proof of Cauchy residue theorem

Being $f$ holomorphic by Cauchy-Riemann equations the differential form $f(z)dz$ is closed. So by the lemma about closed differential forms on a simple connected domain we know that the integral ${\int }_{C}f(z)𝑑z$ is equal to ${\int }_{C^{\prime }}f(z)𝑑z$ if $C^{\prime }$ is any curve which is homotopic to $C$.In particular we can consider a curve $C^{\prime }$ which turns around the points $a_j$ along small circles and join these small circles with segments. Since the curve $C^{\prime }$ follows each segment two times with opposite orientation it is enough to sumthe integrals of $f$ around the small circles.

So letting $z=a_j+\rho e^{i\theta }$ be a parameterization of the curve around the point $a_j$, we have $dz=\rho i{e}^{i\theta }d\theta$ and hence

$${\int }_{C}f(z)𝑑z={\int }_{C^{\prime }}f(z)𝑑z=\sum _j\eta (C,a_j){\int }_{\partial B_{\rho }(a_j)}f(z)𝑑z$$

$$=\sum _j\eta (C,a_j){\int }_0^{2\pi }f(a_j+\rho e^{i\theta })\rho i{e}^{i\theta }𝑑\theta$$

where $\rho >0$ is chosen so small that the balls $B_{\rho }(a_j)$ are all disjoint and all contained in the domain $U$. So by linearity, it is enough to prove thatfor all $j$

$$i{\int }_0^{2\pi }f(a_j+e^{i\theta })\rho e^{i\theta }𝑑\theta =2\pi i\mathrm{Res}(f,a_j).$$

Let now $j$ be fixed andconsider now the Laurent series for $f$ in $a_j$:

$$f(z)=\sum _{k\in \mathbb{Z}}{c}_k{(z-a_j)}^k$$

so that $\mathrm{Res}(f,a_j)=c_{-1}$. We have

$${\int }_0^{2\pi }f(a_j+e^{i\theta })\rho e^{i\theta }𝑑\theta =\sum _k{\int }_0^{2\pi }{c}_k{(\rho e^{i\theta })}^k\rho e^{i\theta }𝑑\theta ={\rho }^{k+1}\sum _k{c}_k{\int }_0^{2\pi }{e}^{i(k+1)\theta }𝑑\theta .$$

Notice now that if $k=-1$ we have

$${\rho }^{k+1}{c}_k{\int }_0^{2\pi }{e}^{i(k+1)\theta }𝑑\theta =c_{-1}{\int }_0^{2\pi }𝑑\theta =2\pi c_{-1}=2\pi \mathrm{Res}(f,a_j)$$

while for $k\ne -1$ we have

$${\int }_0^{2\pi }{e}^{i(k+1)\theta }𝑑\theta ={\left[\frac{e^{i(k+1)\theta }}{i(k+1)}\right]}_0^{2\pi }=0.$$

Hence the result follows.