# Roots of complex polynomials

1. Aug 6, 2005

### complexhuman

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!!!

2. Aug 6, 2005

### lurflurf

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: Aug 6, 2005
3. Aug 7, 2005

### AKG

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: Aug 7, 2005
4. Aug 7, 2005

### HallsofIvy

Staff Emeritus
Since ei&theta;= cos(&theta;)+ i sin(&theta;), AKG and lurflurf are saying exactly the same thing!

5. Aug 7, 2005

### AKG

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.

6. Aug 7, 2005

### HallsofIvy

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

7. Aug 8, 2005

### AKG

Oops, my mistake. Then whatever form my post was in is easier to understand. What would it be called? Exponential form? Euler form?

8. Aug 8, 2005

### HallsofIvy

Staff Emeritus
"Exponential" form would do. Engineering texts talk about "Cis &theta;" which is shorthand for "cos(&theta;)+ i sin(&theta;)" but, yes, exponential form is a lot simpler (though more sophisticated) than "cis".