k is an integer in the result.
Since you are so much in to complex numbers I would like to ask a general question.
I get the idea of the complex number system initiated as a result of square rooting a negative number and how it helped making the engineering calculations so much easier.
However if there are no real solutions to an equation does it imply there are complex solutions? Would it be possible to have no real or complex solutions to some ridiculous equations like 1^x=2 or even Sin x =2?
I have seen people solving those but there is no method to check the answer besides reverse calculation which means nothing.
I don’t see any such equation being generated by an engineering application.
x + 1 = x has no solution in the set of complex numbers.
Principal solution: (1 + i)^i = (√2)^i * e^(-π/4) = e^(i ln(√2) - π/4) = e^(i ½ln(2) - π/4) = (cos(ln(2)/2) + i sin(½ln(2))) / e^(π/4)
General solution: (1 + i)^i = (cos(½ln(2)) + i sin(½ln(2))) / e^(π/4 + 2 k π), k ∈ ℤ determines the branch of the complex log.
I wasn't so happy with the multivalue solution and multiplicity is not always available for complex exponents,
but the point, where we get multiple values, is the complex log when we transform (1 + i) to e^(log(1 + i))...
(1 + i) = e^(log(1 + i)) = e^(ln(|1 + i|) + i (π/4 + 2 k π)). ln() is the real log, thus legal! ln(|1 + i|) = ln(√2) = ½ ln(2).
I got e^(i * (ln(root2)) - pi/4 + 2*n*pi).
Steps:
(1 + i) = root(2i) = root(2) * e^(pi*i/4)
Raised to the i:
root(2)^i * e^(pi*i/4)^i
or
(e^ln(root2))^i * (e^(pi*i/4))^i
Therefore power rule
e^(i * ln(root2)) * e^(i * pi * i/4)
Simplify
e^(i * ln(root 2)) * e^(-pi/4)
Power rules
e^(i * ln(root 2) - pi/4)
Since 2*n*pi for any n e Z = some multiple of 360, we must add it on
so answer is: exp(i * ln(root(2)) - (pi/4) + 2*n*pi) Right?
+2kπ because the argument of 1+i is π/4+2kπ when k is an integer since adding 2kπ to an angle makes it go back to itself
@@michas.2240 I know -- I originally got this in my answer, but forgot to clarify. Thank you
e^(iln(sqrt(2)-pi/4+2npi) where n in an integer
Nice work!
I would compute
log(1+i) = ln(sqrt(2))+ i*(pi/4+2*pi*n) for some integer n
i*log(1+i) = -(pi/4+2*pi*n) + i*ln(sqrt(2))
(1+i)^i=exp(i*log(1+i))=
exp(-pi/4+2*pi*n) exp(i*ln(sqrt(2)))
What about k? Is (1+i)^i indeterminate with in finite number of values?
k is just an integer wth
@@gdtargetvn2418 yeah but k is in exp(-pi/4 +2*k*pi) so it seems to me as well that there's an infinite set of possible solutions (because the number of rotations k in (i+1) can be any integer regardless... If you know the answer do tell, but don't act like this because we know k is an integer
@@pascalanema3377 Exactly!
Any complex number has an infinite number of _representations_ in polar coordinates, since any number multiplied by 1 retains the same value. Note that 1 = e^2nπi.
Let z = r.e^iθ. Then z * 1 = r.e^iθ * e^2nπi = r.e^i(θ + 2nπ).
I think the _value_ of z is the same, although that value may be expressed in an infinite number of ways by picking different values of n.
@@pascalanema3377 ... Then k is still an integer after all
What would the answer be in the form "a + bi"?
0.4288+0.15487i
If my maths is correct, it should have infinitely many solutions! These would have the form: (e^(-2*pi*n))(0.4288+i*0.1549). Note that the "n" is just a positive or negative integer (or 0 if you want the principle value!). The negative integers signify a rotation clockwise within the complex plane, and the positive integers signify a positive rotation (with discrete steps of 2pi)! Hope this was helpful!!
Edit: This can also take the form (e^(2*pi*n))(0.4288+i*0.1549) in which the 2pi factor has lost the minus sign, this would just mean that the rotations provided by the integer "n" would travel in the opposite direction!
