Converting to polar form 
Write each of the given numbers in the polar form $re^{i\theta}$.
a.) $\frac{1-i}{3}$
b.) $-8\pi (1+\sqrt 3 i)$

For a, I got: 
r = $\frac{\sqrt 2}{3}$ and $e^{i7\pi /2}$ since $\frac{\frac{1}{3}}{\frac{\sqrt 2}{3}} = \frac{\sqrt 2}{2}$ and $\frac{-\frac{1}{3}}{\frac{\sqrt 2}{3}} = -\frac{\sqrt 2}{2}$, but the answer for this is $\frac{\sqrt 2}{3}e^{-i\frac{\pi}{4}}$, how did they get that?
For b, I got:
$-16\pi e^{i\frac{\pi}{3}}$ but the answer for this is $16\pi e^{-i\frac{2\pi}{3}}$.
 A: Following two points need to be used here:
(1) The principal value of $\arctan \frac yx$ lies in $[-\frac\pi2,\frac\pi2]$
(2)From this, we need to identify the principal value of the argument of a complex number.
Let $$\frac{1-i}3=re^{i\theta}=r(\cos\theta+ i\sin\theta)$$ where $r>0$
Equating the real & the imaginary parts, 
$r\cos\theta=\frac13$ and $r\sin\theta=-\frac13 $
Squaring and adding we get, $r^2=(\frac13)^2+(-\frac13)^2=\frac29$
So, $r=\frac{\sqrt2}3$ and $\cos\theta=\frac{\frac13}{\frac{\sqrt2}3}=\frac1{\sqrt2}$ and $\sin\theta=\frac{-\frac13}{\frac{\sqrt2}3}=-\frac1{\sqrt2}$ 
so $\tan\theta=-1$ and $\theta$ lies in the 4th quadrant hence $\theta=-\frac\pi4$
Similarly, if we put $-8\pi(1+\sqrt3i)=r(\cos\theta+ i\sin\theta)$ where $r>0$
Applying the same method, $r^2=(-8\pi)^2+(-8\pi\sqrt3)^2=(8\pi)^24\implies r=16\pi$
So, $\cos\theta=\frac{-8\pi}{16\pi}=-\frac12$ and  $\sin\theta=\frac{-8\pi\sqrt3}{16\pi}=-\frac{\sqrt3}2$ 
so, $\tan\theta=\sqrt3$ and $\theta$ lies in the 3th quadrant hence $\theta=-\pi+\frac\pi3=-\frac{2\pi}3$
