Complex number problem

**Shaun Gill** · March 22nd 2006, 04:38 AM

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

**CaptainBlack** · March 22nd 2006, 05:23 AM

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:

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

RonL

**Shaun Gill** · March 22nd 2006, 05:25 AM

thanx a lot. genius.

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