Recent content by cleannj

Then you are saying (z+2)* \frac{1}{(z+2)(z-4)} is what is necessary then 1/ (z-4) plug in -2 and get the residue to be 1 / (-2-4) = -1/6

Very well, please explain (even if you need to use another rational function example) EXACTLY how to compute the residue... Your short answers are not helping -- the limit of (z +2 ) * anything if z approaches -2 will be zero since -2 + 2 = 0.

Cyosis continue For r = 1 the integral must be zero right? For r = 3 only z = -2 lies in the interior So the Res f = lim (z--2) as z --> -2 so this is zero also? AND for r = 5 z = -2 and z = 4 lie in the interior so lastly Res f = lim (z-4) as z --> 4 is zero? This all...

Poles and Residues Quite honestly the only thing I can make out from the theorem is the constant 2\pii is multiplied by the sum of residues... I cannot make out the right hand side of the formula to determine the residue itself. I seem to remember reading somewhere that the coefficient of...

Find integral over counterclockwise circle Much like the problem with residues I seem to be good with the big stuff just not the details. Another question would be: Let \gamma_{r} be the counterclockwie circle w/ center at 0 and radius r. Find \intdz/(z^2-2z-8) for r = 1,3,5 I did...

Thank you for your help.

First of all I am a first time user this evening... So bear with me. The question was: Find a Laurent series for (z-2)/(z+1) centered at z = -1 and specify the region in which it converges.Laurent Series will be employed 1/z \Sigma\frac{alpha}{z}^{k}So far I have done long division and...