Show winding number is non-zero over conformal equivalence

Question

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?

Answer 1

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 $n(\gamma,0)=0$.

Answer 2

This is an old thread, but an answer using the Argument Principle seemed relevant. Technically I do this in the more general setting of cycles/homology as opposed to paths/homotopy.

$g:=f^{-1}$
$\Gamma:= f\circ \gamma$
thus $\Gamma$ and $\gamma$ are both (integrable) cycles.

Suppose that $n\big(\Gamma,0\big) =0$
This allows us to (fatally) conclude that for all $w \not \in D'$ we have $n\big(\Gamma,w\big) =0$ since if $\vert w\vert \leq s$ then $\delta \cdot w$ for $\delta \in [0,1]$ does not meet the image $[\Gamma]$ hence $0$ and $w$ are in the same connected component and the (continuous and integer valued) winding number is constant on a component. For $\vert w\vert \geq 1$ then $w$ is in the unbounded component and the winding number must be zero.

By the Argument Principle
$0\neq n\big(\gamma,0\big) =n\big(g\circ f\circ\gamma,0\big) =n\big(g\circ \Gamma,0\big) = \sum_{z, g(z)=0} v_g(z)\cdot n\big(\Gamma, z\big) = 0$
since $0\not \in g\big(D'\big)$.