Is there a generalization for the conversion cosh(z)=cos(iz)?

Question

I have been comfortable using the conversion $cosh(z)=cos(iz)$

I have come across situations where I need to convert equations such as $cosh(z^2)$ or $cosh(2z)$ to $cos(f(z))$ and I am not sure how to proceed.

Can this conversion be generalized to $cosh(f(z))=cos(if(z))$ or $cosh(f(z))=cos(f(iz))$ or does that not hold?

Answer 1

Can this conversion be generalized to $\cosh(f(z))=\cos(if(z))$

This one is correct. If you start with the given identity $\cosh(\color{blue}{z})=\cos(i\,\color{blue}{z})$ and substitute $\color{blue}{z} \mapsto \color{red}{f(z)}$ you get $\cosh(\color{red}{f(z)}) = \cos(i\, \color{red}{f(z)})$.

In terms of functions, let $T(z) = iz$ then $\cosh(z)=\cos(iz) \iff \cosh = \cos \circ \,T$. Compose with $f$ on the right, then $\,\cosh \circ\, f = \cos \circ \,T \circ\,f\,$ which is equivalent to $\cosh(f(z)) = \cos(i f(z))$.

or $\cosh(f(z))=\cos(f(iz))$

This one is wrong. It would be equivalent to $\,\cosh \circ\, f = \cos \circ\,f \circ \,T\,$, which does not follow since $\,f \circ \,T \ne T \circ \,f\,$, and does not hold true. A correct identity would be $\,\cos\left(f(iz)\right) = \cosh\left(i f(iz)\right)\,$.

Answer 2

Yes, that holds in general. The symbol $f(z)$ just stands for a complex number, so \begin{align} \cosh(f(z))&=\cos(if(z)). \end{align} Why? Because $\cosh(z)=\cos(iz)$ holds for all complex numbers $z$. So, I can say that if $@$ is a complex number then $\cosh(@)=\cos(i@)$ as well. There’s nothing special about $f(z)$… it just just a good old complex number, so nothing stops me from using the formula I already know.