Question

Residue Calculus....need help!

Answer 1

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

Answer 2

Okay, here's what I got. If someone can verify this for me, I'd greatly appreciate it.
My substitutions should be\[\cos2 \theta=z^2+z^{-2}/2\]\[\sin \theta=z-z^{-1}/2i\]\[d \theta=dz/iz\]We then have \[I= (\int\limits\limits)_{|z|=1} [(z^2-z^{-2}/2)/[5-(2/1)(z-z^{-1})] (dz/iz)\]\[=(\int\limits)_{|z|=1} [(z^4+1)dz]/[2iz^2(2iz^2+5z-2i)]\]There is a second order pole at z=0. Solving 2iz²+5-2i=0, we find simple poles at i/2 and 2i. The pole at 2i is outside the circle |z|=1 and can be ignored. Therfore,\[I=2 \pi i \Sigma _{RES} [(z^4+1)/2iz^2(2iz^2+5z-2i)]|_{z=0}^{2i}\]For the residue at z=0,\[(i/2i)\left\{ [d(z^4+1)/(2iz^2+5z+2i)]/dz \right\}|_{z=0}=(-5/8)i\]and from the residue at i/2,\[\left\{ [1/[(2i)(i/2)]^2\right\}\left\{[(i/2)^4+1]/[d(2iz^2+5z+2i)/dz\right\}]\|_{z=i/2}=17i/24\]I know that at this point, it's a bit much to look at, especially cause the formating for these equations sucks on this site, but after all the simplification we should get\[I=2 \pi i[(-5i/8)+(17i/24)]=-\pi/6\]

Can someone please verify my answer?

Answer 3

Sorry, I really can't read this. You should really use more fractions, so not just ()/() but \frac{}{}

Answer 4

Is that how you get fractions on here?

Answer 5

Okay, I'm rewriting from the last two residues down to the final one. For the residue at z=0, I get\[\frac{i}{2i}\frac{d}{dz}[\frac{z^4+1}{2iz^2+5z+2i}]|_{z=0}=\frac{-5}{8}i\] and for the residue at i/2, I get\[\frac{1}{(2i)(\frac{i}{2})^2}\frac{(\frac{i}{2})^4+1}{\frac{d}{dz}[2iz^2+5z+2i]}_{z=i/2}=\frac{17i}{24}\]Simplifying all this should get me\[I=2 \pi i(\frac{-5i}{8}+\frac{17i}{24})=\frac{-\pi}{6}\]

Thanks, nowhereman, for showing me how to make the fractions. It's soooo much easier to read now.

Answer 6

Now can someone please verify my answer?