conformal maps between annulus

Question

The modulus of an annulus $\{ a < \vert z - z_0\vert < b\}$ with inner radius $a$ and outer radius $b$ is defined to be

$$\frac1{2\pi} \log\left(\frac ba\right)$$

(a) Show that any conformal map from one annulus centred at the origin to another such annulus extends to a conformal self-map of the punctured plane.

(b) Show that there is a conformal map of one annulus onto another if and only if the annuli have the same moduli.

(c) Show that any automorphism of the annulus $ \{a < \vert z \vert < b\}$ is either a rotation $z \to e^{i\theta }z$ or a rotation followed by the inversion $z \to ab/z$

I'm totally stuck on this question

Answer 1

Some feasible hints that you may need:

1. $\displaystyle{\left|\Re\left[\iint_A\dfrac{\partial f/\partial\rho}{f\cdot \rho}\right]\right|=\left|\Re\left[\int_0^{2\pi}\int_a^b\dfrac{\partial\log f}{\partial\rho}\mathrm{d}\rho\mathrm{d}\theta\right]\right|=\dfrac{1}{2\pi}\cdot\log\dfrac{b}{a}}$.
2. $\displaystyle{\left|\Im\left[\iint_A\dfrac{\partial f/\rho\partial\theta}{f\cdot \rho}\right]\right|}=\dfrac{1}{2\pi}\cdot\log\dfrac{b}{a}$.
3. $\displaystyle{\iint_A\dfrac{\mathrm{d}z}{|z|^2}=\dfrac{1}{2\pi}\cdot\log\dfrac{b}{a}}$.
4. Denote the Jacobian determinant of $f$ as $\displaystyle{J_f}$. We can easily deduce that $\displaystyle{J_f=\left|\dfrac{\partial f}{\partial\rho}\right|^2=\left|\dfrac{\partial f}{\rho\cdot\partial\theta}\right|^2}$
5. Let $\displaystyle{f}$ be any conformal mapping from one annulus centred at the origin to another such annulus, e.g. $A(0;a;b)$, then $\displaystyle{\iint_A\dfrac{J_f\cdot\mathrm{d}z}{|f(z)|^2}=\dfrac{1}{2\pi}\cdot\log\dfrac{b}{a}}$.

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Via the inequality $\displaystyle{\iint_A\dfrac{\mathrm{d}z}{|z|^2}\cdot\iint_A\dfrac{J_f\cdot\mathrm{d}z}{|f(z)|^2}\geqslant\left(\iint_A\dfrac{\sqrt{J_f}\cdot\mathrm{d}z}{|z|\cdot|f(z)|}\right)^2}$ and hints above, RHS $=\displaystyle{\left(\dfrac{1}{2\pi}\log\dfrac{b}{a}\right)^2}$, where equality holds.

It reveals that

$$ \left\{\begin{align}\dfrac{\partial f}{\partial \rho}&=C_0\cdot\dfrac{f}{\rho}\\\dfrac{\partial f}{\partial\theta}&=C_0\cdot i\cdot f\end{align}\right. $$

Therefore, all self-maps of $A(0;a;b)$ take the form $\tilde{C}\cdot z$ or $\tilde{C}'\cdot z^{-1}$

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There is also a theorem asserting that:

THEOREM (F.H. Schottky, 1877).

An annulus $A(0;r;R)$ can be mapped conformally onto the annulus $A_0(0;r_0;R_0)$ if and only if $r/R = r_0/R_0$.

Moreover, every conformal mapping $f:A\to A_0$ takes the form $f(z) = \lambda z^{\pm 1}$, where $|\lambda| = r_ 0/r$ or $|\lambda| = r_0 R$ as the case may be.