How can I find a conformal map from one domain onto another? I have the following problem:
Problem. Find a conformal mapping from $\{|z-8|<16\}\setminus\{|z-3|<9\}$ onto $\{t<|z|<1\}$. Find $t$.
I know a conformal mapping is one whose derivative doesn't vanish, but nothing more. How should I be thinking about this problem? Is there a standard way of solving it? And what does "find $t$" mean? Is it only possible to find the conformal mapping with one particular value of $t$ and not with other values?

A conformal mapping of a domain $\Omega$ in the $z$-plane onto a domain $\Omega'$ in the $w$-plane is provided by an analytic function $z\mapsto w:=f(z)$ defined in $\Omega$ and mapping $\Omega$ bijectively onto $\Omega'$. From general properties of such functions it follows that $f'(z)\ne0$ in $\Omega$.
In the example at hand there are several circles involved, in the $z$-plane as well as in the $w$-plane. Therefore we are lead to the conjecture that the required mapping can be realized as a Moebius transformation. So let's use that as a working hypothesis. Such transformations have the form
$$f(z)={az+b\over cz+d}$$ and  map  circles (:= circles or lines) onto  circles to start with. The real axis in the $z$-plane is a circle that intersects both circles $|z-8|=16$ and $|z-3|=9$ orthogononally; therefore its image is a circle that intersects both circles $|w|=1$ and $|w|=t>0$ orthogonally, i.e., is a line through $0$ in the $w$-plane.  We may as well assume that this image is the real axis; so $a$, $\ldots$, $d$ can be assumed to be real. One more point: Since the given circles are not concentric, but the image circles are, the map $f$ cannot be a similarity. Therefore $c\ne0$, and we may as well assume $c=1$. 
So we are now looking for a function
$$f(z)={az+b\over z+d}\ , \qquad a,b,d\in{\mathbb R}$$
satisfying
$$f(-8)=-1,\quad f(24)=1,\quad f(-6)=- f(12)\ .$$
Solving for $a$, $b$, $d$ we find two such functions, only one of which could possibly solve our problem, namely
$$f(z)={2z\over 24+z}\ .$$
It follows from general properties of Moebius transformations that this is indeed the solution of the given problem and that the inner radius $t$ of the image annulus is given by $t=f(12)={2\over3}$.
Is this map essentially the only solution of the given problem? If there were another function $g$ mapping the given domain $\Omega$ onto an annulus $t'<|w|<1$ with $t' > t$ then $h:=g\circ f^{-1}$ would be a conformal map of an annulus onto an inherently thinner annulus. It is a deep theorem of complex analysis that this is impossible. Similarly, if $t'=t$, one can prove that $h$ has to be a rotation $w\mapsto e^{i\alpha} w$. This means that $f$ and $g$ are essentially the same.
