Expressing complex numbers in exponential form I do not know why I am having so many issues with complex numbers. I am basically trying to teach myself, but keep doubting myself and getting very frustrated. I've no way of checking my answers so I'm just lost. So I want to write the following in polar form $re^{i\theta}$ (although I thought that was called exponential form?) and $-\pi < \theta \leq \pi$.
My first question is expressing $(\cos(\frac{2\pi}{9}) + i\sin(\frac{2\pi}{9}))^3$. Using De Moivre, this is $$\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3})$$ so it is $e^{i(\frac{2\pi}{3})}$ yes?
My next one is to express $\frac{2+2i}{-\sqrt{3} +i}$. So my attempt is to multiply by the complex conjugate and we get that this is $\frac{-\sqrt{3} +1}{2} + i\frac{-1 - \sqrt{3}}{2}$. Find $r$ which is the modulus which is $\sqrt2$ and $\theta$ is just something I cannot seem to find. I know if $z = a+ib$ then $\theta = \tan^{-1}(\frac{b}{a})$ but I'm stuck even with the $(\frac{b}{a})$ and am confusing myself of what quadrant to look in.
Lastly, I have no idea how to express the next one which is $\frac{4i}{3e^{(4+i)}}$.
Any help and direction at all is appreciated.

You have $\frac {2+2i}{-\sqrt 3 + i}$ rather than rationalizing the denominator, and converting, convert numerator and denominator.
$2+2i = 2\sqrt2e^{\frac {\pi}{4}i}\\
-\sqrt 3 + i = 2e^{\frac {5\pi}{6}i}\\ 
\frac {2+2i}{-\sqrt 3 + i} = \sqrt 2\frac {e{\frac {\pi}{4}i}}{e{\frac {5\pi}{6}i}} = \sqrt2e^{(\frac {\pi}{4}i-\frac{5\pi}{6}i)}= \sqrt2e^{-\frac {7\pi}{12}i}$
Since you did get as far as: 
$z = \frac {-\sqrt 3 + 1}{2} + \frac {-\sqrt 3 - 1}{2}i$
After doing it enough times in trig class I came to memorize $\cos \frac \pi{12} = \frac {\sqrt 6 + \sqrt 2}{4}, \sin \frac \pi{12} = \frac {\sqrt 6 - \sqrt 2}{4}$
Which might have helped.
