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Write a Möbius transformation that maps the region $D = \{ z \in \mathbb{C} \mid |z| \geq 1 \land |z + i| \leq \sqrt{2} \}$ to the region $D' = \{ z \in \mathbb{C} \mid \frac{\pi}{4} \leq \arg(z) \leq \frac{\pi}{2} \}$.

Attempt: The region $ D = \{ z \in \mathbb{C} \mid |z| \geq 1 \land |z + i| \leq \sqrt{2} \} $ is defined by points outside or on the unit circle and inside or on a circle centered at $ -i $ with radius $ \sqrt{2} $. To find a Möbius transformation mapping $ D $ to the region $ D' = \{ z \in \mathbb{C} \mid \frac{\pi}{4} \leq \arg(z) \leq \frac{\pi}{2} \} $, where $ D' $ consists of points with argument between $ \frac{\pi}{4} $ and $ \frac{\pi}{2} $ measured counterclockwise from the positive real axis, I consider the transformation $ T(z) = \frac{z + i}{z - i} $. This Möbius transformation maps the unit circle to the real axis and the circle $ |z + i| = \sqrt{2} $ to a line segment along the imaginary axis. Specifically, it maps $ |z| \geq 1 $ (outside the unit circle) to angles $ \arg(w) = \frac{\pi}{2} $ (points on the positive imaginary axis) and $ |z + i| \leq \sqrt{2} $ (inside the circle centered at $ -i $) to angles $ \arg(w) \leq \frac{\pi}{2} $. Therefore, is this the correct Möbius transformation?

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Your analysis is incorrect, and $T(z)$ is not the transformation you want. $T$ maps the unit circle to the imaginary axis, and the circle $|z + i| = \sqrt{2}$ to the circle $|z + 1| = \sqrt{2}$.

Also, because $\arg()$ is not defined for $0$ or $\infty$, we should redefine $D'$ as $D'=\{ z \in \mathbb{C} \mid \frac{\pi}{4} \leq \arg(z) \leq \frac{\pi}{2} \}\cup \{0,\infty\}.$

You can build the Möbius transformation $f$ you want using a bit of trial and error. The the two circles intersect at $z_1=-1,z_2=1$. We want $f$ to take $z_1$ to $0$, so the numerator of $f$ should contain a factor $(z+1)$. We want $f$ to take $z_2$ to $\infty$, so the denominator of $f$ should contain a factor $(z-1)$. We end up with a trial version $f(z)=\dfrac{z+1}{z-1}$ which takes $D$ to a region bounded by two straight lines meeting at the origin at an angle $\pi/4$, inherited from the angle of $D$ at $z_1$. We may have to multiply $f$ by a constant to rotate the region so that it matches $D'$, but it turns out that we don't need this .. the region is already identical to $D'$.

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    $\begingroup$ I got $f(z)=\frac{z+1}{z-1}$ too, but I realized that for $z=1$, $f(z)=\frac{z+1}{z-1}$ is not defined. Isn't this a problem? $\endgroup$ Commented Jul 12, 2024 at 6:49
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    $\begingroup$ @mathlove Not a problem. Möbius transformations are defined in the so-called extended complex plane $\mathbb C \cup \{\infty\}.$ So, in your example, $f(1)=\infty$. See en.wikipedia.org/wiki/M%C3%B6bius_transformation. $\endgroup$ Commented Jul 12, 2024 at 13:33
  • $\begingroup$ @brainjam ...but $\infty \notin D'$. You should probably address this explicitly. $\endgroup$ Commented Jul 12, 2024 at 13:38
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    $\begingroup$ @BenSteffan Good point. While we're at it, do you consider $0 \notin D'$? Because $\arg()$ is not defined for either $0$ or $\infty$? $\endgroup$ Commented Jul 12, 2024 at 13:48
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    $\begingroup$ @BenSteffan I guess the easiest way to address it is to define $D'$ more explicitly as $D' = \{ z \in \mathbb{C} \mid \frac{\pi}{4} \leq \arg(z) \leq \frac{\pi}{2} \}\cup \{0,\infty\}$ $\endgroup$ Commented Jul 12, 2024 at 14:19

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