Question

If $z=-1-\sqrt{3}i$ and $z^{6}=a+bi$, then $a$ = _ and $b$ = _

Answer 1

Rewriting $z$ in polar form makes this trivial.

$z=|z|e^{i\mathrm{arg}(z)}$

We have

$|z|=\sqrt{(-1)^2+(-\sqrt3)^2}=2$

$\mathrm{arg}(z)=\tan^{-1}(-1,-\sqrt3)=-\dfrac{2\pi}3$

(not to be confused with the standard inverse tangent function $\tan^{-1}x$. Here $\tan^{-1}(x,y)$ is the inverse tangent function that takes into account position in the coordinate plane; look up "atan2" for more information)

So we have

$z=-1-\sqrt3\,i=2e^{-2\pi/3\,i}$

Then

$z^6=2^6\left(e^{-2\pi/3\,i}\right)^6=64e^{-4\pi\,i}=64$

so that $a=64$ and $b=0$.