Complex trig identity proof

Question

Given that $\cos(z)=\cos(x)\cosh(y)-i\sin(x)\sinh(y)$, show that $$|\cos(z)|^2 =\cos^2(x)+\sinh^2(y)$$

Well, $|\cos(z)| = \sqrt {\cos^2(x)\cosh^2(y)+\sin^2(x)\sinh^2(y)}$

$\implies |\cos(z)|^2 =\cos^2(x)\cosh^2(y)+\sin^2(x)\sinh^2(y)$

but I can't quite see where to go next.

Answer 1

$$|\cos(z)|^2=\cos^2x\cosh^2y+\sin^2x\sinh^2y$$

Use $\sin^2x=1-\cos^2x$

$\cosh^2y-\sinh^2y=1$