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..