On page 131 (Chapter 4 section 3.3) of Ahlfors Complex Analysis he states:
The function w = f(z) maps [the closed curve] $\gamma$ onto a closed curve $\Gamma$ in the w-plane, and we find $\int_{\Gamma} \frac{dw}{w} = \int_{\gamma} \frac{f'(z)}{f(z)} dz$
I am having trouble seeing why this equality must hold. I suppose heuristically you could say let w = f(z), then $\frac{dw}{dz}$ = f '(z) so dw = f '(z)dz and since f(z) maps $\gamma$ into $\Gamma$ we get $\int_{\Gamma} \frac{dw}{w} = \int_{\gamma} \frac{f'(z)}{f(z)} dz$. Can somebody please supply me with a rigorous argument for this or explain why my heuristic reasoning is justified. Thanks.
There is a condition for this: The map $f:\gamma \rightarrow \Gamma$ should be 1-1 and orientation preserving. For example, $z\mapsto z^2$ maps the unit-circle to itself but winds around twice; the integrals are not the same. Similarly for $z\mapsto 1/z$. When the condition is verified it amounts to a reparametrization of the closed curve $\Gamma$.