Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Problem

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.

share|improve this question
2  
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
1  
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
2  
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.