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

  • $\begingroup$ This is a very clear topological argument and is much more intuitive than my solution. I think I'm going to stick with the argument I came up using the branch of log since it uses theorems we explicitly proved in class, but I'll accept your answer. :) $\endgroup$ – luthien Nov 15 '17 at 23:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.