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.
 A: *

*Yes, $\left(\cos\left(\frac{2\pi}9\right)+i\sin\left(\frac{2\pi}9\right)\right)^3=\left(\cos\left(\frac{2\pi}3\right)+i\sin\left(\frac{2\pi}3\right)\right)$.

*Note that$$\frac{2+2i}{-\sqrt3+i}=\frac{1+i}{-\frac{\sqrt3}2+\frac12i}.$$Furthermore, $1+i=\sqrt2\left(\cos\left(\frac{\pi}4\right)+i\sin\left(\frac{\pi}4\right)\right)$ and $-\frac{\sqrt3}2+\frac12i=\cos\left(\frac{5\pi}6\right)+i\sin\left(\frac{5\pi}6\right)$. So, the quotient is equal to$$\sqrt2\left(\cos\left(\frac\pi4-\frac{5\pi}6\right)+i\sin\left(\frac\pi4-\frac{5\pi}6\right)\right)=\sqrt2\left(\cos\left(-\frac{7\pi}{12}\right)+i\sin\left(-\frac{7\pi}{12}\right)\right).$$

*Note that$$\frac{4i}{3e^{4+i}}=\frac43\times\frac{e^{\frac{\pi i}2}}{e^{4+i}}=\frac43e^{-4+\frac{\pi i}2-i}.$$

A: 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.
A: $\frac{2+2i}{-\sqrt{3} +i} =$
$\frac {(2+2i)(-\sqrt{3} -i)}{(-\sqrt{3} +i)(-\sqrt{3} -i)}=$
$\frac {(-2\sqrt{3} +2)- (2\sqrt 3+2)i}{3+1}=$
$\frac {-\sqrt3 + 1}2 -\frac{\sqrt 3 + 1}2i=$
$r(\cos \theta + i\sin \theta)$
So we need to solve for $r\cos \theta = \frac {-\sqrt3 + 1}2$ and $r\sin \theta =  -\frac{\sqrt 3 + 1}2$.
That's all.
We solve for $r$ by squaring and adding both terms:
$r^2\cos^2 \theta + r^2 \sin^2 \theta = (\frac {-\sqrt3 + 1}2)^2 + ( -\frac{\sqrt 3 + 1}2)^2$
$r^2(\cos^2 \theta + \sin^2 \theta) = \frac {3 - 2\sqrt 3 + 1}4 + \frac {3+2\sqrt 3 + 1}4$.
$r^2 = 1 - \frac {\sqrt 3}2 + 1+\frac {\sqrt 3}2$
$r^2 = 2$
$r = \sqrt 2$.
And we solve for $\theta$ by dividing both terms and taking the arctangent:
$\tan \theta = \frac {\sin\theta}{\cos\theta} = \frac {r*\sin\theta}{r*\cos\theta} = \frac {-\frac{\sqrt 3 + 1}2}{\frac {-\sqrt3 + 1}2}$
$=\frac {\sqrt 3 + 1}{-\sqrt3 + 1} =\frac {(\sqrt 3 + 1)(-\sqrt 3- 1}{(-\sqrt3 + 1)}=\frac {-4 - 2\sqrt 3}{2} = -2-\sqrt 3$
So $\theta = \arctan  -2-\sqrt 3= -\frac 5{12}\pi$
S $\frac{2+2i}{-\sqrt{3} +i} = \sqrt 2*e^{-\frac 5{12}\pi i}$.
