I have a pretty straight forward question.
Change $z = (-1+i\sqrt3)^{2017}$ to $a+bi$ $\;$ form & polar form. Where $i = \sqrt{-1}$.
So i want to change it to $z = re^{iv}$.
$r$ is easy to calculate. $r = \sqrt4 = 2$.
However the angle is where im struggeling.
I know all the "standard" angles with components like: $\frac{\sqrt3}2, \frac12, \frac1{\sqrt2}$.
However now we have $\frac{\sqrt3}{-1}$. How do you tackle this type of question?
Hint:
Using this, arg$\displaystyle(-1+i\sqrt 3)=\pi+\arctan\dfrac{\sqrt3}{-1}=\pi-\dfrac\pi3$
$\implies\displaystyle -1+\sqrt3i=2e^{2\pi i /3}$
See also: Euler's formula, is this true?
Hint: The point $-1+\sqrt3i$ makes an equilateral triangle together with $0$ and $-2$.
$z = (-1+i\sqrt3)^{2017} = (2\omega)^{2017}$, where $\omega = \dfrac{-1+i\sqrt3}{2}$ is the third root of unity. ($v = \arctan(\sqrt3/(-1)) = \dfrac{2\pi}{3}$), so $$z = 2^{2017} \omega^{2017} = 2^{2017} = 2^{2017} e^{2\pi/3} \\ = 2^{2017} \, \frac{-1+i\sqrt3}{2} = -2^{2016} + i \, 2^{2016}\sqrt3.$$
