$w=\varphi(z)$ is bi-analytic. $\varphi: D=\{z \in C: r_1 < |z| < r_2\} \to G=\{w \in C: R_1 < |w| < R_2\}$. $\varphi$ can be written as the following Laurent series $\varphi(z)=\sum_{n=-\infty}^\infty a_n z^n$.

Prove that:

  1. The area of G is $S(G)=\pi \sum_{n=-\infty}^\infty n|a_n|^2 (r_2^{2n}-r_1^{2n}) $

  2. $\frac{r_1}{r_2} = \frac{R_1}{R_2}$ and $\varphi(z)=e^{i\theta} \frac{R_1}{r_1}z$

I have no problem with the first question and I get that it's meant to be a hint for the second question. But I'm still stuck at Question 2.


1 Answer 1


First of all point 2 is not quite true as $\phi$ can be an inversion too; it is true if we assume $|\phi(z)| \to R_2, |z| \to r_2$ and we will show how this and $1$ imply the claimed result; it will be clear from the proof how one can modify it to show that if $|\phi(z)| \to R_1, |z| \to r_2$, then $\phi(z)=\alpha \frac{r_1R_2}{z}, |\alpha|=1$

we first note that $S(G)=\pi (R_2^2-R_1^2)$ so we get:

$\sum_{n=-\infty}^{-1} |n||a_n|^2 (r_1^{2n}-r_2^{2n})+\sum_{n=1}^\infty n|a_n|^2 (r_2^{2n}-r_1^{2n})=R_2^2-R_1^2$ and in particular

$\sum_{n=1}^\infty n|a_n|^2 (r_2^{2n}-r_1^{2n}) \le R_2^2-R_1^2$ with equality iff $a_n=0, n <0$

But now $|\phi(z)| \to R_2, |z| \to r_2$ means $\frac{1}{2\pi}\int_{|z|=r}| \phi(z)|^2|dz| \to R_2^2, r \to r_2$ so expliciting the integral we get

$R_2^2=|a_0|^2+\sum_{|n| \ge 1}|a_n|^2r_2^{2n}$ and the similar result for $r_1, R_1$, namely:

$R_1^2=|a_0|^2+\sum_{|n| \ge 1}|a_n|^2r_1^{2n}$

Subtracting we get:

$|a_1|^2(r_2^2-r_1^2) +\sum_{n=2}^\infty |a_n|^2 (r_2^{2n}-r_1^{2n}) + \text {non-positive term} \text = R_2^2-R_1^2$

and since $n|a_n|^2 >|a_n|^2, n \ge 2$ unless $a_n=0$ we get:

$\sum_{n=1}^\infty n|a_n|^2 (r_2^{2n}-r_1^{2n}) > |a_1|^2(r_2^2-r_1^2) +\sum_{n=2}^\infty |a_n|^2 (r_2^{2n}-r_1^{2n}) + \text {non-positive term} \text =$

$= R_2^2-R_1^2 \ge \sum_{n=1}^\infty n|a_n|^2 (r_2^{2n}-r_1^{2n})$

and that is a contradiction unless we have $a_n=0, n <0, n \ge 2$ and it follows that $|a_1|^2(r_2^2-r_1^2)=R_2^2-R_1^2, \phi(z)=a_0+a_1z$ as well as $|a_0|^2+|a_1|^2r_{1,2}^2=R_{1,2}^2$

since $|\phi(z)| \to R_1, |z|=r \to r_1$ it immediately follows that $a_0=0, a_1 \ne 0$ and $|a_1|=\frac{R_1}{r_1}=\frac{R_2}{r_2}$ so writing $a_1=e^{i\theta}|a_1|$ we are done!

In the other case, we similarly show that now $a_n=0, n>0, n \le -2$, $\phi(z)=a_0+a_{-1}/z$ and get $a_0=0, a_{-1} \ne 0, |a_{-1}|=r_1R_2=r_2R_1$ and the result follows too!


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