October 21st, 2017, 03:07 AM  #1 
Newbie Joined: Sep 2017 From: Sydney Posts: 17 Thanks: 0  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? Last edited by skipjack; October 21st, 2017 at 06:38 AM. 
October 21st, 2017, 05:43 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 
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 06:16 AM. 
October 21st, 2017, 06:53 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,274 Thanks: 1959 
Robotboyx9 had already found the poles, but hadn't given the residues correctly.

October 21st, 2017, 07:42 AM  #4 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,040 Thanks: 815 Math Focus: Wibbly wobbly timeywimey stuff.  
October 21st, 2017, 03:40 PM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 2,299 Thanks: 1220 
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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