Solving $e^{iz} = 3^{1/2} -i$ The question says find all $z \in {\mathbb C}$ that satisfy $e^{iz} = 3^{1/2} -i$. Write your answer(s) in standard form.
First thing I did was change the RHS to polar form ($2e^{-i\frac{\pi}{6}}$), and then with the LHS I simplified and manipulated it to $e^{-y}(\cos x + i\sin x)$ using Euler's equation. Im not too sure where to go from here. Can anyone help me?
 A: Hint: Begin by showing that $$e^z = 1 \iff z = 2n\pi i$$ for some integer $n$. Now note that $2 = e^{\ln{2}}$, so you have
$$e^{iz} = e^{\ln{2} - i\pi/6}$$ which can be rearranged to state 
$$e^{iz - \ln{2} + i\pi/6} = 1$$
A: $z = x + {\rm i}y$.
$$
{\rm e}^{{\rm i}x}{\rm e}^{-y} = \sqrt{3\,} - {\rm i}\,,
\quad
{\rm e}^{-y}\cos\left(x\right) = \sqrt{3\,}\,,
\quad
{\rm e}^{-y}\sin\left(x\right) = -1
$$
$$
\tan\left(x\right) = -\,{\sqrt{3\,} \over 3}\,,
\quad
{\rm e}^{-2y} = 4\,,
\quad
x = {5 \over 6}\,\pi + n\pi\,,
\quad
n \in {\mathbb Z}\,,
\quad
y = -\ln\left(2\right)
$$
$$\color{#ff0000}{\large%
z_{n}
=
\left(n + {5 \over 6}\right)\pi
-
{\rm i}\ln\left(2\right)\,,
\qquad
n \in {\mathbb Z}}
$$
A: Hint: $|e^{i(x+iy)}|=e^{-y}$ implies $y=-\ln 2$.
A: Putting $\;z=x+iy\;,\;\;x,y\in\Bbb R\;$ :
$$e^{iz}=e^{-y+ix}=e^{-y}\cos x+ie^{-y}\sin x=\sqrt3-i\iff$$
$$e^{-y}\cos x=\sqrt3\;,\;\;e^{-y}\sin x=-1\implies\frac{\sqrt3}{\cos x}=-\frac1{\sin x}\iff$$
$$\tan x=-\frac1{\sqrt3}\iff x=\frac{5\pi}6+k\pi\;,\;\;k\in\Bbb Z$$
and then
$$e^{-y}\cos\left(\frac{5\pi}6+k\pi\right)=\sqrt3\implies e^{-y}=\pm2$$
and since the minus sign in the last equality is impossible, we get
$$e^{-y}=2\implies y=\log\frac12=-\log2$$
Thus
$$z=\left(\frac{5\pi}6+k\pi\right)-i\log2$$
