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I want to show that $\coth=\frac{e^{2z}+1}{e^{2z}-1}$ is a conformal mapping of the horizontal strip $S=\{z\in C: \pi/4<\text{Im}(z)<3\pi/4\}$ onto the unit disc U, but I can't seem to get the right idea?

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1 Answer 1

First of all, notice that $\coth$ is the composition of the maps $f(z)=e^{2z}$ and $g(z)=\dfrac{z+1}{z-1}$, i.e. $\coth=g\circ f:\{z\in C: \pi/4<\text{Im}(z)<3\pi/4\}\longrightarrow \mathbb C $

Now, where does $f$ map the strip $S=\{z\in C: \pi/4<\text{Im}(z)<3\pi/4\} $?

First $z$ is mapped conformally to $2z$, i.e. the strip is mapped to $$S'=\{z\in C: \pi/2<\text{Im}(z)<3\pi/2\}$$ Then $S'$ is mapped via $e^z$ conformally to the left half plane $$H=\{z\in\mathbb C:\mathrm{Re}(z)<0\}\mathbf{\text { (Check it!) }}$$ Finally, where, does $g$ map the left half plane H?

$g$ is a Möbius transform, which maps conformally the boundary of $H$ to the boundary of the unit disk, and the interior of $H$ to the interior of the disk (Check it!!)

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I cannot see how $S^{'}$ is mapped via $e^z$ conformally to H? Don't we want to show that S can be mapped to U? –  Maja May 19 '13 at 11:00
    
Take a horizonal line $y=c, c\in(\pi/2,3\pi/2)$. Where is it mapped through $e^z$?(keep $y$ constant and $x$ takes values from $-\infty $ to $+\infty$) $$e^z=e^{x+iy}=e^x\cdot e^{iy}$$ Note that $|e^z|=e^x\in(0,\infty),\forall z\in S'$, and $\arg(z)\in (\pi/2,3\pi/2)$ . Can you see now what is the image of $S'$? –  Dimitris Nt May 19 '13 at 11:15
    
No, I think I get confused when we introduce $S^{'}$. We want to send f composed with g: $S\longrightarrow H \longrightarrow U$, and so I get confused when we have $S^{'}$. –  Maja May 19 '13 at 11:42
    
Look how it is : $$S\overset{2z}{\longrightarrow}S'\overset{e^z}{\longrightarrow}H\overset{\frac{‌​z+1}{z-1}}{\longrightarrow}U$$ We just use $S'$ as the image of $S$ through the function $2z$, in order to make it simpler –  Dimitris Nt May 19 '13 at 11:44

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