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I am having trouble doing conformal map problems. Any suggestions on how to do this problem? Thanks.

Find a conformal bijection from $\{z : |z| > 1, Im$ $z < 5\}$ onto an annulus centered at the origin.

  • Start by mapping ${z:\operatorname{Im} z < 5}$ conformally to a disk (centered at the origin). Can you see how you could then proceed? – Daniel Fischer Jun 22 '14 at 17:12
  • Find the points (there are 2 of them) whose reflection wrt the unit circle is equal to the reflection wrt the line $Im\ z=5$. Then send one of them to $0$ and the other to $\infty$ with $(z-a)/(z-b)$ ($a,b$ are the 2 points). That's it. – user8268 Jun 22 '14 at 17:28
  • What does reflection wrt the unit circle mean? – kingkongdonutguy Jun 22 '14 at 17:34
  • Inversion, i.e. $z\mapsto 1/\bar z$ – user8268 Jun 22 '14 at 17:36
  • okay...so how do I know the map you described works? – kingkongdonutguy Jun 22 '14 at 17:51
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1 Answers1

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The boundary already consists of two circles (counting a line as a circle). We just have to make them concentric with a suitable Möbius transformation. The inner circle $|z|=1$ looks good already; let's keep it that way. Also, since both circles are symmetric about the imaginary axis, our transformation can be symmetric too: it will map the imaginary axis to itself. So, it will be $$f(z) = \frac{z-ib}{1+ibz}$$ where $b\in(-1,1)$ is a real number to be determined. The image of $\operatorname{Im}z=5$ is symmetric about the imaginary axis; to find it, we just need to know where it crosses the axis. This is where: $$f(5i)=\frac{5i-ib}{1-5b} = \frac{5-b}{1-5b} i, \quad f(\infty) = \frac{1}{ib} = -\frac{1}{b}i$$ These two points are endpoints of a diameter of the image. And we want the center of the image to be zero, so
$$\frac{5-b}{1-5b} = \frac{1}{b}$$ from where $b=5-2\sqrt{6}$.

  • Here is another question in complex analysis that has been answered (and awarded with half a bounty, therefore I conclude) but not in a satisfactory manner: a problem in application of conformal mappings . The gist of the question IMO is how to find a heuristics. As the OP says: "Obtain the complex potential". I wasn't capable to answer the question in that sense and I wonder if you are. – Han de Bruijn Jun 30 '14 at 11:07