$\cosh(iz) -\cosh(z)=0$

Question

$\cosh(iz) -\cosh(z)=0$

Apparently $iz=z$ and $z=0$ is a solution.

How do I proceed next?

Do I need to convert $\cosh$ into $\exp$ form? I tried that I get $e$ to complex and real power and I don't know how to continue.

Please help!

Answer 1

We have

$$w_1 + \frac{1}{w_1} = w_2 + \frac{1}{w_2} \iff (w_1 - w_2)(1 - w_1w_2) = 0.$$

Thus we have $\cosh (iz) = \cosh z$ if and only if $e^{z} = e^{iz}$ or $e^{z}\cdot e^{iz} = 1$. That is, if and only if $e^{(1-i)z} = 1$ or $e^{(1+i)z} = 1$. Since $e^w = 1$ if and only if $w = 2k\pi i$ for some $k\in\mathbb{Z}$, that means

$$\cosh (iz) = \cosh z \iff z = \frac{2k\pi i}{1\pm i} = k\pi i(1\mp i)$$

for some $k\in\mathbb{Z}$.