# Roots of complex polynomials

## Main Question or Discussion Point

Hi all

Jut had a question. How do I go about finding the general formula for roots of the complex poly $${z}^{n}-a$$ where a is another complex number.

Do I just go $${z}^{n}=a$$??? :S so complicated this things!!!

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lurflurf
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complexhuman said:
Hi all

Jut had a question. How do I go about finding the general formula for roots of the complex poly $${z}^{n}-a$$ where a is another complex number.

Do I just go $${z}^{n}=a$$??? :S so complicated this things!!!

It is easier if z is written in geometric form
let a=u+i v where u and v are real
z=r(cos(t)+i sin(t)) where 0<=r
then
z^n-a becomes
(r^n)(cos(n t)+i sin(n t))-a
so r should be chosen by
r^n=|a|=sqrt(a^2+b^2)
cos(n t)+i sin(n t)=a/|a|
so
cos(n t)=u/|a|
sin(n t)=v/|a|
tan(n t)=v/u
t=(1/n)Arctan(v/u)
this is one solution but n can be found using the usual methods of solving trigonometric equations.
in fact
t=(1/n)Arctan(v/u)+2pi k/n
k=0,1,2,...,n-2,n-1,
thus
z=r(cos(t)+i sin(t))
=|a|^(1/n)(cos((1/n)Arctan(v/u)+2pi k/n)+i sin((1/n)Arctan(v/u)+2pi k/n
))
k=0,1,2,...,n-2,n-1

Last edited:
AKG
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It's actually easiest to understand in polar co-ordinates. $a = |a|\exp (i\theta )$. So simply take |z| to be |a|1/n. Take angles $\theta _k = \frac{\theta + 2k\pi}{n}$ for values k = 0, 1, ..., n-1. Then your roots are the complex numbers $|z|\exp (i\theta _k)$ for the n values for k.

Last edited:
HallsofIvy
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Since ei&theta;= cos(&theta;)+ i sin(&theta;), AKG and lurflurf are saying exactly the same thing!

AKG
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We're saying very different things. lurflurf is saying that it's easiest to see it in geometric form, I'm saying it's easiest to see it in polar form.

HallsofIvy
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Writing a+ bi as r(cos(θ)+ i sin(θ)) is writing the complex number in polar form.

AKG