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.
 A: 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 $0$.
A: 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.
