sqrt(a+bi), how to get the square roots of a complex number
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- Опубліковано 17 жов 2024
- How can we get the square root of a complex number? I.e. we want a formula to simplify sqrt(a+bi). This tutorial will show the algebraic way to do so!
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At last I found the hidden algebra channel after all these challenges and many equation I am finally here for the great teacher's techniques to solve problems!
You could also just convert into polar coordinates which gives you: sqrt(r)sqrt(e^(ia)) -> sqrt(r)e^(ia/2) where r is the radius and a the angle.
You can then generalize this even more and get the formula:
nth-root(a+bi) = nth-root(r)e^(i(a+2kπ)/n) in which k is an integer
I think this is cleaner than getting a form with an arctangent involved, but yes polar is much simpler and easier to deal with exponents
Using the the modulus of a+bi, you get three equations :
x²+y²=sqrt(a²+b²)
x²-y²=a
2xy=b
the two first equations give us : x² =1/2(a+sqrt(a²+b²))and y²=1/2(a-sqrt(a²+b²)) and you can conclude with the sign of b.
nice!!
I don't see clear the first equation: is the modulus of a square the square of the modulus?
@@rmlu9767
(x+iy)²=a+ib
Thus : |x+iy|²=|a+ib|
This gives : x²+y²=sqrt(a²+b²)
Is it more clear?
@@dulot2001 Yes, I completely forgot that property. Thank you.
Fabulous teacher, got me loving math again!
These "just x" channels are really cool. You always go into more detail on some things that we might not have seen in a typical "x" course.
Technically x≠0 for b=0 since y=b/(2x) which makes sense since for b=0 you would need y=0 i guess
No, it depends on the value of a. If a
Can't we convert the complex number into trigonometric form and apply De Moivres theorem?
not all numbers can be converted to trig forms by, we can only do multiples of 15 and try to use sin(a+b) formulas but much easier using this formula
@@AayushSrivastava0307 You can use half angle formula, it should be equivalent
Why not just convert to polar coordinates? (a + bi)^(1/2) = ((a^2 + b^2)e^itaninv(b/a))^(1/2) = +/- (a^2 + b^2)^(1/4)e^i(taninv(b/a))/2
We're assuming that we don't have a calculator or the brainpower to work out an inverse tangent. After all, there's no fun in that.
Computing square roots are much easier than computing complex exponentials and inverse tangents.
@@he-man4076 but you can easily convert that to square roots bc a complex exponential is just sin and cos
So we have an angle equal to the inverse tangent, and that angle is inputed into sin and cos, and we can easily find the value of that just by using triangles.
7:51 "…and hopefully everything is clear." Crystal.
My first thought was to use polar or trig form: a+bi = re^(it) so by DeMoivre's, √(a+bi) = √r•e^(i•t/2) =
√r(cos(t/2) + isin(t/2) or √r(cos((t/2)+π)+isin((t/2)+π)). Here r = √(a²+b²), cos(t/2)=
±√((1+cos t)/2), sin(t/2) = ±√( (1-cos t)/2). cos(t) = a/√(a²+b²)
After simplification and accounting for ± signs, results are equivalent.
If the argument is a nice fraction of π, converting to modulus-argument form makes sense, but sometimes you'll get a horrible decimal for the argument
Didn't really understand why we have to put only plus on 3:07
If sqrt(16(a^2+b^2)) is greater than 4a it's ok? Isn't it
If sqrt(16(a^2+b^2))>4a, then 4a " - " sqrt(16(a^2+b^2)) would be negative, and x^2 can't end up with a negative because x is real
It's my first comment in this channel!
Congratulations to myself!
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If b=0 it's y that is = 0, just plug in b = 0 in both of these and y will reduce to 0
if b = 0 and a is negative, y will not be 0 since sqrt(a) would be a complex number. x will be 0 though.
@@koenmazereeuw4672 but x and y are defined to be real
@@plislegalineu3005 y is the coefficient of i, still real
And then there is the similar identity:
sqrt(a + b) = (S + b / S) / sqrt(2)
where S = sqrt( sqrt(a^2 - b^2) + a)
It is not really a simplification (unless it is a denestable nested radical), but it is an interesting identity nonetheless. And for b = i * c , it can be used to derive the identity in the video.
How does that simplification help at all? Also what’s this identity called?
@@Firefly256 What simplification?
@@XJWill1 “it is not really a simplification” so it kind of is a simplification but not really at the same time?
@@Firefly256 Consider a = 2 and b = sqrt(3)
@@XJWill1 I typed this in wolframalpha and it said this is not always true (a=-1, b=1/2). Is it only true when a and b are positive?
Edit: I replaced a with |a| and b with |b| but it still says not always true, why?
Actually, if b=0, then y=0. The x might be 0 or non-0. In this case, the x value depends solely on a. Right?
what about \sqrt(z^2)? Is it always equal to z that is \sqrt{z^2}=z and it can be -z?
2:08 if b=0 and a
7:40 its only y equal to 0, x is the root of a
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Both graph y = sin x and y = sin^2 x are similar?
Is there any formula I can use sir
woah another channel! glad i found this
Great work!!
