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Does anyone have a recommendation as how to go about solving this problem?

I want a conformal from G to H where $$ G = \{ z \in \Bbb C \ | \ |z|<1, |z+i|>\sqrt{2} \}, S = \{ z \in \Bbb C \ | \ \Bbb Re(z) \in (-\pi,\pi) \}. $$

The previous part of the question was to do with branches of the log. Any advice would be most appreciated!

The sort of answer that I'm looking for is this:

Use f(z) = $ \alpha i \log(z) + \beta \arg(z)$

then show how to find $\alpha$ and $\beta$ using the boundaries.

Thanks!

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Consider $$ f(z)=-5\pi+8 i \mathop{\rm Log}\left(\frac{z+1}{z-1}\right) $$ The boundary of $G$ consists of two arcs : $\gamma_0$ that has the parametrization $\phi(t)=e^{it}$ for $t\in[0,\pi]$, followed by $\gamma_1$ that has the parametrization $\psi(t)=-(i+(1-i)e^{-i t})$ for $t\in[0,\pi/2]$.

Now, $f(\phi(t))=-\pi+8i\log(\cot(t/2))$ and as $t$ varies from $0$ to $\pi$, the point $f(\phi(t))$ varies on the line $x=-\pi$ from $i(+\infty)$ to $i(-\infty)$.

Similarly, $f(\psi(t))=\pi-8i\log\left(\dfrac{\cot(t/2)-1}{\sqrt{2}}\right)$ and as $t$ varies from $0$ to $\pi/2$, the point $f(\psi(t))$ varies on the line $x=\pi$ from $i(-\infty)$ to $i(+\infty)$.

Thus, $f(G)=S$ as desired.

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  • $\begingroup$ Thank you - this is the sort of answer that I was looking for. Can you explain how you chose your parameters in the equation for $f$? Did you use $$f(z) = \alpha \pi + i(\beta \log (z+1) + \gamma \log (z-1) ) $$ the determine what the constants should be when you went through it or what? $\endgroup$ – Sam T Apr 7 '14 at 8:15
  • $\begingroup$ @Smiley. The tow circular arcs has common points $-1$ and $1$, so I sent one to zero and the other on to infinity using the homographic transformation $z\mapsto \dfrac{z+1}{z-1}$, this gives you an angular sector in the plane, and the Logarithm transforms this to a strip. The final step is to use the right affine transformation namely $z\mapsto 8iz-5\pi$ to get the desired one. $\endgroup$ – Omran Kouba Apr 7 '14 at 9:02
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$G$ is bounded by two circular arcs, thus it can be mapped by a Möbius transformation to an angular sector $A_\alpha = \{z : 0 < \arg z < \alpha\}$.

That angular sector can be mapped to a half-plane or a slit plane by a power function, and finally, the half-plane or slit plane can be mapped to a strip using a logarithm. Scalings and/or rotations at some steps may be helpful.

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