Solve $\sin(z) = 2$ There are a number of solutions to this problem online that use identities I have not been taught. Here is where I am in relation to my own coursework:
$ \sin(z) = 2 $
$ \exp(iz) - \exp(-iz) = 4i $
$ \exp(2iz) - 1 = 4i \cdot \exp (iz) $
Then, setting $w = \exp(iz),$ I get:
$ w^2 - 4iw -1 = 0$
I can then use the quadratic equation to find:
$ w = i(2 \pm \sqrt 3 )$
So therefore,
$\exp(iz) = w = i(2 \pm \sqrt 3 ) $ implies
$ e^{-y}\cos(x) = 0   $, thus $ x = \frac{\pi}{2} $
$ ie^{-y}\sin(x) = i(2 \pm \sqrt 3 ) $ so $ y = -\ln( 2 \pm \sqrt 3 ) $
So I have come up with $ z = \frac{\pi}{2}  - i  \ln( 2 \pm \sqrt 3 )$ 
But the back of the book has $ z = \frac{\pi}{2}  \pm i  \ln( 2 + \sqrt 3 ) +2n\pi$ 
Now, the $+2n\pi$ I understand because sin is periodic, but how did the plus/minus come out of the natural log? There is no identity for $\ln(a+b)$ that I am aware of. I believe I screwed up something in the calculations, but for the life of me cannot figure out what. If someone could point me in the right direction, I would appreciate it.
 A: Your answer is actually identical to the book's solution (after accounting for $2n\pi$).
The key identity:
$$ \frac{1}{2+\sqrt{3}} = 2 - \sqrt{3}.$$
A: Setting $w=e^{iz},$ we need to solve the equation $w^2-4iw-1=0.$ The solutions to this quadratic equation are $w=i(2+\sqrt 3)$ and $w=i(2-\sqrt 3).$
Let's deal with the first solution. We need to find $z=x+iy$ such that $e^{iz}= e^{ix}e^{-y}= i(2+\sqrt 3).$ This implies $\cos x =0.$ As you point out, that has solution set $\pi/2 + n\pi, n\in \mathbb Z.$ But there is another implication: In order to get $2+\sqrt 3$ as the imaginary part, we have to delete all $\pi/2 + n\pi$ for $n$ odd, as they lead to negative imaginary values. This is why we end up with $\pi/2 + 2n\pi.$
At this point, let's say goodbye to the original post for inspiration, as things are a little cloudy there. The easiest way to do this is write $e^{ix}e^{-y}=i(2+\sqrt 3)= e^{i\pi/2}(2+\sqrt 3).$ This tells us that $x= \pi/2 +2n\pi,$ and $-y= \ln(2+\sqrt 3).$
Solving for $z$ in the case $w=i(2-\sqrt 3)$ is the same. So in all, the solutions to the original problem are $z = (\pi/2 +2n\pi) -i\ln(2\pm\sqrt 3),n\in \mathbb Z.$
A: Let us solve 
$$e^{-y+ix}=i(2\pm\sqrt3),\quad (x,y)\in \mathbb R^2.$$
One can get
\begin{cases}
e^{-y}\cos x = 0\\
e^{-y}\sin x = 2\pm\sqrt3. 
\end{cases}
It seems that $\sin x=\pm1.$ What's happened?
$$\cos x = 0\rightarrow x=k\pi+\frac\pi2 = 2n\pi\color{red}\pm\frac\pi2.$$
$$(e^{-y}\sin x = 2\pm\sqrt3 >0) \wedge (e^{-y} > 0)\rightarrow \sin x >0 \rightarrow x = 2\pi\color{red}+\frac\pi2,$$
$$e^y = \color{red}+(2\pm\sqrt3),$$
$$y= \ln(2\pm\sqrt3) =\pm \ln(2+\sqrt3)=\pm \ln(2-\sqrt3),$$
$$\boxed{x+iy = \frac\pi2+2\pi n \pm \ln(2+\sqrt3)}$$
or
$$\boxed{x+iy = \frac\pi2+2\pi n \pm \ln(2-\sqrt3)}.$$
