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 [tex]{z}^{n}-a[/tex] where a is another complex number.

Do I just go [tex]{z}^{n}=a[/tex]??? :S so complicated this things!!!

Thanks in advance!
 

Answers and Replies

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 [tex]{z}^{n}-a[/tex] where a is another complex number.

Do I just go [tex]{z}^{n}=a[/tex]??? :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
 
Last edited:
AKG
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It's actually easiest to understand in polar co-ordinates. [itex]a = |a|\exp (i\theta )[/itex]. So simply take |z| to be |a|1/n. Take angles [itex]\theta _k = \frac{\theta + 2k\pi}{n}[/itex] for values k = 0, 1, ..., n-1. Then your roots are the complex numbers [itex]|z|\exp (i\theta _k)[/itex] 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
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Oops, my mistake. Then whatever form my post was in is easier to understand. What would it be called? Exponential form? Euler form?
 
HallsofIvy
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"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".
 

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