Let $f:A(r,1) \rightarrow \mathbb{C}$ be holomorphic, injective, bounded and non-zero for some fixed $0<r<1$ (where $A(r,1)$ is the annulus with radii $r$ and $1$). How can we see that $$\int_\gamma \dfrac{f'(z)}{f(z)} dz = 2 \pi i$$ for $\gamma(t) = \rho e^{it}$ and $r < \rho < 1, \ t \in [-\pi, \pi]$. Or is this not true in general?
I know that this integral is the winding number. More precisely Im pretty sure its the winding number of $f \circ \gamma$ around $0$. Thats because of $\int_\gamma \dfrac{f'(z)}{f(z)} dz = \int_{-\pi}^\pi\dfrac{(f \circ \gamma)'(t)}{(f \circ \gamma)(t)} dt = \int_{f \circ \gamma} \dfrac{1}{z} dz = 2\pi i\operatorname{Ind}_{f \circ \gamma}(0)$.
Obviously $\operatorname{Ind}_{\gamma}(0) = 1$. But for the statement above we would need to follow $\operatorname{Ind}_{f \circ \gamma}(0) = 1$.
I know this is pretty much out of context but do you have any hints or ideas?
Not true. If $f(z)=e^{z}$ then $f$ is injective bounded, non-zero and holomorphic on the annulus and the integral is $0$.