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Find a conformal equivalence between the following domains:

the strip $ S = \{ z \in \Bbb C \ | \ 0 < \Bbb Im(z) < 1 \} $ and the quadrant $ Q = \{z \in \Bbb C \ | \ \Bbb Re(z) > 0, \Bbb Im(z) > 0 \} $.

By considering a suitable bounded solution of Laplace's equation $ u_{xx} + u_{yy} = 0 $ on the strip S, find a non-constant, harmonic function on Q which is constant on each of the two boundaries of the quadrant.

I can manage the first part - use a variation on the conformal mapping $ z \mapsto e^z $, but then I am unable to determine the final bit.

This is one of my example sheet questions, which is completely non-examinable. I have already had the supervision on it, but was just hoping for a bit more detail (the supervisor wasn't able to explain it fully himself).

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Set $$ f(z)=\exp(\pi z/2), $$ Then $f'(z)\ne 0$, $f$ is one-to-one on $$ S=\{z\in\mathbb C : 0<\mathrm{Im}\, z<1\}, $$ and $$ f[S]=Q=\{z\in\mathbb C: \mathrm{Re}\,z>0,\,\mathrm{Im}\,z>0\} $$

The harmonic function on $S$ is $v=\frac{\pi}{2}\mathrm{Im}\, z$ and $u=v\circ f^{-1}$ which is constant on the boundaries of $Q$.

In particular, $$u(x,y)=\mathrm{Im}\,\log z$$

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  • $\begingroup$ Hello Yiorgos. That's the bit that I was ok on (I couldn't remember the exact form, but I knew that it was $ z \mapsto exp(kz) $ for some $k$. $\endgroup$
    – Sam OT
    Mar 6, 2014 at 22:24
  • $\begingroup$ It's the next bit that I'm having trouble with: relating the solution to Laplace's equation on S to one on Q. $\endgroup$
    – Sam OT
    Mar 6, 2014 at 22:25
  • $\begingroup$ It asks me to consider a suitable solution on $S$, then (I assume by using the conformal mapping) translate this into a solution on $Q$. You've just given the equation on Q, not originating on S. Are you able to elaborate a bit more? Thanks very much! :) $\endgroup$
    – Sam OT
    Mar 6, 2014 at 23:07

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