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I'm new in the complex analysis and I'm stuck with this integral :

$I=\displaystyle \int_{|z|=4} \frac{\mathrm{d}z}{(z^2+9)(z+9)} $

the exercise is about Cauchy integral, I don't want the whole solution, just give me a hint (Please don't post fully worked solutions)

what I have done : using partial fraction we get :

$I=\displaystyle \frac{1}{90} \int_{|z|=4} \frac{1}{9+z} + \frac{(9-z)}{9+z^2} \mathrm{d}z$

I'm trying to do this : $\displaystyle \int_{|z|=4} \frac{\mathrm{d}z}{9+z} $.

but $|9|>4$, how to do the integration in this case ?

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You are not expected to integrate in the classical sense but to use the Cauchy Integral formula to compute the integral without looking for the primitive. For that purpose, look at the poles of your function and what the formula says about them. – busman Feb 3 '13 at 11:17
the poles are :$-9,-9i,9i$ but they are out of the disk $|z|\leq 4$, can you explain more ? – aziiri Feb 3 '13 at 11:18
Well, if they are outside, it means they are not inside, and the Cauchy Integral Formula talks about the sum of the inside poles so, how many inside poles are there and hence what is the integral? – busman Feb 3 '13 at 11:26
By the way, check the poles you've calculated, you got two of them wrong. – busman Feb 3 '13 at 11:28
I sorry bushman, the poles are : $-3i,3i,-9$ I'm not focusing too much, I must understand this chapter these days. – aziiri Feb 3 '13 at 11:32
up vote 2 down vote accepted

Note that $(z^2 + 9)(z+9) = (z-3i)(z+3i)(z+9)$, so the integrand has simple poles at $3i, -3i,$ and $-9$. Figure out which of these are in the region $|z| \leq 4$ and then apply Cauchy's Integral Formula.

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both $-3i,3i$ are in the region, I can choose what I want, right ? – aziiri Feb 3 '13 at 11:35
You don't choose them, you just take them both. Cauchy's Integral formula is about the sum of all the poles inside the integration domain. Just compute the residues first. – busman Feb 3 '13 at 11:42
What do you mean by choose? I would advise looking at the example in the Wikipedia article. In that situation you have two simple poles of your integrand, just as you do here. – Michael Albanese Feb 3 '13 at 11:42
is the result is $\displaystyle -\frac{1}{45} (i \pi )$ ? – aziiri Feb 3 '13 at 15:26
That's what I got. – Michael Albanese Feb 4 '13 at 7:35

The function $1/(9+z)$ is a holomorph function on the integration domain $\{|z| = 4\}$, hence by Cauchy integral formula its inegral vanishes: $$\int_{|z|=4} \frac{dz}{9 + z} = 0 $$

How would you proceed with the second term?

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