Prove that $\cos^2(z)+\sin^2(z)=1$ Can anybody show me how to prove that $\cos^2(z)+\sin^2(z)=1$ for $z\in \mathbb{C}$? I can prove it for the case where $z$ is real, but can't seem to find a way to prove it for complex numbers in general.

Let $x,y\in\mathbb{R}$ a real number, so :
$$
\cos(y+ix)=\frac{e^{i(y+ix)}+e^{-i(y+ix)}}{2}=\frac{e^{iy}e^{-x}+e^{-iy}e^{x}}{2}
$$
and the same for $\sin$ :
$$
\sin(y+ix)=\frac{e^{i(y+ix)}-e^{-i(y+ix)}}{2i}=\frac{e^{iy}e^{-x}-e^{-iy}e^{x}}{2i}
$$
if we put $z=y+ix$ :
\begin{eqnarray}
4(\cos^2(z)+\sin^2(z))&=&(e^{iy}e^{-x}+e^{-iy}e^{x})^2 -(e^{iy}e^{-x}-e^{-iy}e^{x})^2\\
&=& e^{-2x+2iy}+e^{2x-2iy}+2e^{-x+iy+x-iy}-(e^{-2x+2iy}+e^{2x-2iy}-2e^{-x+iy+x-iy})\\
&=&4
\end{eqnarray}
so :
$$
\cos(z)^2+\sin(z)^2=1 \qquad \forall z \in \mathbb{C}
$$
