# Calcuating $\frac{1}{2 \pi i} \oint_{\Gamma_{2}} \frac{\zeta^{2} + 5 \zeta}{\zeta -2}d \zeta$ subtracted from itself?

In the text theory of Functions of a Complex Varible i'm attempting to calculate the following integrals in $(1)$

$(1)$

$$\frac{1}{2 \pi i} \oint_{\Gamma_{2}} \frac{\zeta^{2} + 5 \zeta}{\zeta -2}d \zeta - \frac{1}{2 \pi i} \oint_{\Gamma_{1}} \frac{\zeta^{2} + 5 \zeta}{\zeta - 2}d\zeta$$

$\text{Remark}$:

Let $\Gamma_{1}$ be the curve $\partial D(0,1)$ and let $\Gamma_{2}$ the curve $\partial D(0,3)$ both be equipped with counterclockwise orientation. The two curves from the boundary of an annulus $\Phi$. Rigsourly speaking $\Psi \subset \Phi \subset \mathbb{C}$, note $\Psi$ is our open subset.

Our integrad: $\frac{\zeta^{2} + 5 \zeta}{\zeta -2}$, can be factored as follows and one gets the developments in $(2)$

$(2)$

$$\frac{1}{2 \pi i} \oint_{r_{1} < |z-0| < 1} \frac{ \zeta( \zeta^{} + 5 \zeta)}{\zeta -2}d \zeta - \oint_{r_{1} < |z-0| < 3}\frac{ \zeta( \zeta^{} + 5 \zeta)}{\zeta -2} d \zeta$$

In order to achieve calculations for our integrals in $(2)$, one will need to represent $f(z)=\frac{ \zeta( \zeta^{} + 5 \zeta)}{\zeta -2}$ as a Laurent series as formally discussed in $\text{Lemma (1.2)}$

$\text{Lemma (1.2)}$

$$\text{Theorem 7.19} \, \, \, \text{(Laurent Series)}$$

If $f(z)$ is analytic throughout the annular region: $A: r_{1} < |z-z_{o}| < r_{2}$ there is a series expansion as follows in $(1)$

$(1)$

$$f(z) = \sum_{}^{}a_{k}(z-a)^{k} + \sum_{}^{}b_{k}(z-a)^{-k}$$

where,

$$a_{k}= \frac{1}{2 \pi i} \oint_{\Gamma}\frac{f(\zeta)d \zeta}{(\zeta - a)^{k+1}}$$ $$b_{k} = \frac{1}{2 \pi i } \oint_{\Gamma}(\zeta - a)^{k - 1}f(\zeta)d \zeta$$

$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,$ $\text{Remark}:\Gamma \, \text{is any circle} |z-z_{o}| = r \, \text{inside} \, \Phi$

Applying $\text{Theorem 7.19}$, to $(2)$ one can notion the following results in $(3)$

$(3)$

$$\sum \frac{1}{2 \pi i}\oint_{r_{1} < |z-0| < 1} \frac{ \zeta( \zeta^{} + 5 \zeta)d \zeta}{(\zeta -2)^{k+1}}(z-a)^k + \sum \frac{1}{2 \pi i } \oint_{r_{1} < |z-0| < 1}(\zeta -z)^{k+1}\zeta(\zeta + 5 \zeta) - \sum \frac{1}{2 \pi i}\oint_{r_{1} < |z-0| < 3} \frac{ \zeta( \zeta^{} + 5 \zeta)d \zeta}{(\zeta -2)^{k+1}}(z-a)^k + \sum \frac{1}{2 \pi i } \oint_{r_{1} < |z-0| < 3}(\zeta -z)^{k+1}\zeta(\zeta + 5 \zeta)$$

Concerning the recent developments in $(3)$, may I have hints on how to attack $(3)$ ?

• You realise that the integrand has a simple pole at $\zeta=2$, to which you can apply the Residue Theorem? Jul 21, 2017 at 20:20
• I mean... are there poles in the circle of interest? Naively I would suggest residues. Jul 21, 2017 at 20:20

My take would be to use residues: reverse the orientation of $\Gamma_1$ and regroup the integrals. Now consider the integral of that same function over the arc $\sigma(t) = 1+2t$ for $t \in [0,1]$. Let $\gamma = \Gamma_2 + \sigma - \Gamma_1 - \sigma$ (changing the starting points of your curves, but who cares). This is an integral over a close path, thus we have: $$\frac{1}{2\pi i} \int_\gamma \frac{\zeta^2-5\zeta}{\zeta-2}d\zeta = Res_{\zeta=2}\frac{\zeta^2-5\zeta}{\zeta-2} = 14$$
• For the conclusion $$\frac{1}{2\pi i} \int_\gamma \frac{\zeta^2-5\zeta}{\zeta-2}d\zeta = Res_{\zeta=2}\frac{\zeta^2-5\zeta}{\zeta-2} = 14$$ you looked at if $f(z)$ had simple poles, but how did you "regroup the integrals" to get the reach your "hint" ? Jul 21, 2017 at 20:47
• here's what I meant: $$\int_{\Gamma_2} - \int_{\Gamma_1} = \int_{\Gamma_2} + \int_{-\Gamma_1} = \int_{\Gamma_2 - \Gamma_1}$$ Jul 21, 2017 at 20:57
• So from the what you've recently stated:$\int_{\Gamma_2} - \int_{\Gamma_1} = \int_{\Gamma_2} + \int_{-\Gamma_1} = \int_{\Gamma_2 - \Gamma_1}$ to take the residues one would not need to interchange the integral and series to take apply the residue theorem ? Jul 21, 2017 at 23:02
• No, the residue theorem has nothing to do with Laurent series, the fact that you're integrating over a path which inscribes an annulus I think is best exploited by the residue Theorem rather than laurent developements. If you insisted on wanting to use Laurent series the problem would become somewhat longer to solve, and you'd find the same result due two the fact that the residue is actually the coefficient $B_1$ of the Laurent developement. Jul 22, 2017 at 11:13
• How would one begin applying the Residue Theorem, would look at the functions $f(z)=\frac{ \zeta( \zeta^{} + 5 \zeta)d \zeta}{(\zeta -2)^{k+1}}$, $f(z)=(\zeta -z)^{k+1}\zeta(\zeta + 5 \zeta), f(z)=\frac{ \zeta( \zeta^{} + 5 \zeta)d \zeta}{(\zeta -2)^{k+1}}$ , $f(z)=(\zeta -z)^{k+1}\zeta(\zeta + 5 \zeta)$ and begin to see if those functions have simple poles. But this brings another key question what contour would one use ? PS: Sorry for the dumb questions this is my first time with the tools of Residue Theory :(. Jul 22, 2017 at 13:15