complex integration

• Sep 8th 2007, 10:50 AM
galactus
complex integration
I am in the process of studying a little complex analysis when I have time.

What is the best way to approach something like this?:

$\displaystyle \int\frac{z^{2}+sinh(2z)}{z^{3}-31z^{2}+4z-12i}$

First, I thought about using $\displaystyle sinh(2z)=\frac{1}{2}(e^{2z}-e^{-2z})$

Do the roots of the denominator have a lot to do with it?.

Just like some input from those in the know.
• Sep 8th 2007, 01:53 PM
topsquark
Quote:

Originally Posted by galactus
I am in the process of studying a little complex analysis when I have time.

What is the best way to approach something like this?:

$\displaystyle \int\frac{z^{2}+sinh(2z)}{z^{3}-31z^{2}+4z-12i}$

First, I thought about using $\displaystyle sinh(2z)=\frac{1}{2}(e^{2z}-e^{-2z})$

Do the roots of the denominator have a lot to do with it?.

Just like some input from those in the know.

(I was wondering what happened to that little "dz" thingy in there. :) )

But seriously, I can't be sure of what you are integrating over here. The real line, a curve, an area... The best method is going to depend heavily on what you are integrating over. (Warning: not that I am likely to be able to show you how once you specify it.)

-Dan
• Sep 8th 2007, 02:01 PM
CaptainBlack
Quote:

Originally Posted by galactus
I am in the process of studying a little complex analysis when I have time.

What is the best way to approach something like this?:

$\displaystyle \int\frac{z^{2}+sinh(2z)}{z^{3}-31z^{2}+4z-12i}$

First, I thought about using $\displaystyle sinh(2z)=\frac{1}{2}(e^{2z}-e^{-2z})$

Do the roots of the denominator have a lot to do with it?.

Just like some input from those in the know.

If the integral is around a well behaved closed curve then the residue
theorem will give the integral.

RonL
• Sep 8th 2007, 02:53 PM
galactus
Thanks fellas. Yes, it's a closed curve. And very well-behaved. Not a brat.
• Sep 8th 2007, 03:02 PM
topsquark
Quote:

Originally Posted by galactus
Thanks fellas. Yes, it's a closed curve. And very well-behaved. Not a brat.

The residue theorem is going to be a pain in the arse here. galactus: you are going to have to supply the curve as the three poles may or may not be included inside the contour.

-Dan
• Sep 8th 2007, 03:11 PM
galactus
I see what you mean. I knew there were three poles, but this is all that was given. I will try to gather more info regarding the curve. Thanks for your help.
• Sep 8th 2007, 04:55 PM
ThePerfectHacker
To galactus: Use $\displaystyle \oint$ instead to signify around a closed curve.

It is possible that Galactus is integrating in an open set where the function is holomorphic. In that case, Cauchy's Theorem will say the integral is zero.

Where is this problem from?
• Sep 8th 2007, 05:44 PM
galactus
I seen it on another forum. I thought it was an unusually advanced and interesting problem. I have been studying complex analysis a bit and was curious also.