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Problem: Find a conformal map $f$ from $ A = \left\{ z \in \mathbb{C} \mid \text{Im}(z) > 0, |z| > 1 \right\}$ into the unit disk.

Attempt: I started off with a map $F_1: z \mapsto z + \frac{1}{z}$. I was trying to visualize what this mapping actually does to the region in the complex plane. I know that the semi-circle in the upper half plane will get mapped to the interval $[-2, 2]$ on the real line. Also, a point like $2i$ gets mapped to $3i/2$.

I wish to map the region $A$ into the whole upper half plane (if that is possible), then I can easily find a map into the unit disc. Do I need to use a dilatation to rescale the $|z| > 1$ condition?

Help is appreciated.

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  • $\begingroup$ Just use the map $z\mapsto 1/z$. It maps $A$ conformally into the standard unit disk. As stated, the map need not be surjective. $\endgroup$
    – quasi
    Commented Aug 19, 2018 at 9:23
  • $\begingroup$ In our book we defined a conformal map as 'holomorphic bijection'. $\endgroup$
    – Kamil
    Commented Aug 19, 2018 at 9:26
  • $\begingroup$ What book are you using? $\endgroup$
    – quasi
    Commented Aug 19, 2018 at 9:27
  • $\begingroup$ 'Complex Analysis', by Stein and Shakarchi. Princeton lectures in Analysis $\endgroup$
    – Kamil
    Commented Aug 19, 2018 at 9:29
  • $\begingroup$ It's not always defined that way. For example: en.wikipedia.org/wiki/Conformal_map Perhaps edit the question to require $f$ to be bijective. $\endgroup$
    – quasi
    Commented Aug 19, 2018 at 9:43

2 Answers 2

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Hint: With conformal mapping $$w=\left(\dfrac{z+1}{z-1}\right)^2$$ you map region $A=\{z\in\mathbb{C}:{\bf Im}\ z>0,\ |z|>1\}$ to lower half plane $\{z\in\mathbb{C}:{\bf Im}\ z<0\}$.

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  • $\begingroup$ Great. Could you explain to me how you arrived at this map? What was your logic? $\endgroup$
    – Kamil
    Commented Aug 19, 2018 at 11:02
  • $\begingroup$ what does mapping $\frac{1+z}{1-z}$ do with upper unit circle. then square it and let $z\to\frac1z$ to bring inner points to outer of semicircle. $\endgroup$
    – Nosrati
    Commented Aug 19, 2018 at 11:06
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    $\begingroup$ The function $(1+z)/(1-z)$ maps the upper unit circle to the positive imaginary axis. Then I think under this mapping, the whole domain $A$ gets mapped to the quadrant $\left\{w = u + iv: u < 0, v > 0 \right\}$. Is this correct? $\endgroup$
    – Kamil
    Commented Aug 19, 2018 at 11:39
  • $\begingroup$ yes, it is..... $\endgroup$
    – Nosrati
    Commented Aug 19, 2018 at 11:45
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It seems that your region $A$ is the limit as $R \to + \infty$ of $$ A_R=\{z \in \mathbb C, \Im(z)>0, 1<|z|<R \}=\{r\exp(i\theta), \theta \in (0,\pi), r \in (1,R) \}. $$ I think that such a "half-annulus" can be conformally mapped into the upper plane for large $R$. Then you may follow the lines of Conformal maps from the upper half-plane to the unit disc has the form

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