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 $\displaystyle z^3$ and use the quadraticOriginally 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