<s>[INST] complex number written in $re^{ix}$ form where $r<0$ 
The complex number $\left(\sqrt{\frac{1}{2} + \frac{\sqrt{3}}{2}i}\right)^{1/3}$ can be written in the polar form as $r(\cos x + i\sin x)$. If $r<0$, find the smallest positive value of $x$ in degrees.

$\frac{1}{2} + \frac{\sqrt{3}}{2}i = e^{i\pi/3}$. Thus $\left(\sqrt{\frac{1}{2} + \frac{\sqrt{3}}{2}i}\right)^{1/3} = e^{i\pi/18}$. At this point, I feel like the answer is $\pi/18$ or 20 degrees, but I'm not sure about the $r<0$ part. How does that affect the answer (if it does)?
 [/INST] Just do it.
$z= re^{i\theta} = r(\cos \theta + i\sin \theta)$ where $r = \pm |z|$.
As $z = e^{i\frac {\pi}{18}}$ and $|z| = 1$ then $r = -1$ and
$e^{i\frac{\pi}{18}} = -1(\cos \frac{\pi}{18} + i \sin \frac{\pi}{18})$.
We are asked to find the smallest positive $x$ so that -$1(\cos \frac{\pi}{18} + i \sin \frac{\pi}{18})=\cos x + i\sin x$.
In other words words $-\cos x = \cos \frac{\pi}{18}$.  And $-\sin x = \cos \frac{\pi}{18}$.
Now basic trig identities $-\cos x = \cos(\pm\pi \pm x)$ and $-sin x = \sin(-x) = \sin(x \pm \pi)$. (All modulo $2\pi$ of course)
So the only case were $-\cos x =  \cos y$ and $-\sin x = \sin y$ is when $y = x \pm \pi$, or vice versa, $x = y \pm \pi$. (All modulo $2\pi$ of course)
So $x = \frac{\pi}{18} \pm \pi$  (All modulo $2\pi$ of course).
At for $x$ to be the least positive value $x = \frac {\pi}{18} + \pi = \frac {19}{18}\pi$.
.....
Note:  This reasoning leads to the identity that $e^{i\theta + i2k\pi} = -e^{i(\theta + \pi)+ i2k\pi}$ which is very easy to prove.  [$e^{i(\theta + \pi)} = \cos (\theta + \pi) + i\sin(\theta + \pi) = -\cos\theta - i\sin \theta = -e^{i\theta}$]
Has we simply kept that in mind the solution would be simply:
$e^{i\frac {\pi}{18}}= -e^{i(\frac {\pi}{18}+\pi)} = \cos x + i\sin x$ so $x = \frac {\pi}{18}+\pi$.
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