Using Möbius transformation to change $B\left(a;R\right)$ to halfplane In Conway's Functions of One Complex Variable, there is a proposition which says:
Let $f$ be analytic in the disk $B\left(a;R\right)$ and suppose
that $\gamma$ is a closed rectifiable curve in $B\left(a;R\right)$.
Then
$$\int_{\gamma}f=0.$$
Now, we have a problem which asks us to use a Möbius transformation
to change $B\left(a;R\right)$ into a half-plane which preserves the integral. How do I proceed?
 A: It's simple change of variables. I'll do it in a fancier language since it is easier but gives essentially the same idea: let $f$ be analytic in, say, upper half plane $P$. We can find a Mobius transformation $T$ which sends $B(0;1)$ to $B(a;R)$. Let $\gamma$ be a closed, rectifiable curve in $P$. We need to show that
$$\int _ {\gamma} f = 0 $$
Observe that we can have some open set $U$ in $P$ such that $\partial U = \gamma$. $$\int _ {\gamma} f = \int _ {\partial U} f = \int _ {U} df = \int _ {T^{-1}U} T^*df $$ This last integral is the integral in Proposition 2.15.
A: I don't see what the proposition has to do with the exercise. You should know that Möbius transformations are determined by where they send three points. Take three points on the boundary of your disk to three points on the boundary of the half-plane, preserving order (or else you'll map to the lower half-plane by mistake). I would recommend doing this once and for all for the unit ball centered at the origin. It's easy then to compose with a translation/dilation to get what you need.
