Absolute value of hyperbolic function

Question

Is this statement true, if yes, can anyone show me why?

$$ \cosh(z)\cosh(z^*) = |\cosh(z)|^2 $$

Answer 1

You know that $ww^\ast = |w|^2$ for every $w \in \Bbb C$, right? Just note that: $$\cosh(z^\ast) = \frac{e^{z^\ast}+ e^{-z^\ast}}{2} = \frac{(e^z)^\ast + (e^{-z})^\ast}{2} = \left(\frac{e^z+e^{-z}}{2}\right)^\ast = \cosh(z)^\ast,$$ since: $$e^{(x+iy)^\ast}=e^{x-iy} = e^x(\cos (-y) + i \sin(-y)) = e^x(\cos y - i \sin y) = (e^{x+iy})^\ast.$$

Answer 2

Since $Z Z^*=|Z|^2$, by choosing $Z=\cosh z$ all you need to show is that $\cosh (z^*)=(\cosh z)^*$ which is true. Let $z= x+i y$, where $x,y$ are reals and $i$ is imaginary unit, then $$ (e^{x+iy})^*=e^x(\cos y + i\sin y)^*=e^x(\cos y - i\sin y)=e^x(\cos(- y) + i\sin (-y))=e^{x-iy} $$ and therefore $$(\cosh z)^* =\left( \frac{e^{x+iy} + e^{-(x+iy)}}{2}\right)^*= \frac{(e^{x+iy})^* + (e^{-(x+iy)})^*}{2} = \frac{e^{x-iy} + e^{-(x-iy)}}{2} = \cosh(z^*) $$

Answer 3

If $f$ is any entire function such that $f(x)$ is real for all real $x$, then $f(\bar z) = \overline{f(z)}$ for all $z \in \mathbb{C}$.

(Let $g(z) = \overline{f(\bar z)}$. Then $g$ is entire, and $g = f$ on the real axis, and thus by the identity principle everyhere.)

In particular $f(z)f(\bar z) = |f(z)|^2$, and $\cosh$ is entire (and real valued on the real axis).