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.