Hi can i get a Proof Check please! thanks
Let $f(z)$ be a polynomial and $g(z) = \overset{n}{\underset{k=1}{\Pi}} (z-z_k)$ with $z_j \neq z_k$ for $j \neq k$ and $f(z_k) \neq 0$. for $k \in [1,n]$. Calculate the integral $$\oint_{\gamma}\frac{f(z)}{g(z)}$$ by using the residue theorem
Where $\gamma$ is a closed contour on $\mathbb{C}$
Define $h(z) = \frac{f(z)}{g(z)}$ with $S = \{z_k\}~k \in [1,n]$ to be the set of zero's of $g(z)$ Then assume with out loss of generality that h is reduced (ie f and g share no zeros) then $h$ is analytic on $\mathbb{C} \setminus S$
Since h is a rational function $\infty$ is not an essential singularity
Since $h(z)=\frac{f(z)}{g(z)}$ and g is a polynomial, of order n then its zeros are of finite order and $\underset{z \longrightarrow \infty}{\lim} h(z)$ exists. taking values 0 $\infty$ or C respectively.therefore $\infty$ is not an essential singularity.
Then $h(z)$ has an isolated singularity at $z_0 = \infty$, Define for $0<r<R$ the residue at infinity to be $$Res(h,\infty) = \frac{1}{2 \pi i}\oint_{A^{-}_{o,r}}h(z)dz= -\frac{1}{2 \pi i}\oint_{A^{+}_{o,r}}h(z)dz = -Res\left(\frac{1}{z^2}h\left(\frac{1}{z}\right),0\right)$$ where $A^{-}_{o,r}$ is the negatively orientated anulus with internal radius 0 and external radius r, and $A^{+}_{o,r}$ is the positively orientated version. In this we request that r be sufficiently large
Then Since for a function E analytic on $\mathbb{C} \setminus T$ where T is a finite collection of points $t_i \in T$ where E is not analytic on $\mathbb{C}$ we have $$\sum_{t_i \in T} Res(E,t_i) = -Res(E,\infty)$$ then we have $$\oint_{\gamma^{+}} h(z)dz =2 \pi i \sum_{z_k \in S} W(\gamma^{+},z_k)Res(h,z_k) = -2 \pi i Res(h,\infty)$$
for $\gamma$ Sufficiently large. Otherwise we can default back to $$\oint_{\gamma^{+}} h(z)dz =2 \pi i \sum_{z_k \in S} W(\gamma^{+},z_k)Res(h,z_k)$$ where the singularities outside of the contour has zero contribution to the integral
Thanks for taking the time to read this, appreciate any help given.