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The question: Let $\gamma:[0,2\pi] \rightarrow \mathbb{C}, t \mapsto e^{int} + re^{imt}$, with $n,m \in \mathbb{Z}$ and $0<r \neq 1$. Calculate the winding number.

My attempt: My first try was to calculate the integral $\operatorname{Ind}(\gamma,z) = \int_0^{2\pi} \frac{\gamma'(t)}{\gamma(t) - z}\,\mathrm dz$, but when I solved this integral by subsituting the denominator after some calculation I end up with a logarithm dependent on $r$ and $z$ divided by $2\pi i$ which is obviously wrong since the winding number should be an integer. I suppose there should be a solution by making a homotopy kind of argument without having to calculate the integral but I'm kinda stuck in here.

Greetings a Peaceful Slosh

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Suppose that $0<r<1$. Then the loop $\gamma$ is homotopic to the loop $\gamma^\star\colon[0,2\pi]\longrightarrow\mathbb C$ defined by $\gamma^\star(t)=e^{int}$ in $\mathbb{C}\setminus\{0\}$. Just define$$\begin{array}{rccc}H\colon&[0,2\pi]\times[0,1]&\longrightarrow&\mathbb{C}\setminus\{0\}\\&(t,u)&\mapsto&e^{int}+rue^{imt}.\end{array}$$Clearly, $H(t,0)=\gamma^\star(t)$, whereas $H(t,1)=\gamma(t)$. So, the winding number is $n$.

Now, suppose that $r>1$. The the loop $\gamma$ is homotopic to the loop $\gamma^\star\colon[0,2\pi]\longrightarrow\mathbb C$ defined by $\gamma^\star(t)=e^{imt}$ in $\mathbb{C}\setminus\{0\}$. The argument is similar. This time you shrink that $e^{int}$ part.

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  • $\begingroup$ Right. I've edited my answer. I hope that everything is correct now. $\endgroup$ – José Carlos Santos Jun 27 '18 at 10:46
  • $\begingroup$ Thanks for your answer :) why would you need two separate cases? Isn't the first argument already enough? $\endgroup$ – PeacefulSlosh Jun 27 '18 at 13:23
  • $\begingroup$ The first argument proves that the answer is $n$ when $r\in(0,1)$, whereas the second one proves that the answer is $m$ when $r\in(1,+\infty)$. I don't see how the first argument could prove both cases. $\endgroup$ – José Carlos Santos Jun 27 '18 at 15:04

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