It can be derived also from polar form
a²+b²=c²
2X²=a+c=c+a
2Y²=-a+c=c-a
[a+i×(c²-a²)½]½=X+i×Y
what about applying de moivre’s formula to the polar form of a+bi?
That’s what I thought of as well, I think it’d be easier
@@shadow-ht5gk I’ve done that in my newer video comment. It happens that the use of the quadratic formula involved in the video is basically the same as the use of half-angle formulae that I need. But my calculations reveal more precisely how the signs for the square roots should be considered.
how i would generally do it is convert it to polar form, then half the angle and root the distance
Thx best math teacher now i can caculate fourth root of i and the sin of 22.5° and cos of 22.5°
Write z=a+bi=r*cis(w) where r=|z|=sqrt(a^2+b^2) and w=Arg(z) which here is restricted to lie in (-pi, pi].
Then, by de Moivre’s formula, the two distinct square roots of z are given by:
z_1=sqrt(r)*cis(w/2) and
z_2=sqrt(r)*cis(pi+w/2)=-z_1,
where w/2 lies in (-pi/2, pi/2].
Write r=sqrt(a^2+b^2), c=cos(w)=a/r, s=sin(w)=b/r.
cos(w/2)=sqrt((1+c)/2)
=sqrt((r+a)/(2r)), which is always nonnegative;
sin(w/2)=sgn(b)*sqrt((1-c)/2)
=sgn(b)*sqrt((r-a)/(2r))
So the square roots of z are:
+/-[sqrt((r+a)/2)+sgn(b)*sqrt((r-a)/2)*i].
3:39 you say that we only want the positive root because then when we try to get x it'll be real, but that's only if the discriminate is greater than -b.
What you're saying is true in this case but generally it's not necessarily true.
Or at least I think so lol
That's nice, but this case is the case he's solving.
I like how the modulus of the original number comes into play. Maybe this has some nice geometric visualisation?
Converting to polar coordinates shows how
Taking the sqrt halves the angle to the x axis and square roots the modulus
Wow! Amazing!
Thank you so much tr plz do cube root of complex number
the first thought coming to my mind after seing that minia consist in passed from algebra to exponential form to solve the problem
6:42 This is not andy! 😂😂
@@bprpmathbasics hehe. looking forward to seeing you at the PREMIERE in about 3 hours...😉
Hello
The consequence of the formula is that the complex number will have two distinct root, hmm. Complex numbers sure are complex and they really defy my expectations.
Positive real numbers already have 2 distinct square roots. We just usually pretend one of them doesn't exist unless we're solving a quadratic. Be glad you're not dealing with hypercomplex numbers like quaternions. In those cases there can be infinitely many square roots for certain values.
For complex numbers, x² = y -> √y = ±x.
But sometimes, i² = j², but ±i ≠ ±j.
But in the start, didnt you use (a * b)² = a² * b² ? I thought that doesnt apply for imaginary numbers! Because you have to half the angle? Did i miss something?
a and b are real numbers on their own, b is just the "size" of the imaginary component of the complex number z=a+bi
I'm always wondering. If b is an even number (i.e., b=2b'), x=[-b'±√{(b')^2-ac}]/a is the formula for solving quadratic equations, but I've never seen anything using in other videos of the same kind. I think this is easier and more convenient, but why? In this video, x^2=[2a+√{(-2a)^2-4*(-b^2)}/4, so it seems like there's not much difference, but if "a" of a(x^2)+bx+c=0 is 1, the denominator becomes 1 and feels very easy.
More complex numbers later on this channel?
I don't know English, but it isn't a problem, because the mathematics transcends worlds :)
I solved it
Me right now: 🧠n't
Thank u so much! Its hard but your way of explaining makes it better🤗Your videos are great!!
Sqrt(a²+b²) = |a+bi|
@@antonis12gl48 i think they probably mean for the final expressions for x and y
@@antonis12gl48 somebody has forgotten about || notation and complex numbers
@@proloycodes You 're right, it means distance. I didn't see the i at the end. And I thought he meant sqrt(a^2 +b^2)= |a+b|. Sorry.
I don't undersand one thing, why you have b/2a, bacause it is -b/2a, it is a mistake
And there is me who said: sqrt a+bi=sqrt i
So a+bi=i
2:10 how to simplify: a = x^2 - (b^2 / 4x^2)
cancel the x^2's
a = 1 - b^2 / 4
a = 1 - b^2 / 2^2
cancel the ^2's
a = 1 - b/2
🤜😵
I think I need some sleep after reading this
I don't think that you can cancel the x^2's. Or the ^2's. Better than that, let's stop using the word 'cancel', since it's misleading.
@@star_ms I think it's sarcasm lmao
@@HershO. Math and sarcasm usually doesn't go together in most contexts, so I assumed it wasn't.
This is what makes me want to quit college week 1
Cube roots of a+bi using this method cannot be obtained. Prove me wrong.
Its super easy, either use the exact sqme taylor seiries you would use to do sqrt in the first place.
Or if you have q purely real calculator use qrchtqn to do q complex logarithem divide by two and raise e to it
Maaaaaaaaaaaaaaa
Slightly difficult
just geometry?
first
This is an unnecessary waste of time. Just put this all in polar coordinates.
so you call this is hard ?????