Which of the following is the bijective analytic function?

Question

Let $D :=\{z\in \mathbb{C}:|z|<1\}$ and let $f \colon D\to D$ be a bijective analytic function such that $f(\frac{1}{2})=0$ then :

1. $f(z)=\frac{2z-1}{2-z}$
2. $f(z)=e^{i\theta}\big({\frac{2z-1}{2-z}}\big)$ for some real $\theta$
3. $f(z)=\frac{2z-1}{2+z}$
4. $f(z)=\frac{z-\frac{1}{2}}{1+z}$

I am not getting how to deal with the problem, I was trying to discard the options by using $f(\frac{1}{2})=0$, but all the options satisfies it. Now it remains for me to check which one is bijective and analytic.But I believe that will be too much. And this may not require that much time.Can you please tell me the best way to deal with this problem.

Answer 1

Hint. The functions are all Möbius transformations. To rule out some options, check that $| f(z) | < 1$ for all $z \in \mathbb C$ with $| z | < 1$, that is, that we actually have $f \colon D \to D$.

Rewrite all Möbius in the form $e^{i \theta} \frac{z + b}{c z + d}$. There is an explicit form for the subgroup of bijective Möbius transformations $D \to D$. If you covered this in class, you can just read off the correct one.

They have the form $f(z) = e^{i \phi} \frac{z + b}{\overline{b} z + 1}$ for some $b \in D$ and $\phi \in \mathbb R$. For $b = - \frac{1}{2}$ we obtain $f(z) = e^{i \phi} \frac{z - \frac{1}{2}}{- \frac{1}{2} z + 1} = e^{i \phi} \frac{2 z - 1}{2 - z}$.

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As request in the comments, I show how one can verify $| f(z) | < 1$ for $| z | < 1$ for option 1.

The inequality $| f(z) | < 1$ is equivalent to (where I write $z = x + i y$ with $x, y \in \mathbb R$) \begin{align*} | 2 z - 1 | < |2 - z | & \iff | 2 z - 1 |^2 < |2 - z |^2 \\ & \iff (2 x - 1)^2 + (2 y)^2 < (2 - x)^2 + (- y)^2 \\ & \iff 3 y^2 < 3 - 3 x^2 \\ & \iff x^2 + y^2 < 1 \\ & \iff | z |^2 < 1 \\ & \iff | z | < 1. \end{align*}