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Let $f(x)=\frac{z-3}{z+2}$ be a mobius transformation. I want to find where the set $\Omega = \left\lbrace z\in \mathbb{C}: 0\leq Arg(z)<\frac{\pi}{4} \right\rbrace$ is mapped under the conformal map $f$.

I tried to find the argument of a new variable $w=\frac{z-3}{z+2}$. After some math: $$z = \frac{2w+3}{1-w}.$$

Now I know that $0\leq Arg(z)<\frac{\pi}{4}$ hence

$$0\leq Arg\left(\frac{2w+3}{1-w}\right)<\frac{\pi}{4}$$ $$0\leq Arg(2w+3)-Arg(1-w)<\frac{\pi}{4}$$ But now I'm stuck here since I don't know how to find a good formula for $Arg(w)$. So I tried with a second method.

I considered the straight line $y=0,x\geq 0$, the one representing $Arg(z)=0$, and I mapped that into the straight line $y=0, x\leq 1$. But after that I'm not able to map the other line into a circle(?), I suppose... Any hint? Thank you!

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2 Answers 2

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Hint: the image of a line or circle by a Möbius map is determined by the image of three points.

Take three points on each of the lines ($\arg(z)=0$ and $\arg(z)=\pi/4$) and see to what circles/lines they are mapped to (since Möbius maps send circles/lines to circles/lines).

Then determine which of the regions that appear is the correct one.

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  • $\begingroup$ I've chosen these point for the $Arg(z) =\frac{\pi}{4}$: $$0 \rightarrow -3/2$$ $$1+i \rightarrow -\frac{1}{2}+\frac{1}{2}I$$ $$3-3i\rightarrow -\frac{1}{4}+\frac{5}{12}I$$ Thus I found a circle with equation $$x^2+y^2+\frac{13}{6}x-\frac{5}{6}y+1=0$$. $\endgroup$
    – James Cats
    Commented May 29 at 0:02
  • $\begingroup$ But now: is the region everything except the point inside the circle? $\endgroup$
    – James Cats
    Commented May 29 at 0:07
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You can also do it as follows. This Mobius transformation maps the extended real line to itself, $\infty$ maps to 1, and 0 maps to -3/2. Thus, the half-line $[0,+\infty]$ will map to $[-3/2,1]$. The half-line $[0,i\infty]$ (the ray starting at 0 extending upwards) must therefore map to a circle through $-3/2$ and $1$, and it must intersect the real axis at 45 degree angles since Mobius transformations preserve angles. By the preservation of orientation, it must live in the upper half plane. The equation of the circle should be straightforward from here.

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  • $\begingroup$ Wow, I've never approached the problem this way. Thanks $\endgroup$
    – James Cats
    Commented Jun 2 at 13:36

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