Complex solutions of $\sin z = i \alpha \cos z$ I'm trying to solve the following question:
Let $\alpha$ $\in [-1, 1]$ be a real number. Find all complex numbers $z$ that satisfy the equation: 
$\sin z = i \alpha \cos z$

This is what I've done so far:
I let $z = x + iy$ which makes the the equation equal:
$$\sin (x+iy) = i \alpha \cos(x+iy)$$
I simplified it to:
$$\sin x + i\sin y = a(i\cos x-\cos y)$$
I don't know what to next because I have never encountered anything like this before.
I would appreciate your help!
 A: I'll try to give some insight ...
We know that $\frac {\sin z}{\cos z} = \tan z$. So our equation becomes..
$$\tan z = i\alpha$$
Now use the exponential form of $\sin$ and $\cos$ to find the exponential form of $\tan z$. And what is the exponential form of $i$? Now we can solve the question.
For much general result, you can see next (though I'll strongly recommend doing this yourself)
want solution for $\tan z = z_0 \ \ ;z_0,z \in \mathbb{C}$
Solution:
$$\tan(z) = \frac{e^{iz} - e^{-iz}}{i(e^{iz} + e^{-iz})} = z_0$$
Putting $p= e^{iz} ;\frac 1p = e^{-iz}$ our equation becomes,
$$\frac{p - 1/p}{i(p + 1/p)} = z_0$$
  $$\frac{p^2 - 1}{p^2 + 1} = iz_0$$
  $$p^2 - 1 = iz_0(p^2 + 1)$$
  $$(1 - iz_0)p^2 = 1 + iz_0$$
  $$p^2 = \frac{1 + iz_0}{1 - iz_0}$$
Now plugging in original value, we get,
$$e^{2iz} = \frac{1 + iz_0}{1 - iz_0}$$
Giving the solution to be,
$$z = \frac 1{2i} \ln \frac{1 + iz_0}{1 - iz_0}$$
Disclaimer: Just typed quickly, might be errors in it but method will be same.
EDIT:
As asked in comment by the poster. Put $z_0=i\alpha$ in the solution.
Then we get, $$z=\frac 1{2i} \ln \frac{1 + i \cdot i \alpha}{1 - i \cdot i \alpha}=\frac 1{2i} \ln \frac{1 + i^2 \alpha}{1 - i^2 \alpha}=\frac 1{2i} \ln \frac{1 - \alpha}{1 +  \alpha}$$.
I suppose it is clear now
If not, please reply..
A: Your equation is equivalent to $\tan(z)=i\alpha$. Even for complex $a$ and $b$,
$$\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$$
So let $z=a+b\,i$, with $a,b$ real, and $$\begin{align}\tan(a+bi)&=\frac{\tan(a)+\tan(bi)}{1-\tan(a)\tan(bi)}\\
&=\frac{\tan(a)+i\tanh(b)}{1-i\tan(a)\tanh(b)}\\
&=\frac{(\tan(a)+i\tanh(b))(1+i\tan(a)\tanh(b))}{1+\tan^2(a)\tanh^2(b)}
\end{align}$$
Here we have used that $\tan(ix)=i\tanh(x)$.
You want the real part of this expression to be $0$, since you want it to equal $i\alpha$. So $\tan(a)-\tan(a)\tanh^2(b)=0$. That is, $\tan(a)(1-\tanh^2(b))=0$. Now, there are no real numbers $b$ with $\tanh(b)=\pm1$, so $\tan(a)=0$. Therefore $a$ must be $k\pi$ for some integer $k$. 
Back substituting, $$\begin{align}\frac{(\tan(k\pi)+i\tanh(b))(1+i\tan(k\pi)\tanh(b))}{1+\tan^2(k\pi)\tanh^2(b)}&=\alpha i\\
{i\tanh(b)}&=\alpha i\\
{\tanh(b)}&=\alpha \\
\end{align}$$
So there is only a solution if $\alpha\in(-1,1)$, in which case $b=\tanh^{-1}(\alpha)$, and $z=k\pi+\tanh^{-1}(\alpha)$ for some integer $k$.
