# Conformal mapping of $\cosh 3z$

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.

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Hint: $\cosh(x+iy)= \cos(y) \cosh(x) + i \sin(y) \sinh(x)$ – Fabian Apr 22 '12 at 15:18
Hmm...we have $\cosh 3z = \cos(3y)\cosh(3x)+i\sin(3y)\sinh(3x)$. Since $y$ ranges from $0$ to $\pi$, $3y$ ranges from $0$ to $3\pi$ and hence $\cos(3y)$ and $\sin(3y)$ will range from $-1$ to $1$. Since $\cosh$ and $\sinh$ are basically just exponentials, they are surjective on the positive reals. Therefore $\cosh 3z$ will range through the whole complex plane, so yes? – 192803 Apr 22 '12 at 15:39
You got it .... – Fabian Apr 22 '12 at 17:55