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

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- Mar 22nd 2006, 04:38 AMShaun GillComplex 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 - Mar 22nd 2006, 05:23 AMCaptainBlackQuote:

Originally Posted by**Shaun Gill**

formula.

This will give you two complex roots.

Stage 2:Convert each of the roots into polar form (that is write $\displaystyle w$

as: $\displaystyle w=|w|\ e^{\bold{i}\theta}$, then:

$\displaystyle

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 - Mar 22nd 2006, 05:25 AMShaun Gill
thanx a lot. genius.