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I'm trying to find a Möbius transformation to map the region

$$\{z:|z-i|<\sqrt2 \text{ and } |z+i|<\sqrt{2} \}$$ onto

$$\left\{z=re^{i\theta}:0<\theta<\frac{\pi}{2}, 0<r\right\}$$

Any help would be appreciated. Regards.

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First step: sketch the two regions. Second step: send $1$ to $\infty$ and $-1$ to $0$. Third step: Multiply by $e^{i\theta}$ if necessary. – t.b. Dec 27 '11 at 15:07
@t.b.:Would it be a good idea to first apply the map $z \mapsto \frac{1}{z^2-3}$ to send the points of intersection on the real axis to $\infty$? And why would we multiply by $e^{i\theta}$? Do you mean raise it to a power? – Mathmo Dec 27 '11 at 15:13
I think you'll find the points of intersection of the circles are $\pm 1$, not $\pm \sqrt 3$. – Thomas Andrews Dec 27 '11 at 15:16
I don't think so. You won't end up with a Möbius transformation. Do you see that the angle at which the two circles around $i$ and $-i$ intersect (at $\pm 1$, not $\pm \sqrt{3}$!) is already $\pi/2$? --- By multiply by $e^{i \theta}$ I just meant to rotate if necessary. – t.b. Dec 27 '11 at 15:17
@ThomasAndrews: Haha, what an error. Thanks for pointing that out! – Mathmo Dec 27 '11 at 15:17
up vote 4 down vote accepted

The target set is the upper right quarter-plane - the set of $z$ with positive real and imaginary parts.

Let $S_1=\{z:|z-i|=\sqrt 2\}$ and $S_2=\{z:|z+i|=\sqrt 2\}$, and let $D_1,D_2$ be their respective interiors.

Any Möbius transformation that sends $D_1\cap D_2$ to the upper-right plane must send either $S_1$ or $S_2$ to the real line, and the other to the imaginary line. In particular, it must send the pair $\{-1,1\}$ to $\{0,\infty\}$ since that is the two points on the Riemann sphere where the real and imaginary axes intersect.

We'll try $f(z)=a\frac{z+1}{z-1}$ for some $a$, since $f(1)=\infty$ and $f(-1)=0$.

For $f(z)$ to send $S_1$ to the real numbers, you only need to prove it for one value on $S_1$. We'll pick $w_1=1+2i$. Then we want $a\frac{2+2i}{2i}$ to be real, and, since $w_1$ is not on the boundary of $D_1\cap D_2$, we want $f(w_1)$ to be negative. So let's try to solve:

$$ -1 = f(w_1) = a\frac{2+2i}{2i}=a(1-i)$$ or $$a = \frac{-1-i}2$$

Then you need to show that, for this value of $a$, $f$ send $S_2$ to the imaginary line, and, again, you only need to validate $f(w_2)$ is imaginary for some $w_2$ on $S_2$ not in $\{\pm 1\}$, e.g., $w_2=1-2i$.$f(w_2)$. (You could just prove that $S_1$ and $S_2$ are orthogonal, since their image under $f$ would have to be orthogonal, so the image of $S_2$ would be a line that goes through $0$ and is orthogonal to the real line, hence must be the imaginary line.)

Finally, you need to show that $f(0)$ is in the upper-right quarter-plane.

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