I am trying to prove the identity $|\cos z|^2 = \cos^2 x + \sinh^2 y$, where $z=x+iy$.
I know $|\cos z|^2=(\cos z)(\cos \bar{z})$, since cosine is analytic for all $z \in \mathbb{C}$. Thus,
$$|\cos z|^2=(\cos (x+iy))(\cos (x-iy))=(\cos x \cos iy-\sin x \sin iy)(\cos x \cos iy+\sin x \sin iy)=\cos^2 x \cosh^2 y + \sin^2 x \sinh^2 y$$
The last equality follows (with some algebra) because $\cos iy=\cosh y$ and $\sin iy=i\sinh y$.
I'd like to know if my work is correct so far, if I'm heading in the right direction, and what I could do to finish.
$\endgroup$ 32 Answers
$\begingroup$You are almost there. Note that:
$$\cosh^2(y)=\left(\frac{e^y+e^{-y}}{2}\right)^2=\frac{e^{2y}+e^{-2y}+2}{4}$$ $$\sinh^2(y)=\left(\frac{e^y-e^{-y}}{2}\right)^2=\frac{e^{2y}+e^{-2y}-2}{4}$$ Hence $\cosh^2(y)=1+\sinh^2(y)$. Now since $\sin^2x+\cos^2x=1$ we find: $$\cos^2x\cosh^2y+\sin^2x\sinh^2y=\cos^2x(1+\sinh^2y)+\sin^2x\sinh^2y=\cos^2x+\sinh^2y$$
$\endgroup$ 1 $\begingroup$All fine, continue further using $ \cosh ^2 x -\sinh ^2 x = 1 $.
$\endgroup$
