calculating winding number around zero 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?

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