Image of Half-Strip under mapping $w=\cosh (z)$

complex-analysisconformal-geometry

I'm trying to get the image of a half-strip under the conformal mapping $w=\cosh (z)$.

Here is the half-strip: $\{z=x+iy | x>0, 0<y<\pi\}$ and here is what I have attempted:

  1. Start by setting $x=0$ and let $y$ vary between $0$ and $\pi$:
    $\cosh(iy)=\cos(y) \in [-1,1]$. Therefore we map a vertical line on the Imaginary axis to a segment on the Real axis in the $w$-plane.
  2. Set $y=0$ and let $x>0$: $\cosh(x) \in [1, +\infty[$. So the positive Real axis is mapped on itself.
  3. Set $y=\pi$ and let $x>0$: $-\cosh(x) \in ]-\infty, -1]$. The line is mapped on the negative Real axis in the $w$-plane.

This doesn't feel right, what do you think?

Many thanks!

Best Answer

Note that $\cosh(x+iy) = \cosh(x) \cos(y) + i \sinh(x) \sin(y)$. On a vertical line segment at constant $x$, $0 < y < \pi$, this is the top half (if $x > 0$) or the bottom half (if $x < 0$) of an ellipse with major axis from $-\cosh(x)$ to $\cosh(x)$ and minor axis from $-|\sinh(x)|$ to $|\sinh(x)|$. For $x = 0$, you get the line segment from $-\cosh(1)$ to $\cosh(1)$. Taking all these together, you get the whole complex plane.

EDIT: Oops! The whole plane is the image of the closed full strip $-\infty < x < \infty$, $0 \le y \le \pi$. For $x > 0$ and $0 < y < \pi$, you only get the upper half plane.

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