# How can I use Cauchy integral formula for this integral $g(z)=\int_{C}\frac{s^2+s+1}{s-z}ds$?

Let $C$ be the ellipse $9x^2+4y^2=36$ traversed once in the counterclockwise direction. Define the function $g$ by $$g(z)=\int_{C}\frac{s^2+s+1}{s-z}ds.$$

Find $g(4i)$.

Well I know I must find $g(z)$ (that is the integral) before computing $g(4i)$, so I decided to use Cauchy's integral formula $f(z_{0})=\frac{1}{2\pi i}\int_C\frac{f(z)}{z-z_{0}}dz$. This put me into trouble, because I do not how to start. Please i need a hint.

Thanks.

-
Try putting in the $4i$ first, then evaluating the integral. It's a lot easier this way because then you're working with a specific integral. – Ted Mar 24 '12 at 5:40
@Hassan To make it more familiar, start with changing $z$ with $z_0$ and $s$ with $z$. – Pedro Tamaroff Mar 24 '12 at 5:41
@PeterT.off: I will try it. – Hassan Muhammad Mar 24 '12 at 5:58
Also you should remember if f(z)/(z-z_0) is analytic within and on C, then the integral is 0. – Apprentice Queue Mar 24 '12 at 6:09

Hint:

1. Carefully sketch the curve $C$.
2. Use some theorem of Cauchy.

It might also be interesting to look at other points than $z=4i$, e.g. $z=0$...

-
What do you mean by saying 'use some theorem of cauchy?' More clarifications please. – Hassan Muhammad Mar 24 '12 at 6:15
Use the one used in the proof of the Cauchy integral formula... – AD. Mar 24 '12 at 7:15
Ok, $g(0)=1$, $g(i)=i$ Correct? But I think $4i$ is outside the ellipse. – Hassan Muhammad Mar 24 '12 at 7:47
Yes, $4i$ is outside, hence the curve is contractable to a point.. – AD. Mar 24 '12 at 10:31
Hassan, you should look closer to Cauchy's integral theorem... – AD. Mar 24 '12 at 11:58

$$g(4i)=\int_C \frac{s^2+s+1}{s-4i}ds$$

$\frac{x^2}{4}+\frac{y^2}{9}=1$ is the equation of the vertical ellipse with vertices at $-2, 2, -3i, 3i$. Therefore $4i$ is outside the ellipse and the function $f(s)=\frac{s^2+s+1}{s-4i}$ is analytic within $C$. Then the integral is $0$.

-
What is $g(i)$? – Hassan Muhammad Mar 27 '12 at 6:06
$g(i)=\int_C \frac{s^2+s+1}{s-i}ds$. In this case, $i$ is within the ellipse and we assume $f(s)=s^2+s+1$. using Cauchy Integral Formula, we get $g(i) = 2\pi i f(i) = -2\pi$. – MNos Apr 1 '12 at 14:02