Solving $\sin z = i$ I know that
$$\sin z = \frac{e^{iz}-e^{-iz}}{2i}$$
so:
$$\frac{e^{iz}-e^{-iz}}{2i} = i\implies e^{iz}-e^{-iz} = -2$$
but I can't take anything useful from here. How do I solve such equations?
What about $\tan z = 1$? Is there any solutions?
 A: Set $x=e^{iz}$, we then are solving for $x$ that satisfy
$$x-1/x+2=0$$
Multiplying by $x$ gives
$$x^2+2x-1=0$$
Which means
$$x=-1\pm\sqrt{2}$$
Thus we have that
$$e^{iz}=-1\pm\sqrt{2}$$
and
$$z=\frac{\ln(-1\pm\sqrt{2})}{i}=-i\ln(-1\pm\sqrt{2})$$
Then for your question about $\tan(z)=1$, that is easy. This occurs whenever $\sin(z)=\cos(z)$, implying $z=\frac{\pi}{4}+2n\pi$ for $n\in\mathbb{N}$.
A: using $z=a+ib$ we get
$$ \sin(z)=\sin(a)\cos(ib)+\cos(a)\sin(ib)\\
=\sin(a)\cosh(b)+i\cos(a)\sinh(b)  $$
so we can split the equation into real and imaginary parts to yeild two real equations ...
$$ \sin(a)\cosh(b)=0 \\
\cos(a)\sinh(b) =1
$$
since $\cosh(b)$ is never zero, the only solutions to the firset equation are 
$$ a=n\pi$$
for any integer $n$.
Now use $\cos(n\pi) = (-1)^n$ to get
$$ b=\sinh^{-1}((-1)^n)   $$
following the first answer we can get 
$$ \sinh^{-1}((-1)^n) = \ln(\sqrt2+(-1)^n)   $$
So the final answer ends up being ...
$$z=n\pi+ i\ln(\sqrt2+(-1)^n)\\ =n\pi+i(-1)^n\ln(1+\sqrt2)   $$
A: How about multiplying both sides with $e^{iz}$ then variable substitution and solve a regular second degree polynomial.
A: $\sin z = \cos z $ comes down to ..
$$ e^{iz}-e^{-iz}  = i( e^{iz} + e^{-iz} )  $$
$$  (1-i)e^{iz}= (1+i)e^{-iz} $$
$$  e^{2iz}= \frac{ 1+i}{1-i} =i=e^{i( \frac\pi2+2n\pi )}$$
So $$z=\frac\pi4 +n\pi$$
All solutions are real.
