# Residues

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• Oct 2nd 2012, 01:39 AM
linalg123
Residues
Calculate $\displaystyle \int_{|z|=3}\frac{e^{iz}}{z^2(z-2)(z+5i)}dz$ So the point Z= -5i is not enclosed by the contour.

So am i right that we only calculate the residues at z=0 and z=2 and sum them then multiply by 2i*pi to evaluate the integral?

Residue at z=0 = $\displaystyle \frac{-3}{25} +\frac{i}{20}$

Residue at z=2 = $\displaystyle \frac{e^{2i}}{58} - \frac{5ie^{2i}}{116}$

These just seem like strange answers, have i made a mistake somewhere?
• Oct 2nd 2012, 06:22 AM
FernandoRevilla
Re: Residues
Quote:

Originally Posted by linalg123
So am i right that we only calculate the residues at z=0 and z=2 and sum them then multiply by 2i*pi to evaluate the integral?

Right.

Quote:

These just seem like strange answers, have i made a mistake somewhere?
Why do you think those are strange answers?
• Oct 2nd 2012, 06:57 AM
linalg123
Re: Residues
Quote:

Originally Posted by FernandoRevilla
Right.

Why do you think those are strange answers?

In the ones we have done previously, usually the residues have had similar denominators or something so it could be further simplified. So are these correct?
• Oct 2nd 2012, 07:17 AM
FernandoRevilla
Re: Residues
Quote:

Originally Posted by linalg123
In the ones we have done previously, usually the residues have had similar denominators or something so it could be further simplified. So are these correct?

No problem, show the way you have calculated the residues and we can review it.
• Oct 2nd 2012, 08:27 AM
linalg123
Re: Residues
res(f,2)= lim z-->2 $\displaystyle \frac{e^{iz}}{z^2(z+5i)}$ = $\displaystyle \frac{e^{2i}}{4(2+5i)} = \frac{e^{2i}}{8+20i} \frac{8-20i}{8-20i} = \frac{8e^{2i}-20ie^{2i}}{464} = (\frac{1}{58} - \frac{5i}{116})e^{2i}$

res (f,0) = lim z-->0 $\displaystyle \frac{d}{dz}{\frac{e^{iz}}{(z-2)(z+5i)}} = lim z-->0 \frac{ie^{iz}(z^2-(2-7i)z-(5+12i)}{(z-2)^2(z+5i)^2}= \frac{i(-5-12i)}{-100}$

$\displaystyle =\frac{-12+5i}{100} = \frac{-3}{25} + \frac{i}{20}$
• Oct 2nd 2012, 10:59 PM
FernandoRevilla
Re: Residues
Quote:

Originally Posted by linalg123
res(f,2)= lim z-->2 $\displaystyle \frac{e^{iz}}{z^2(z+5i)}$ = $\displaystyle \frac{e^{2i}}{4(2+5i)} = \frac{e^{2i}}{8+20i} \frac{8-20i}{8-20i} = \frac{8e^{2i}-20ie^{2i}}{464} = (\frac{1}{58} - \frac{5i}{116})e^{2i}$
res (f,0) = lim z-->0 $\displaystyle \frac{d}{dz}{\frac{e^{iz}}{(z-2)(z+5i)}} = lim z-->0 \frac{ie^{iz}(z^2-(2-7i)z-(5+12i)}{(z-2)^2(z+5i)^2}= \frac{i(-5-12i)}{-100}$
$\displaystyle =\frac{-12+5i}{100} = \frac{-3}{25} + \frac{i}{20}$

Right.

P.S. With \lim_{z\to 0}f(z), you'll obtain $\displaystyle \lim_{z\to 0}f(z)$ .