# Mobius transform |z|<1 to the right half plane

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.

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.

• Correct answer. Can you elaborate on why you said "let's have -1 -> i? And how did you choose -1? – Dr.Doofus Jan 27 at 7:31
• $-1$ is another point on $\mid z\mid=1$ (besides $-i$), and I wish to move it onto the $y$-axis. – Chris Custer Jan 27 at 7:34

It is $$\frac {1-iz} {1+iz}$$. Multiply numerator and denominator by $$1-i\overline {z}$$ to prove that this works.

• You said upper half plane in the title and right half plane in the question. I have taken it as right half plane. – Kavi Rama Murthy Jan 27 at 5:11
• Yes, thanks for spotting that. I've edited the question title. In regards to your answer, how do you get that value? – Dr.Doofus Jan 27 at 6:09
• 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. – Kavi Rama Murthy Jan 27 at 6:36