1. ## closed curve, trace

Let $\displaystyle \gamma$ be a closed path in a domain $\displaystyle D$ such that $\displaystyle W(\gamma, \zeta)=0$ for all $\displaystyle \zeta \not \in D$. Suppose that $\displaystyle f(z)$ is analytic on $\displaystyle D$ except possibly at a finite number of isolated singularities $\displaystyle z_1, \ldots, z_m \in D \char\\ \Gamma$. Show that $\displaystyle \int_{\gamma} f(z)dz = 2 \pi i \sum W(\gamma, \zeta) \text{Res}[f, z_k]$.

Hint

Consider the Laurent decomposition at each $\displaystyle z_k$.

I do not see how to get the Laurent decomposition of the $\displaystyle z_k$ to do this problem. By the way, the above notation means the winding number. It looks like perhaps the residue theorem would come into play here somehow too. I just do not see how to proceed now. Any suggestions would be very nice. Thank you.

2. Originally Posted by poincare4223
Let $\displaystyle \gamma$ be a closed path in a domain $\displaystyle D$ such that $\displaystyle W(\gamma, \zeta)=0$ for all $\displaystyle \zeta \not \in D$. Suppose that $\displaystyle f(z)$ is analytic on $\displaystyle D$ except possibly at a finite number of isolated singularities $\displaystyle z_1, \ldots, z_m \in D \char\\ \Gamma$. Show that $\displaystyle \int_{\gamma} f(z)dz = 2 \pi i \sum W(\gamma, \zeta) \text{Res}[f, z_k]$.

Hint

Consider the Laurent decomposition at each $\displaystyle z_k$.

I do not see how to get the Laurent decomposition of the $\displaystyle z_k$ to do this problem. By the way, the above notation means the winding number. It looks like perhaps the residue theorem would come into play here somehow too. I just do not see how to proceed now. Any suggestions would be very nice. Thank you.
For each $\displaystyle z_j$ consider the set $\displaystyle D_j= \{ z\in \mathbb{C} : 0<|z-z_j|<r_j \}$ where $\displaystyle r_j$ is such that $\displaystyle D_j\cap D_i =\emptyset$ if $\displaystyle i\neq j$. $\displaystyle f$ has a Laurent expansion in each of these (shrink them if necessary) so $\displaystyle f= \sum_{n=0}^{\infty } a_n(z-z_j)^n+ \sum_{n=1}^{\infty } \frac{b_n}{(z-z_j)^n}$. Now define $\displaystyle S_j(z)=\sum_{n=1}^{\infty } \frac{b_n}{(z-z_j)^n}$ then using a dual to Abel's lemma we get that $\displaystyle S_j$ is uniformly convergent in sets of the form $\displaystyle \{ z\in \mathbb{C} : |z-z_j|\geq \epsilon \}$ for all $\displaystyle \epsilon >0$. Now define $\displaystyle g(z)=f(z)-\sum_{k=1}^{m} S_j(z)$ then it's clear that $\displaystyle g$ is holomorphic in $\displaystyle D$ (the $\displaystyle z_i$'s are removable singularities) then $\displaystyle \int_{\gamma } g=0$ (because of Cauchy's theorem and the fact that $\displaystyle \gamma$ is homotopic to a point in $\displaystyle D$) from which $\displaystyle \int_{\gamma } f = \sum_{k=1}^{m} \int_{ \gamma } S_j$. Now to conclude just calculate $\displaystyle \int_{ \gamma } S_j$ where you need to remember this http://www.mathhelpforum.com/math-he...tml#post446504 and that $\displaystyle \int_{\gamma } \frac{1}{z-z_j} = 2\pi i W(\gamma ,z_j)$