The question is:

Map the domain between the two circles $(x−1)^2+y^2=1$ and $(x−2)^2+y^2=4$ conformally onto the upper half plane. Note that both circles are tangent to the y-axis at the origin.

With hint given: show that the mapping $w=\large \frac{1}{z}$ takes the region between the circles in the z plane to a strip in the w plane.

My attempt,

I tested 3 distinct points from the boundaries, which are the two circles, and under the mapping $\frac{1}{z}$, I see that the larger circle maps to a vertical line x = 1/4, and the smaller circle maps to a vertical line x = 1/2.

To line up this vertical strip with the y-axis, I subtracted off 1/4. Now my mapping is

$$ z \to f(z) = (\frac {1}{z} - \frac{1}{4}) $$

Now the image is a vertical strip with its left boundary line coinciding with the y-axis, and its right boundary line is x= (1/2 - 1/4), i.e., the line x = 1/4.

I then rotated this vertical strip of width 1/4 90 degrees, so that it becomes a horizontal strip of height 1/4, sitting on the real axis now.

$$ z \to f(z) = i(\frac {1}{z} - \frac{1}{4}) $$

I'm stuck with this strip in the upper half plane. What can I do now?


  • $\begingroup$ Say you have a strip $S = \{ z : 0 < \operatorname{Im} z < \pi\}$. Any idea how to transform that into a half-plane? $\endgroup$ – Daniel Fischer Oct 21 '15 at 9:55
  • $\begingroup$ Hi @DanielFischer, if I consider the exponential map and look at its action on the boundary lines of the strip that you suggested, then it looks like $e^z$ maps the real line to itself, and the boundary line $x+i\pi$ to the negative real axis. How can we interpret this? I am guessing the answer is that the strip has mapped to the upper half plane, which is what we wanted. But I'm not sure what kind of argument to give to justify it. Is it a connectedness argument, by the continuity of $e^z$? Thanks, $\endgroup$ – user282456 Oct 21 '15 at 10:07
  • $\begingroup$ Oh, I think I see it @DanielFischer - it actually maps the real line to the positive real line. $\endgroup$ – user282456 Oct 21 '15 at 10:11
  • $\begingroup$ Actually, a silly question @DanielFischer: now that I see the strip's boundary lines map to the real line, how would I know that the area in between indeed fills up all of the UHP? $\endgroup$ – user282456 Oct 21 '15 at 10:15
  • $\begingroup$ Right. It's also useful to recall $e^z = e^{x+iy} = e^x \cdot e^{iy} = e^x(\cos y + i \sin y)$ to see that. $\endgroup$ – Daniel Fischer Oct 21 '15 at 10:15

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