@@birb1947 That's not correct. The way you wrote it, your first term is a real number that scales your second term, which is a complex number. Each 'n' results in a different complex number that is longer but points in the same direction in the complex plane. But there is only one answer: 0.4288+0.15487i. In polar notation a complex number has multiple representations as r*e^(i*2*pi*n). So in rectangular form your answer should look like r*cos(2*pi*n)+r*i*sin(2*pi*n)
@@MushookieMan Oh! I think I somewhat understand what you mean? Although I'm a little confused? My first step was to convert the inside of (1+i)^i to exponential form re^it (t = theta) where it takes on the value (sqrt(2)^i) * (e^(it+2pi*n))^i (this is where I got the "n" term) this can then simplify to become (sqrt(2)^i) * (e^-t)*(e^-2pi*n) in which case the exponential containing the "n" term becomes real. The rest is just converting the remaining complex numbers to polar form which we already did in the previous comments. I'd appreciate it if you could show me the step I messed up! The working out is a bit oversimplified as I didn't want to bombard you with pages of typed maths shorthand haha! Although I can go more rigorous if you need me to! Thank you!
isn't it easier to call our number z, tranform everything ith exponential notation, take the log on both sides (at this point you should have an equation looking like logz=(ipi/4)exp(ipi/2)log(sqrt2)). Converting ipi/4 into its exponential notation you get (pi/4)exp(ipi/2). Combine the exponentials in the equation and ouu get logz=pi/4(exp(ipi))log(sqrt2)=-log(2)pi/8 which implies that z=exp(-log(2)pi/8).
Why is pi over 4, the angle?
I love your problems and your explanations.
Thank you
Since ln(√2) = (ln2)/2, we could write (i + i)^i = e^( π(2n - 1/4) + (i.ln2)/2 ), perhaps we could even write it as e^( (π(8n-1) + 2i.ln2) / 4 )
This math represents a Big Crunch event, where matter is breaking up and more still stable matter is piling on due to inertia generated by gravity.
When you wrote 2npi in the answer's exponent, does n refer to whole numbers too? (-1,-2, etc), asking because the captions werent clear enough... (:
e^-(π/4+2πn)*cis(½ln2) for some integer n
The rule (a^b)^c = a^bc … IS NOT VALID FOR IMAGINARY NUMBERS. If it were, then it is easy to show that -1 = 1: Simply replace 1 by e^(2pi*i) in the right part of the equation 1 = sqrt(1).
Nice video syber
Thanks!
TMHO there are 2 questions in 1:
(1) let the function f(Z)=Z^i, Z in C, evaluate : f(1+i)= ?
(2) solve for Z in C the following equation : Z - (1+i)^i = 0
The video gives the answer for the second formulation, but for the first, we must modify it because the complex Log function (given Z=R e^(i Theta)) has the following restrictions in C:
- R is non zero (complex log is defined for C*)
- 0
The modulus for the final answer has a "+2kπ" in the exponent. What does that mean? How can there be an infinite number of moduli?
It's not the modulus, it's the argument, so the angle, since it's an angle, all measures that are in the form - pi/4 (mod 2pi), are the same angle, that's why cos and sin are 2pi-periodic functions
@@thecrazzxz3383 In the answer in the video the argument is given as "ln(√2)" (with the modulus, that also includes "e" being on the right side of the first "e" instead of the left.)
I think maybe the answer just isn't completely simplified and technically needs more work.
@@JefiKnight Oh sorry i was completely missing it, yes there should be only one solution ! Actually, the whole video doesn't make any sense, nothing defines raising a number to a complex exponent, it doesn't make any sense !
How about P(a, 2) = a ^ a = 1 + i ; a = ?
The calculation is cool. But can the answer be written as a+bi? The answer doesn't look satisfying. I have no idea how to comprehend this number at all...
Thanks! I think we can put it in the exact a+bi form using arctan but it will be a little painful 😂
I'm wondering if Euler's formula is valid for complex powers? Meaning raised to the power n; where n is complex
Can anyone justify this?
I haven't watched the video yet.
First, convert the complex number on the base into polar form, where r²=x²+y² and tanθ=y/x. x=1 and y=1, so r²=1²+1²=2 and r=√2. tanθ=y/x=1/1, so θ=π/4+2πn, where n is an arbitrary integer. That means that 1+i=√2·e^{πi/4+2πin}.
Raising that to the power of i multiples the exponent by i, giving us √2·e^{-π/4-2πn}.
(1+i)^i has infinite real solutions, following a geometric progression. Its principal root is √2·e^{-π/4}.
What tells you that raising to the power of i multiplies the exponent by i? Can you prove it? By definition, all we know is that e^ix = cosx + isinx for all real numbers x...
e^(iθ)=cosθ+i sinθ can be proven from a much simpler claim, that d(e^x)=e^x dx. If d(e^x)=e^x dx, then d[e^(ix)]=e^(ix) i dx. d(cos x + i sin x)=-sin x + i cos x = i² sin x + i cos x = i (cos x + i sin x). As well, e^0=1, and cos 0 + i sin 0 =1. With these two facts established, we now know that e^(ix)=cos x + i sin x.
