Consider the semi-infinite strip $\{z = x + iy: x > 0, 0 < y < \pi\}$. What is the image of this strip under $\cosh 3z$? Is it just the whole complex plane?
My reasoning is as follows: Note that $\cosh 3z = \sin(i3z+\pi/2)$. Then the strip can be mapped conformally to $R:=\{z = x + iy: -5\pi/2 < x < \pi/2, y > 0\}$. To get our final answer, we need to see where $\sin z$ sends $R$. But $\{z = x + iy: -3\pi/2 < x < \pi/2, y > 0\}$ gets sent to the whole complex plane by $\sin z$ and hence $R$ gets sent to the whole complex plane.