Math Help - Residues

1. Residues

Calculate $\int_{|z|=3}\frac{e^{iz}}{z^2(z-2)(z+5i)}dz$ 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 = $\frac{-3}{25} +\frac{i}{20}$

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

These just seem like strange answers, have i made a mistake somewhere?

2. Re: Residues

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.

These just seem like strange answers, have i made a mistake somewhere?
Why do you think those are strange answers?

3. Re: Residues

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?

4. Re: Residues

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.

5. Re: Residues

res(f,2)= lim z-->2 $\frac{e^{iz}}{z^2(z+5i)}$ = $\frac{e^{2i}}{4(2+5i)} = \frac{e^{2i}}{8+20i} \frac{8-20i}{8-20i} = \frac{8e^{2i}-20ie^{2i}}{464} = (\frac{1}{58} - \frac{5i}{116})e^{2i}$

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

$=\frac{-12+5i}{100} = \frac{-3}{25} + \frac{i}{20}$

6. Re: Residues

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

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