Finding Roots of Complex Polynomials: General Formula and Exponential Form

[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!

Thanks in advance!

 

[lurflurf]

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!

Thanks in advance!

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

 

[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.

 

[HallsofIvy]

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

 

[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.

 

[HallsofIvy]

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

 

[AKG]

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

 

[HallsofIvy]

"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".