Solve the complex equations I have a question from complex calculus. How to solve this two equations: 
a)
$$
\sin(z)=2015
$$
I know that $\sin(z)$ equals to 
$$
\frac{e^{iz}-e^{-iz}}{2i}
$$
And i don't know whats next.

b)
$$
e^{z^{2}}=1
$$
I know that
$$
e^{z}=e^{x}(\cos(y)+i\sin(y))
$$
Can you help me to solve this equations? 
 A: $$\sin z = \sin(x + iy) = \sin x \cos (iy) + \cos x\sin(iy) = \sin x \cosh y + i\cos x\sinh y$$
$$\Rightarrow \left\{\begin{array}{l} \sin x \cosh y = 2015 \\ \cos x \sinh y = 0\end{array} \right.$$
For the second equation to be true, either $y = 0$ or $x = \displaystyle\frac{\pi}{2} + n\pi$. 
$y = 0$ gives $\sin x = 2015$ which has no solution, so $\displaystyle x = \frac{\pi}{2} + n\pi$ 
Since $\cosh y > 0$, we expect $\sin x > 0$, so $n$ has to be an even number
Thus $\displaystyle x = \frac{\pi}{2} + 2k\pi$. This gives $\cosh y = 2015$ or $y = \cosh^{-1} (2015)$
A: $$sin(z)=2015 <=> z=sin^{-1}(2015)=\frac{\pi}{2}-i*ln(2014+24\sqrt{7049})$$
It gives us (for Z is the set of integers, n element of Z):
$$z=2\pi n+\pi-sin^{-1}(2015)=2\pi n+\pi-(\frac{\pi}{2}-i*ln(2014+24\sqrt{7049}))$$
Or:
$$z=2\pi n+\pi+sin^{-1}(2015)=2\pi n+\pi+(\frac{\pi}{2}-i*ln(2014+24\sqrt{7049}))$$
And:
$$(e^z)^2=1$$
Integer solution is:
$$z=0$$
It gives us (for Z is the set of integers, n element of Z):
$$z=\sqrt{2\pi}(-\sqrt{i*n})$$
Or:
$$z=\sqrt{2\pi}\sqrt{i*n}$$
