# If z is a complex function and z^{3}=8i, what are

If z is a complex function and ${z}^{3}=8i$, what are all the values of z?
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Vivian Soares
${z}^{3}=8i$
${z}^{3}-8i=0$
${z}^{3}+{\left(2i\right)}^{3}=0$
$\left(z+2i\right)\left({z}^{2}+{\left(2i\right)}^{2}-2iz\right)=0$
$\left(z+2i\right)\left({z}^{2}-2iz-4\right)=0$
$z=-2i$
or ${z}^{2}-2iz-4=0$
$z=\frac{2i±\sqrt{-4+16}}{2}$
$z=i±\sqrt{3}$
So roots are $-2i,i+\sqrt{3},i-\sqrt{3}$
###### Not exactly what you’re looking for?
aquariump9
${z}^{3}=8i=8{e}^{i\left(\frac{\pi }{2}+2n\pi \right)}$
where $n\in Z$
$z=2{e}^{i\left(\frac{\pi }{6}+2n\frac{\pi }{3}\right)}$
$z=2{e}^{i\frac{\pi }{6}},2{e}^{i\left(5\frac{\pi }{6}\right)},2{e}^{i\left(3\frac{\pi }{2}\right)}$
$z=\sqrt{3}+i,-\sqrt{3}+i,-2i$
###### Not exactly what you’re looking for?
alenahelenash
Let $\omega ={e}^{\frac{2i\pi }{3}}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$ is a root of unit. So the roots of We have: ${z}^{3}=8i⇔{z}^{3}=\left(-2i{\right)}^{3}⇔\left(\frac{z}{-2i}{\right)}^{3}=1$ So: $\frac{z}{-2i}={\omega }^{k}k\in 0;1;2$. Thus The root of the equation ${z}^{3}=8i$ are: