Question

I asked this question last night and posted my answer. I wanted someone please verify my answer or show me if I did something wrong.

Answer 1

Posting now...gimme a few minutes....

Answer 2

Find\[I=\int\limits_{0}^{2\pi}\frac{\cos2\theta}{5-4\sin \theta}d \theta\]using residues.

Answer 3

sin(theta) =t

Answer 4

With substitutions I have\[\cos 2\theta=\frac{z^2+z^{-2}}{2},\]\[\sin \theta=\frac{z-z^{-1}}{2i},\]\[d \theta=\frac{dz}{iz}\]With that, I get\[I=(\int\limits)_{|z|=1}\frac{\frac{z^2+z^{-2}}{2}}{5-\frac{2}{i}(z-z^{-1})}(\frac{dz}{iz})=(\int\limits)_{|z|=1}\frac{(z^4+1)dz}{2iz^2(2iz^2+5z+2i)}\]There is a second order pole at z=o. Solving 2iz²+5x+2i=0, I found simple poles at i/2 and 2i. The pole at 2i is outside th circle and can be ignored (I think). So then I rewrote for the residue:\[I=2 \pi i \Sigma_{Res}\left| \frac{z^4+1}{2iz^2[2iz^2+5z+2i]} \right| _{z=0}^{i/2}\] The residue at z=0, I get\[\frac{i}{2i}\frac{d}{dz}\left[ \frac{z^4+1}{2iz^2+5z+2i} \right]_{z=0}=\frac{-5}{8}i\]and at the residue at i/2, I get\[\left|\frac{1}{(2i)(\frac{i}{2})^2}\frac{(\frac{i}{2})^4+1}{\frac{d}{dz}[2iz^2+5z+2i]} \right| _{i/2}=\frac{17i}{24}\]Simplifying everything, I get\[I=2 \pi i\left( \frac{5i}{8} +\frac{17i}{24}\right)=\frac{-\pi}{6}\]

Answer 5

Note:\[(\int\limits)\] are representative of line integrals.

Answer 6

From your explanation and what you did, it makes sense. However, let me run over my complex analysis book really quick.

Answer 7

Seems right to me :)

Answer 8

Malevolence, you seem to be the only one here who knows complex analysis. And you do it with a raisin face. It's excellent.

Answer 9

Haha, why thank you sir /bows/ What are you studying that involves complex analysis? (The reason I ask is I know a lot of engineers take it)

Answer 10

I'm a physics major but it invokes an engineering concentration so along side getting to do the cool physics stuff I get to take other courses like control systems, signals & systems, circuit, theory, electronics, microprocessing, etc...much fun. Next semester I take my Electromagnetic Theory course and I have to be pretty proficient with anything and everything Cauchy.

Answer 11

Sweet :) I'm a physics and math major myself. I have to say, I went in wanting to do alot with physics but my concentration is slowly shifting to math. That sounds awesome though. I'm taking a junior level E&M set and just started quantum. I love it.

Answer 12

Oh man, Quantum is the best!

Answer 13

I just finished up the first semester of it. I like it a lot. "Applied linear algebra" as my instructor calls it xD.

Answer 14

Haha...yup.

Answer 15

You do a lot of "applied linear algebra" in circuit analysis as well.

Answer 16

Anything yet?