# Use Laurent Series to Determine Bi-Analytic Function

$$w=\varphi(z)$$ 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

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!