Question

For z = 2cis Pi/6 , find z5 in rectangular form.

Answer 1

$z=2\mathrm{cis}\dfrac\pi6=2\left(\cos\dfrac\pi6+i\sin\dfrac\pi6\right)$

By DeMoivre's theorem, you have

$z^5=2^5\mathrm{cis}\dfrac{5\pi}6=32\left(\cos\dfrac{5\pi}6+i\sin\dfrac{5\pi}6\right)$
$\implies z^5=32\left(-\dfrac{\sqrt3}2+\dfrac12i\right)=-16\sqrt3+16i$