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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.
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October 21st, 2017, 04:43 AM   #2
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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.
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October 21st, 2017, 05:53 AM   #3
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Robotboyx9 had already found the poles, but hadn't given the residues correctly.
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October 21st, 2017, 06:42 AM   #4
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Quote:
Originally Posted by Country Boy View Post
$\displaystyle z^2+ 4z+ 29= z^2+ 4z+ 4+ 25= (z+ 2)^2+ 25$ is never 0.
What about $\displaystyle z = -2 \pm 5i$?

-Dan
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October 21st, 2017, 02:40 PM   #5
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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.
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