# Show winding number is non-zero over conformal equivalence

Suppose $D = \{ z: r< |z| < 1\}$ and $D' = \{z: s < |z| < 1\}$. Then let $f:D\to D'$ be a conformal bijection. Now suppose $\gamma$ is a closed piecewise smooth path in $D$ with winding number $n(\gamma,0) \neq 0$. Then $\beta = f\circ \gamma$ is a path in $D'$. I need to show that $n(\beta,0)\neq 0$.

So let's say that $\gamma:[a,b]\to D$, so $\beta : [a,b] \to D'$. Then \begin{align} n(\beta,0) &= \frac{1}{2\pi i} \int_{\beta} \frac{dz}{z} \\ &= \frac{1}{2\pi i} \int_a^b \frac{\beta'(t))}{\beta(t)}dt \\ &= \frac{1}{2\pi i} \int_a^b \frac{f'(\gamma(t)\gamma'(t)}{f(\gamma(t))}dt \\ &= \frac{1}{2\pi i} \int_{\gamma} \frac{f'(z)}{f(z)}dz. \end{align}

Edit: Here are my current thoughts. Suppose for contradiction that $n(\beta,0) = 0$. This means $\int_{\gamma} \frac{f'(z)}{f(z)}dz = 0$ for any piecewise smooth pat with winding number greater than 0. We also know by Cauchy's Theorem that $\int_{\gamma} \frac{f'(z)}{f(z)}dz = 0$ if $\gamma$ has winding number $0$. Thus we know that $\log(f(z))$ has a branch in $D$. But this then implies that $D$ is simply connected. However, $D$ is not simply connected because there exist closed paths in $D$ that are not contractible. Hence, we have a contradiction, and it must follow that $n(\beta,0) \neq 0$. Does this make sense?

Let $g=f^{-1}$, which is also a conformal mapping and $\alpha=f\circ \gamma$, and assume that $n(\alpha,0)=0$. Assume, WLOG that $[a,b]=[0,1]$. This mean that there exist a continuous mapping $A:[0,1]^2\to \{s<|z|<1\}$, such that $A(\cdot,0)=\alpha$ and $A(1,t)=z_0\in\{s<|z|<1\}$. This defines an also continuous mapping $\Gamma:[0,1]^2\to \{s<|z|<1\}$, with $\Gamma=g\circ A$, satisfying $\Gamma(\cdot,0)=\gamma$ and $\Gamma(1,t)=g(z_0)\in\{r<|z|<1\}$, which means that $n(\gamma,0)=0$.