A question on complex numbers We are given
If $\cos(a+ib)$=$r (\cos\theta +i\sin\theta)$
then prove that $e^{2b} = \sin(a-\theta)/­\sin(a+\theta)$
I just tried and got $b = 0$ such that $\cos(a) = ra$. Will there be other solutions?
 A: You know that $$\frac{e^{ia-b}+e^{b-ia}}2=re^{i\theta},$$
so we can rearrange this into
$$e^{-b}e^{i(a-\theta)}+e^be^{-i(a+\theta)}=2r$$
and it's conjugate,
$$e^be^{i(a+\theta)}+e^{-b}e^{-i(a-\theta)}=2r.$$
If we match these equations, we get
$$e^{-b}e^{i(a-\theta)}+e^be^{-i(a+\theta)}=e^be^{i(a+\theta)}+e^{-b}e^{-i(a-\theta)}$$
$$e^{-b}(e^{i(a-\theta)}-e^{-i(a-\theta)})=e^b(e^{i(a+\theta)}-e^{-i(a+\theta)})$$
$$e^{-b}\sin(a-\theta)=e^b\sin(a+\theta)$$
$$e^{2b}=\frac{\sin(a-\theta)}{\sin(a+\theta)}$$
and we're done.
A: $\displaystyle \cos (a+ib) = \cos a \cos ib - \sin a \sin ib = \cos a \cosh b - i \sin a \sinh b = r (\cos \theta + i \sin \theta) \\
\displaystyle r \cos \theta = \cos a \cosh b = \cos a \frac {e^{2b} + 1}{2e^b}\\
\displaystyle r \sin \theta = -\sin a \sinh b = -\sin a \frac {e^{2b} - 1}{2e^b} \\
\displaystyle \tan \theta = -\tan a \frac {e^{2b}-1}{e^{2b}+1} \\
\displaystyle \tan \theta (e^{2b}+1) = -\tan a(e^{2b}-1)\\
\displaystyle e^{2b}(\tan a + \tan \theta) = \tan a - \tan \theta \\
\displaystyle e^{2b} = \frac {\tan a - \tan \theta}{\tan a + \tan \theta} = \frac{\frac {\sin (a - \theta)}{\cos a \cos \theta}}{\frac{\sin(a + \theta)}{\cos a \cos \theta}} = \frac{\sin(a - \theta)}{\sin (a + \theta)}$
