Prove that $\sin(z) =\frac{e^{iz}-e^{-iz}}{2i} =\sin(x)\cosh(y) +i\cos(x)\sinh(y)$ For complex $z=x+iy$. 
How to prove that 
$$\sin(z) =\frac{e^{iz}-e^{-iz}}{2i} =\sin(x)\cosh(y) +i\cos(x)\sinh(y)$$ 
by using the power series definition 
$$\sin(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\ldots$$ 
and Euler's formula $$e^z =e^x (\cos (y) +i\sin(y))?$$
 A: Is it really necessary to use the power series definition of sine? The left equality is easier to prove right-to-left with the exp. power expansion.
\begin{align}
\sin z &=\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)!} z^{2n+1} =\sum_{n=0}^\infty \frac{i^{2n}}{(2n+1)!} z^{2n+1}\\
&=\frac{1}{i} \sum_{n=0}^\infty \frac{i^{2n+1}}{(2n+1)!} z^{2n+1} =\frac{1}{i} \sum_{n=0}^\infty \frac{(iz)^{2n+1}}{(2n+1)!}.
\end{align}
Consider $\dfrac{1-(-1)^k}{2}$ which equals $1$ for odd $k$ and $0$ for even $k$.
\begin{align}
\sin z &=\frac{1}{i} \sum_{k=0}^\infty \left(\frac{1-(-1)^k}{2}\right) \frac{(iz)^k}{k!}
=\frac{1}{2i} \sum_{k=0}^\infty \frac{1-(-1)^k}{k!} (iz)^k\\
&=\frac{1}{2i} \sum_{k=0}^\infty \frac{(iz)^k}{k!} -\frac{1}{2i} \sum_{k=0}^\infty \frac{(-iz)^k}{k!} \\
&=\frac{e^{zi} -e^{-zi}}{2i}\\
&=\frac{e^{(x+yi)i} -e^{-(x+yi)i}}{2i}\\
&=\frac{e^{-y+xi} -e^{y-xi}}{2i}\\
&=\frac{e^{-y} e^{xi} -e^{y} e^{-xi}}{2i}\\
&=\frac{e^{-y} (\cos x +i\sin x) -e^{y} (\cos x -i\sin x)}{2i}\\
&=\frac{i}{i} \cdot \sin x \cdot \frac{e^{y} +e^{-y}}{2} -\frac{1}{i} \cdot \cos x \cdot \frac{e^{y} -e^{-y}}{2}\\
&=\sin x \cosh y +i \cos x \sinh y.
\end{align}
