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Let $\epsilon>0$. I was asked to find a conformal mapping from

$(\mathbb{R}\times(0,2))-((-\infty,i-\epsilon ] \cup[i+\epsilon,i+\infty))$

(An infinite horizontal strip but chopped a fine strip symmetrically)

to $\mathbb{H}$, the upper half plane.

The main obstacle is that I do not know how to deal with the chopped part. Does inversion help? But this is just a bounded strip instead of a "unbouneded" strip.

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What do you mean by $$(-\infty,i-\epsilon ] \cup[i+\epsilon,i+\infty)\ ?$$ –  Christian Blatter May 13 '12 at 19:49
    
Sorry. this is an abuse of notation. For example, $(\infty,i-\epsilon]$ means the horizontal line extends from $i-\epsilon$ to the left. –  iloveinna May 13 '12 at 19:57
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1 Answer 1

up vote 1 down vote accepted

Hint: Start with $z \to w = e^{z \pi}$, mapping your region to ${\mathbb C}$ with two rays removed. Next, invert around an endpoint of one of those rays.

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Isn't the image of $e^{\pi z}$ equal to $\mathbb{C}$ with one ray and one line segment removed? –  froggie May 14 '12 at 0:36
    
The real line and $\text{Im}\ z = 2$ go to the positive real axis; the ray from $i-\infty$ to $i - \epsilon$ to the line segment from $0$ to $-e^{-\pi \epsilon}$; the ray from $i+\epsilon$ to $i+\infty$ to the ray from $-e^{+\pi \epsilon}$ to $-\infty$. And of course $0$ is not in the image either. –  Robert Israel May 14 '12 at 4:45
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