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)?
 A: First: I think the question is badly posed. Unless I'm mistaken, there's no reason to assume (unless you've a stated convention) that taking the square root and then the cube root must lead to the complex number with the minimum argument. All the roots are equally correct. 
That said, your calculation of $e^{i \pi /18}$ is correct. Now note that
$$
-e^{i \pi /18} = e^{i \pi}e^{i \pi /18}
$$
to compute the argument when "$r$ is negative".
A: You want the $e^{i \pi/18}$ in the form of $r*e^{ix}$, where $r$ is negative and $x$ is minimal?
$-1=e^{i\pi}$, so by multiplying it with $-e^{i \pi}$, you won't change it's value, so:
$$e^{i \pi/18}=-e^{i \pi}e^{i \pi/18}=-e^{i\pi+i\pi/18}=-e^{\frac{19}{18}*i\pi}$$
So $\frac{19}{18}\pi$ is the lowest positive $x$, since the period is $2 \pi$.
A: 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 $x = \frac {\pi}{18}+\pi$.
