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 March 8th, 2017, 01:50 PM #1 Newbie   Joined: Mar 2017 From: Glasgow Posts: 4 Thanks: 0 Contour Integration Problem I'm struggling to evaluate this integral: $\displaystyle \int_{0}^{2\pi} {\frac{d\theta}{7-2\cos \theta}}$ I have used the substitution $\displaystyle z=e^{i\theta}$ and obtained $\displaystyle d\theta = -i\frac{dz}{z}$ and substituted it in. I'm then left with the integrand equal to $\displaystyle \frac{1}{z^2 -7z +1}$ after taking the constant $\displaystyle i$ out the front of the integral. This left me with a pole inside the unit circle equal to $\displaystyle \frac{7-3\sqrt{5}}{2}$ and I'm not quite sure how to proceed after this. I tried using the residue formula for a simple pole but it's not giving me the correct answer which is given in the question and is $\displaystyle \frac{2\sqrt{5}\pi}{15}$. So I think my method is wrong. Any help would be appreciated.
 March 8th, 2017, 02:40 PM #2 Member   Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35 First, what is your contour integral and what is your contour?
 March 8th, 2017, 02:49 PM #3 Newbie   Joined: Mar 2017 From: Glasgow Posts: 4 Thanks: 0 I'm not given any more information. I thought I could use the contour as the unit circle centred at 0 since f(z) equal to the integrand given would be analytic inside the unit circle except at the pole calculated. Just generally a bit confused with the method.
 March 8th, 2017, 03:45 PM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 2,100 Thanks: 1093 did you multiply the residue by $2\pi i$ ?
 March 8th, 2017, 04:35 PM #5 Newbie   Joined: Mar 2017 From: Glasgow Posts: 4 Thanks: 0 I think I am calculating the residue wrongly. I'm using Res(f,c) for the calculated simple pole c is equal to the limit as z tends to c of (z-c)f(z) however I am getting 0 as an answer.
March 8th, 2017, 05:09 PM   #6
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 March 8th, 2017, 11:27 PM #7 Newbie   Joined: Mar 2017 From: Glasgow Posts: 4 Thanks: 0 Thanks a lot!

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