calculating winding number around zero

Question

let $m, n \in \mathbb{Z}$ be fixed and let $ 0 < r \neq 1$, determine the winding number of the closed curve $\gamma(t) : = e^{imt} + re^{int}$ $\gamma: [0, 2\pi] \to \mathbb{C} \backslash 0$ around zero

so we know that $w_\gamma (0) = \frac{1}{2\pi i} \int_{\gamma} \frac{1}{z-0} dz$ would I then be having

$$w_\gamma (0) = \frac{1}{2\pi i} \int_{0}^{2\pi} \frac{ime^{imt} + in re^{int}}{e^{imt} + re^{int}} dt$$ ?

is my approach correct until now and how could I proceed?

Answer 1

$\newcommand{\d}{\,\mathrm{d}}$Your work is correct. There is probably a slick homotopy theoretic method for this but I don't see it, it's hard for me to visualise this contour.

Let's say $0<r<1$. Then: $$\begin{align}w_{\gamma}(0)&=\frac{1}{2\pi}\int_0^{2\pi }\frac{m+rn\cdot e^{i(n-m)\phi}}{1+r\cdot e^{i(n-m)\phi}}\d\phi\\&=\frac{1}{2\pi}\sum_{k\ge0}(-r)^k\int_0^{2\pi}(m+rn\cdot e^{i(n-m)\phi})\cdot e^{ik(n-m)\phi}\d\phi\\&=\frac{1}{2\pi}\sum_{k\ge0}(-r)^k\int_0^{2\pi}(me^{ik(n-m)\phi}+rn\cdot e^{i(k+1)(n-m)\phi})\d\phi\\&=m+\sum_{k\ge1}\frac{(-r)^k}{2\pi}\left(m\cdot\int_0^{2\pi}e^{ik(n-m)\phi}\d\phi-n\cdot\int_0^{2\pi}e^{ik(n-m)\phi}\d\phi\right)\\&=m+\sum_{k\ge1}(-r)^k(m\cdot\delta_{n,m}-n\cdot\delta_{n,m})\\&=m-\delta_{n,m}(m-n)\cdot\frac{r}{1+r}\\&=m\end{align}$$

Similarly you can brute-force compute $w_\gamma(0)=n$ if $r>1$. Here I use $\delta_{i,j}$ the Kronecker delta notation. It just means "$0$", if $i\neq j$, and "$1$" if $i=j$. Notice that $\delta_{i,j}\cdot(i-j)$ is always zero.

The case $r=1$ should be handled with care, and this is based on Mark Viola's comments.

In light of: $$e^{im\phi}+e^{in\phi}=e^{i(m+n)\phi/2}(e^{i(m-n)\phi/2}+e^{-i(m-n)\phi/2})=2e^{i(m+n)\phi/2}\cos\left(\frac{m-n}{2}\phi\right)$$

And the observation that, whenever $m,n$ are distinct integers, there is at least one integer $k$ with: $$\frac{k+\frac{1}{2}}{m-n}\in[0,1]$$Shows that there is at least one root of $\cos\left(\frac{m-n}{2}\phi\right)$ in the interval $[0,2\pi]$, so our contour passes through the origin and its winding number about the origin is not defined. So the question only becomes sensible if $m=n$, and then we can write: $$\begin{align}w_\gamma(0)&=\frac{1}{2\pi}\left(\frac{m+n}{2}\int_0^{2\pi}\frac{e^{i(m-n)\phi/2}+e^{-i(m-n)\phi/2}}{e^{i(m-n)\phi/2}+e^{-i(m-n)\phi/2}}\d\phi\\+\frac{m-n}{2}\int_0^{2\pi}\frac{e^{i(m-n)\phi/2}-e^{-i(m-n)\phi/2}}{e^{i(m-n)\phi/2}+e^{-i(m-n)\phi/2}}\d\phi\right)\\&=\frac{n+m}{2}\cdot\frac{1}{2\pi}\int_0^{2\pi}(1)\d\phi+0\\&=n=m\end{align}$$

But this is actually obvious, since $\gamma$ is the curve $t\mapsto2e^{int}$.

Answer 2

Here is a simpler approach that is easy to visualize. First consider the winding number around zero of
$\sigma(t) = \exp\big(i\cdot t\big)$ for $t\in [0,2\pi]$. A standard calculation is that $n\big(\sigma,0\big)=1$; finish by using the fact that winding numbers (around zero) of products split into sums of winding numbers. Note this implies $n\big(\sigma^k,0\big)=k$ for $k\in \mathbb Z$. The original problem reads

$n\big(\gamma,0\big)= n\big(\sigma^m\cdot(1+r\cdot \sigma^{n-m}),0\big)=n\big(\sigma^m,0\big)+n\big(1+r\cdot \sigma^{n-m},0\big)=m+n\big(r\cdot \sigma^{n-m},-1\big)$

(i) if $r\lt 1$ then $-1$ is in the unbounded component so $n\big(r\cdot \sigma^{n-m},-1\big)=0\implies n\big(\gamma,0\big)=m$
(ii) if $r\gt 1$, then $-1$ is in the same component as zero (they are path connected on the real line) so
$n\big(r\cdot \sigma^{n-m},-1\big)=n\big(r\cdot \sigma^{n-m},0\big)=n\big(\sigma^{n-m},0\big)=n-m\implies n\big(\gamma,0\big)=n$