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October 21st, 2017, 02:07 AM   #1
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Complex Numbers Integrals

Hey guys.

I have a problem with part b of this question. I should calculate the residue. Can someone guide me how to correct this and what should the final solution be?
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Last edited by skipjack; October 21st, 2017 at 05:38 AM.

 October 21st, 2017, 04:43 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 First this is very difficult to read! Am I correct that the integral is $\displaystyle \oint \frac{e^{tz}}{(z- 3)^2(z^2+ 4z+ 29)}$? And that the integration is around the circle $|z|= 7$? $z= 3$, which makes $z- 3$ and so the denominator 0, is clearly inside that circle, while $\displaystyle z^2+ 4z+ 29= z^2+ 4z+ 4+ 25= (z+ 2)^2+ 25$ is never 0. So the only pole inside that circle is at $z= 3$ and has order 2. Now, the whole point of "residue" at a pole is this: we can expand a function in a power series at a pole with a finite number of negative power terms. If a function, f, has "a pole of order 2 at x= 3" then we can write $\displaystyle f(x)= \frac{a_{-2}}{(x- 3)^2}+ \frac{a_{-1}}{x- 3}+ a_0+ a_1(x- 3)+ \cdot\cdot\cdot$ Last edited by skipjack; October 21st, 2017 at 05:16 AM.
 October 21st, 2017, 05:53 AM #3 Global Moderator   Joined: Dec 2006 Posts: 19,511 Thanks: 1743 Robotboyx9 had already found the poles, but hadn't given the residues correctly.
October 21st, 2017, 06:42 AM   #4
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Quote:
 Originally Posted by Country Boy $\displaystyle z^2+ 4z+ 29= z^2+ 4z+ 4+ 25= (z+ 2)^2+ 25$ is never 0.
What about $\displaystyle z = -2 \pm 5i$?

-Dan

 October 21st, 2017, 02:40 PM #5 Senior Member     Joined: Sep 2015 From: USA Posts: 2,091 Thanks: 1087 You are correct Dan. Those two poles, $z = -2 \pm 5i$, both lie within $|z| \leq 7$ Fortunately their residues are both zero. The second order residue at $z=3$ is the only pole that contributes to the integral. Thanks from topsquark and Country Boy

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