Does there exist an analytic function whose domain is the complement of the closed unit disk and whose range is the open unit disk?

I am studying John B. Conway's book Functions of One Complex Variable for my own amusement. One of the exercises following Chapter III, Section 3, asks whether the open unit disk can be mapped conformally onto the punctured open unit disk. I solved that one (it's true), and then asked myself whether the converse were true as well. I immediately realized that that question is equivalent to whether the complement of the closed unit disk can be mapped conformally onto the open unit disk. Then I realized that I didn't even know the answer to the simpler question asked above.


1 Answer 1


Take $g(z)$ any Blaschke product of degree two that is not $cz^2$; then $g$ maps the punctured unit disc holomorphically onto the unit disc since $g(0)=g(w)$ for some $w \ne 0$. Then $f(z)=g(1/z)$ will map the exterior of the closed unit disc holomorphically onto the open unit disc.

For example $g(z)=\frac{(1-2z)^2}{(2-z)^2}$ is one such Blaschke product, giving


It is considerably more difficult (though doable) to find a conformal such $f$ (in other words with the additional condition that $f'(z) \ne 0, |z|>1$)

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    $\begingroup$ This question goes against our quality standard policy. Instead of posting an answer, you should encourage the person who posted the question to improve it. $\endgroup$ Jul 27, 2021 at 21:49
  • $\begingroup$ Nice solution. Can you direct me to a source for the conformal function that does the trick? $\endgroup$ Jul 29, 2021 at 22:27
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    $\begingroup$ Here there is a neat solution which is fairly explicit ocf.berkeley.edu/~wwu/cgi-bin/yabb/… $\endgroup$
    – Conrad
    Jul 29, 2021 at 22:33
  • $\begingroup$ Thanks very much. $\endgroup$ Jul 30, 2021 at 19:30

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