# conformal maps between annulus

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

• This material is covered in many complex analysis textbooks. I have no idea what you know about complex analysis, I don't know what you tried or what your starting point is. So it is very hard to give you any suggestions other than go look it up in a book. Apr 7, 2017 at 12:56
• Im trying to prove it using maybe Schwarz reflection principle. Its in the complex analysis in Schwarz reflection principle. Apr 7, 2017 at 15:16
• no help :( ? !! Apr 8, 2017 at 5:24

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 $$\displaystyle{\iint_A\dfrac{J_f\cdot\mathrm{d}z}{|f(z)|^2}=\dfrac{1}{2\pi}\cdot\log\dfrac{b}{a}}$$.

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 $$\tilde{C}'\cdot z^{-1}$$

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.