How to show $|\sin(x+iy)|^2=\sin^2x+\sinh^2y$ How would I show that $|\sin(x+iy)|^2=\sin^2x+\sinh^2y$? 
Im not sure how to begin, does it involve using $\sinh z=\frac{e^{z}-e^{-z}}{2}$ and $\sin z=\frac{e^{iz}-e^{-iz}}{2i}$?
 A: If you use the sine addition formula, the pythagorean identities, and the fact that $\sin(ix)=i\sinh (x)$ and $\cos(ix) = \cosh(x)$, then you get this:
$$
\begin{align}
\sin(x+iy) &= \sin x \cos (iy)+\cos x \sin(iy) \\
&= \sin x \cosh y + i \cos x \sinh y
\end{align}
$$
$$
\begin{align}
|\sin(x+iy)|^2 &= (\sin x \cosh y)^2 + (\cos x \sinh y)^2
\end{align}
$$
Now you can get rid of the cosines knowing that $\cos^2 x + \sin^2 x = 1$ and that $\cosh^2 x - \sinh ^2 x = 1$. You can take it from there.
By the way, to get the sine addition formula and the sine and cosine of imaginary numbers, convert them to exponential form:
$$\sin x = \frac{e^{ix}-e^{-ix}}{2i}$$
$$\cos x = \frac{e^{ix}+e^{-ix}}{2}$$
$$\sinh x = \frac{e^x-e^{-x}}{2}$$
$$\cosh x = \frac{e^x+e^{-x}}{2}$$
Plug in what you want to find out; the derivation of the identities is straightforward.
A: $$z=x+iy\Longrightarrow \sin z=\frac{e^{iz}-e^{-iz}}{2i}=\frac{e^{-y+ix}-e^{y-ix}}{2i}=$$$$=\frac{e^{-y}(\cos x+i\sin x)-e^y(\cos x-i\sin x)}{2i}=\frac{1}{2i}\left[i\sin x\left(e^y+e^{-y}\right)-\cos x\left(e^y-e^{-y}\right)\right]=$$$$=\sin x\cosh y+i\cos x\sinh y\Longrightarrow ...$$
A: \begin{align}
\sin(z)^2 &= (\sin x \cos (iy))^2 +(\cos x \sin(iy))^2 \\
&=\sin^2x \cosh^2y+\cos^2x \sinh^2y \\
&= \sin^2x (1+\sinh^2y )+(1-\sin^2x ) \sinh^2y \\
&=\sin^2 x+\sin^2 x \sinh^2y+ \sinh^2y-\sin^2x \sinh^2y \\
&=\sin^2x+\sinh^2y 
\end{align}
