Dirichlet Problem on an annulus. Having found the solution for the Dirichlet problem in the region $A=\{x+iy: 0\leq y\leq 1\}$ such that $u(x,0)=0$ and $u(x,1)=1$ to be $u(x,y)=y$, I am asked to find, using conformal maps, the solution in $B=\{z:r_1\leq|z| \leq r_2 \}$ such that $u(z)=0$ on the internal disc and $u(z)=1$ on the external one.
Now, I could find a conformal map from $A$ onto $B$ to be $z \rightarrow e^{i((z-i)\log r_1- z\log r_2)}$
But I think I need a map from $B$ onto $A$ instead and this one is obviously not invertible..
Once I find this conformal map I would be done as the solution wound simply be the composition of the solution in the strip and the conformal map.
EDIT1: I can find a solution quite easily which is $u(x,y)= \frac{1}{\log\frac{r_2}{r1}}\log(\frac{\sqrt{x^2+y^2}}{r_1})$ but I would like to use the conformal map method!
EDIT2: continuing on mrs's hint: having found the solution, we have that if there is a conformal map $f$ from $B$ onto $A$ then $f(x+iy)=u_1(x,y)+i\ u_2(x,y)$ where $u_2=u$ the solution we found, then we use C-R equations to work out the harmonic conjugate of $u$ and we find that $f(x,y)=\frac{1}{\log\frac{r_2}{r1}}(\frac{r_2}{r_1}\tan^{-1}\frac{x}{y}+ i\log\frac{\sqrt{x^2+y^2}}{r_1})$ 
EDIT3: the $f$ I found, sadly, has two problems: it is
possibly not holomorphic when $y=0$ and the image of the annulus under it is only a rectangle...
 A: It is not important that the mapping is one to one. Being a solution of the Dirichlet problem is a local property, independent whether the mapping involves multiple sheets. Also the fact that the mapping is not holomorphic at in the image at $r=0$ is important as this is outside the region $B$. What is important however is to check whether the mapped function is unique, i.e., whether the function value on the different sheets give the same result.
The mapping $f\colon A\mapsto B$ is given by your expression
$$f(z) =  e^{(i z+1 )\log r_1- iz\log r_2} = r_1 \left(\frac{r_1}{r_2}\right)^{iz}.$$
As you have noticed the inverse function is multivalued
$$f^{-1}(w)= i\frac{ \log(w/r_1)}{\log(r_2/r_1)} $$
but this will not hinder us from mapping the solution $u$ from $A$ to $B$.
Starting with $u(x,y)=y$ defined on $A$, we map it via $f$ to $B$ to find the solution $v= \mathop{\rm Im} f^{-1}(z)$ on $B$; more explicitly
$$v(r=|w|) 
= \frac{ \log( r/r_1)}{\log (r_2/r_1)}.$$
Note that even though the inverse was initially multivalued the solution $v$ is uniquely defined and in fact independent of $\text{arg} w$!
