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Question

Let $R$ be the complex plane with the non positive real axis taken out.

Find explicitly a conformal mapping $f$ of $R$ onto the unit disc $U$ such that $f(1)=0$ and $f'(1)\gt 0$


solution:

We know that all automorphism of the unit disk are of the form:

$$f(z)=e^{i\theta}\frac{z-\alpha}{1-\overline{\alpha}}$$ for $|\alpha|<1$

Let $g(z)$ be a conformal map. And let's define a conformal map $$f(z)=e^{i\theta}(\frac{g(z)-g(z_0)}{1-\overline{g(z_0)}g(z)})$$

Then $f(1)=0 \iff g(1)=g(z_0)$

And $$f'(z)=\frac{g'(1).e^{i\theta}}{1-|g(1)|^2}$$

From here, I dont know how to reach its result? Is this true way to solve this question? If it is true, how to continue? Thanks.

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A hint:

The map $\psi:\ z\mapsto {\rm pv}(z^{1/2})$ ("taking the positive square root") maps $R$ conformally onto the right half-plane $H$, and $\psi(1)=1$.

Now find a map $\phi$ from $H$ to $U$. By symmetry you can arrange $\phi(1)=0$.

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  • $\begingroup$ Hmm.. This is not the way I thought before. Thus, can you show more clearly? $\endgroup$ – user315 Oct 29 '13 at 12:02
  • $\begingroup$ Please can you teach me this question. I want to learn how to solve this type of questions. Thank you Dear Blatter. $\endgroup$ – user315 Oct 29 '13 at 12:13

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