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Find a Mobius transformation mapping the unit disk {|z| < 1} into the right half-plane and taking z = −i to the origin.

My workings: $\phi(t) = \frac{az+b}{cz+d}$

We map -i to the origin (0) by taking the numerator and equating it to 0, knowing b = +i (since we started at -i).

That is: $az+i=0 \implies i = b, a = 1$, which is correct.

However, how do we map $|z|$ to the upper right half plane (i.e. Re(w) = 0)?

Thanks.

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Watch that first step: $-ai+b=0\implies b=ai$. Let's have $-1\to i$, so that $-a+ai=i(-c+d)$. And $0\to1$, giving $b=d$. So $c=-a$.

So, we get $f(z)=\frac{az+ai}{-az+ai}$ or $\boxed{f(z)=\frac{z+i}{i-z}}$.

By specifying the values at $3$ points, the Möbius transformation is determined.

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  • $\begingroup$ Correct answer. Can you elaborate on why you said "let's have -1 -> i? And how did you choose -1? $\endgroup$ – Dr.Doofus Jan 27 at 7:31
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    $\begingroup$ $-1$ is another point on $\mid z\mid=1$ (besides $-i$), and I wish to move it onto the $y$-axis. $\endgroup$ – Chris Custer Jan 27 at 7:34
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It is $\frac {1-iz} {1+iz}$. Multiply numerator and denominator by $1-i\overline {z}$ to prove that this works.

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    $\begingroup$ You said upper half plane in the title and right half plane in the question. I have taken it as right half plane. $\endgroup$ – Kavi Rama Murthy Jan 27 at 5:11
  • $\begingroup$ Yes, thanks for spotting that. I've edited the question title. In regards to your answer, how do you get that value? $\endgroup$ – Dr.Doofus Jan 27 at 6:09
  • $\begingroup$ You can start with $1-iz$ in the numerator so that it vanishes at $-i$. Take denominator as $a+bz$, multiply numerator and denominator by the conjugate of the denominator. You then want the real part of the numerator to be positive when $|z| <1$. It should be easy to guess what $a$ and $b$ should be. $\endgroup$ – Kavi Rama Murthy Jan 27 at 6:36

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