# Conformal Equivalance of Sets in $\mathbb C$

Are these sets conformally equivalent, That is, is there a conformal bijection between them. $S_1=\{z\in\mathbb C\mid0<|z|<1\}$ and $S_2=\{z\in \mathbb C\mid1<|z|<2\}$?

The only thing that I could find is a theorem of (F.H. Schottky, 1877) see here (the first search result).

I don't know if I could use the theorem for these sets because I have zero in one of them.
And also we did not learn this theorem in class.

The question is could the theorem be applied in this case? How can I answer this question without this theorem? Any ideas?Thank you.

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Suggestion for getting started: Suppose $f:S_1\to S_2$ is analytic. What can you say about the isolated singularity at $0$? – Jonas Meyer Feb 8 '13 at 17:46
@JonasMeyer I'm not sure what do you mean. Are you suggesting they are equivalent? – i.a.m Feb 8 '13 at 17:53
Since $f$ is bounded, the singularity of $f$ at $z=0$ must be removable. Try to take it from there. – mrf Feb 8 '13 at 18:06
@mrf Ok, Than you. – i.a.m Feb 8 '13 at 18:45

I like the suggestion by Meyer. Suppose $f: S_1 \rightarrow S_2$ is biholomorphic. Then it is bounded and has a removable singularity at $0$. Therefore, $f$ can be extended to a analytic map on the unit disk centered at the origin. This is a contradiction as no point near $S_2$ can be the image of $0$.

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So are you saying that if $f$ can't be extended to the open unit disk and still be 1-1? but does that imply the the sets are not equivalent? – i.a.m Feb 8 '13 at 19:03

I believe the mapping which sends $|z|$ to $|z|+1$ and maintains the angle will do. You're just blowing-up the punctured disk into an annulus. I haven't worked out the details, but it seems plausible.

Added 2-9-13 Follow-up on why my conjecture fails:

$$f(z) = (|z|+1)\frac{z}{|z|}$$

is this map I indicated. If $|z| = \sqrt{z\bar{z}}$ then for $z \neq 0$,

$$f(z) = \left( 1+\frac{1}{\sqrt{z\bar{z}}} \right)z$$. Clearly the above expression has nontrivial $\bar{z}$-dependence and as such cannot be holomorphic (hence not conformal). If you prefer, $z=x+iy$ we have $$f=u+iv \ \ \& \ \ u(x,y) = x\left(1+\frac{1}{\sqrt{x^2+y^2}}\right) \ \ \& \ \ v(x,y) = y\left(1+\frac{1}{\sqrt{x^2+y^2}}\right)$$ and $u_y = v_x$ hence the Cauchy-Riemann equations are violated by my mapping (everywhere) and this map is not complex-differentiable, analytic, holomorphic, etc.

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Thank you, I'll try to fill the details. – i.a.m Feb 8 '13 at 17:47
I could be wrong! – James S. Cook Feb 8 '13 at 17:49
Ok Thank you again. – i.a.m Feb 8 '13 at 17:52
This map is not holomorphic. – Jonas Meyer Feb 8 '13 at 17:54
There is no way to work out the details here – mrf Feb 8 '13 at 18:09