# Conditions under which a conformal map cannot be extended holomorphically

Let $f$ be a conformal map from unit disk $|z| <1$ to square $D=\{x+iy \in \mathbb{C}:|x|<1,|y|<1\}.$ Could anyone advise me how to prove $f$ cannot be extended to holomorphic function defined on disk $|z| <R, \forall R>1 \ ?$

Suppose $f$ can be extended to a holomorphic function on $|z|<R,$ where $R>1.$

Consider $z$ where $|z|=1.$ Then $f(z) \not\in D.$

But how do I continue from here? Hints will suffice, thank you very much.

Do you know that $f$ can be continuously extended to a homeomorphism from the closed disk to the closed square? What would happen at the points being mapped to the vertices of the square if an analytic extension existed?

Let $z_0 \in \partial\mathbb{D}$ a point that is mapped to the vertex $1+i$ of the square $\overline{D}$ by the continuous extension of $f$ to $\overline{\mathbb{D}}$. Since that continuous extension is a homeomorphism, the two boundary arcs of $\mathbb{D}$ meeting in $z_0$ - $A_- = \{z_0 e^{i\varphi} : -\delta < \varphi < 0\}$ and $A_+ = \{z_0 e^{i\varphi} : 0 < \varphi < \delta\}$ - are mapped to the two boundary segments $B_- = \{1 + ti : 0 < t < 1\}$ and $B_+ = \{(1-t) + i : 0 < t < 1\}$ of $D$ [for small enough $\delta > 0$], $A_-$ being mapped into $B_-$ and $A_+$ into $B_+$, since $f$ is orientation-preserving.

The two arcs $A_-$ and $A_+$ meet at an angle of $\pi$ (the unit circle is smooth), and the two segments $B_-$ and $B_+$ meet at an angle of $\frac{\pi}{2}$ (or $\frac{3\pi}{2}$ if you look at the exterior angle).

Now, if $f$ had a holomorphic continuation $\tilde{f}$ to a neighbourhood of $z_0$, then $\tilde{f}$ would attain the value $1+i$ with multiplicity $m$ in $z_0$.

Hence $\tilde{f}$ would map the two arcs $A_-$ and $A_+$ to two curves meeting at an angle of - what?

• Thanks for the advice. I'm still unsure of how to proceed. Here is my attempt: Consider $B(z,r) \subset D(0,R),$ where $|z|=1$ and $R>1.$By Opening Mapping theorem, $f[(B(z,r)]$ is open in $\mathbb{C}. \$ Also, $f[B(z,r)]$ contains points that do not lie within and on the square. – Alexy Vincenzo Sep 28 '14 at 13:16
• Look at the behaviour of $f$ at $z_0$, where $z_0$ is a (the) point of $\partial \mathbb{D}$ that is mapped to the vertex $1+i$. What does $f$ do to the boundary curve (the unit circle)? Is that compatible with $f$ being holomorphic in a neighbourhood of $z_0$? – Daniel Fischer Sep 28 '14 at 13:20
• $f$ maps the boundary curve of the unit circle to the boundary of the square. – Alexy Vincenzo Sep 28 '14 at 13:40
• Yes, and at $z_0$ that means that what happens? – Daniel Fischer Sep 28 '14 at 13:42
• I apologize for being unable to make full use of your hints yet. I am thinking of using Open Mapping theorem but the set of boundary points of the circle is obviously not open. – Alexy Vincenzo Sep 28 '14 at 14:03