We can also show, starting from d(e^x) = e^x dx, that (e^a)(e^b)=e^(a+b); and from there, it's trivial to show that (e^a)^b=e^(ab). If b=I, then (e^a)^i=e^(ai).
Despite i's reputation for being exotic, it's actually a really well-behaved number
Beautiful!!! ❤️
Are logarithmic rules applicable for non-real numbers also? Since Ln(z) is defined when z is positive isn't it?
yes.
Wait since when can you even put numbers at complex numbers? Is this defined?
Yes. It’s defined
@@SyberMath Where?
I feel it is enlightening to expose the answer in rectangular form so that the real and imaginary parts are clearly visible.
2kπ - π/4 ____ ____
e . ( cos √ln 2 + %i . sin √ln 2 )
Applying the half-angle identities is perhaps overkill but pretty nevertheless:
2kπ - π/4
e ____________ ___________
------------------ . (√1 + cos ln 2 + %i . √1 - cos ln 2 )
√2
2:15 dark sarcasm in the classroom ? 😅
You can't just multiple powers of i time i since they are complex numbers. That rule doesn't apply to complex numbers, even though you got lucky the answer came out correct. That rule only applies to real numbers. Please re-think is over and make the necessary correction to the steps. You are on the right track.
You're right, complex powers are not defined !! And he gets something really weird, because the for the modulus of complex number he gets at the end, so e^-pi/4+2kpi, there's an infinite number of possibilities for (1+i)^i, which doesn't make sense, raising to complex powers doesn't neither !
But when you say "You are on the right track.", what do you mean ?
You will find that the n should not actually be in your answer, however. (1+i)ⁱ is a number, and has only one value. If you want an infinite number of solutions, solve the equation x⁻ⁱ = (1+i). This is similar to the concept that √49 = 7 has only that one value, whereas solving the equation x² = 49 has two solutions, ±7.
No, having n in the final answer is correct. In polar form there are an infinite number of ways to write a single complex number.
1+i = e^i*pi/4, but also, 1+i = e^i*9pi/4, and e^-i*7pi/4, and so on
Also the last step is still missing with transformation of the result into complex number (but that's probably something obvious for all those who got to that point)
exp(2kπ - π/4) * (cos(ln(2)/2) + i*sin(ln(2)/2))
Nice and useful video
very nice
x^i=e^ln(x)i. Therefore
a^i=cos(ln(a))+isin(ln(a))
Why is x^i this e^ln(x)i, you mean x^i = [e^ln(x)] ^i, but what tells you that [e^ln(x)] ^i is actually, e^(ln(x)i)? Multiplication of powers when you raise a complex to a power is technically only defined for a complex raised to a natural number, or at most a rationnal number, but nothing is defined for a complex to a complex?
（1+愛）の愛情は複雑。愛の愛情は実数。
1 + e^-pi/2
Зачем вообще копаться в комплексных числах? Мы ведь точно знаем, что их не существует в реальности. Причём, в отличие от отрицательных, их себе даже образно не представить.
Passatempo...
(1+i)^i is a transcendental number.
The i became 1 and the 1 became i.(?)
Cool!
The answer is 2 bc "i" am the "one"
Joey Cliffs
2inPI NOT 2nPI
Caleb Tunnel
It's like a meal with no dessert.
What do you mean?
When I wrote it I meant the video ended too soon, before the calculation was completed. But I was wrong, so my comment was stupid. I expected a final result in the form of r . e^iθ and didn't pay enough attention, probably because it's written as e^iθ . r, let's say I was tired ...
@@joluju2375 Oh OK
r=Sqrt(1^2+i^2)=0 haha
if you're gonna turn -n into k might as well turn -2n into k
No, has to be an even integer which is why the 2 is necessary
@@asparkdeity8717 okay well then nπ = k
@@firelow no, u don’t understand the use of the constant k. k represents some arbitrary number restricted to a specific domain to show all possible numerical solutions. In this case, he specified k to be any integer, which gives the correct answer. Changing any other constants to k like u mentioned at your discretion equivalent is like changing the answer of 1+1 to 3, it makes the whole answer incorrect (unless u change the domain of k and explicitly state it)
@@asparkdeity8717 poggers
❤❤❤
🥰❤🥰
Purdy Vista
This video is incorrect. There is a single principal branch of the complex power function, so the answer should be a single complex number, not an infinite number of them.
I see. Thanks!
Super important point!!
Wehner Avenue
?
Before solving the problem, one should specify what is understood by (1+i)^i. For example, if we are interested in the principal branch, then n=0. As presented, this is a semi-literate nonsense.
Hilton Squares
Grade😊
答え 2i