# Complex number problem

• March 22nd 2006, 04:38 AM
Shaun Gill
Complex number problem
Just wondering how i would solve (z^6) -2(z^3) +4 = 0 (where z is a complex number)

any help would be greatly appreciated.

Shaun
• March 22nd 2006, 05:23 AM
CaptainBlack
Quote:

Originally Posted by Shaun Gill
Just wondering how i would solve (z^6) -2(z^3) +4 = 0 (where z is a complex number)

any help would be greatly appreciated.

Shaun

Stage 1: Treat it as a quadratic in $z^3$ and use the quadratic
formula.

This will give you two complex roots.

Stage 2:Convert each of the roots into polar form (that is write $w$
as: $w=|w|\ e^{\bold{i}\theta}$, then:

$
w^{\frac{1}{3}}=|w|^{\frac{1}{3}}\ e^{\bold{i}\theta/3}\ \wedge\ |w|^{\frac{1}{3}}\ e^{\bold{i}(\theta/3+2\pi/3)}\ \wedge\ |w|^{\frac{1}{3}}\ e^{\bold{i}(\theta/3+4\pi/3)}
$

RonL
• March 22nd 2006, 05:25 AM
Shaun Gill
thanx a lot. genius.