Evaluate:

∫C e^z (z-1)(z-2)^2 / sin^2 pi z

where C is the circle │z│= 2.5

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- Dec 20th 2006, 07:16 AMtaypezComplex Analysis-- Integral
Evaluate:

∫C e^z (z-1)(z-2)^2 / sin^2 pi z

where C is the circle │z│= 2.5 - Dec 20th 2006, 10:27 AMCaptainBlack
- Dec 20th 2006, 10:52 AMTD!
Are we talking about:

$\displaystyle

\int\limits_C {\frac{{e^z \left( {z - 1} \right)\left( {z - 2} \right)^2 }}{{\sin ^2 \left( {\pi z} \right)}}} dz

$

Because then I find non-zero residues for all the poles (-2,-1,0,1).

Also, from the residue theorem, aren't we forgetting a factor 2pi.i ? - Dec 20th 2006, 11:50 AMCaptainBlack
The factor of 2 pi i helps, and I would be happry to be convinced about the

residues at the other points as I still don't have agreement with my numerical

integral (he said giving away the secret of why he was not happy with his previous

post).

But this is the problem I have. At 0 for instance:

$\displaystyle z^2{\frac{{e^z \left( {z - 1} \right)\left( {z - 2} \right)^2 }}{{\sin ^2 \left( {\pi z} \right)}}} $

is analytic so the residue is (2 pi i times) the coefficient of x in the taylor series

of this about $\displaystyle z=0$, which of course now that I compute it I find it non zero

(where before it seemed to be 0:mad: )

So now I am with you.

RonL - Dec 20th 2006, 11:53 AMTD!
I found the residues to be:

- at z = -2: -8/(eČ.piČ)

- at z = -1: 3/(e.piČ)

- at z = 0: 4/piČ

- at z = 1: e/piČ - Dec 20th 2006, 01:01 PMCaptainBlack
Now that agrees with my numerical integration :)

Code:`This is EULER, Version 2.3 RL-06.`

Type help(Return) for help.

Enter command: (20971520 Bytes free.)

Processing configuration file.

Done.

>

>

>dthe=0.001*pi;

>

>the=dthe/2:dthe:2*pi;

>rc=2.5;

>z=rc*exp(I*the);

>

>fz=exp(z)*(z-1)*(z-2)^2/(sin(pi*z))^2;

>

>

>..numerical integral:

>

>II=sum(fz*rc*I*exp(I*the))*dthe

7.26249e-014+4.29033i

>

>

>e=exp(1)

2.71828

>

>..TD!'s residues

>

>(-8/e^2/pi^2+3/e/pi^2+4/pi^2+e/pi^2)*2*pi*I

0+4.29033i

>

- Dec 21st 2006, 07:15 AMTD!
Isn't it wonderful when numerical and analytical results appear to match :)

- Dec 21st 2006, 08:21 AMCaptainBlack
- Dec 21st 2006, 08:36 AMThePerfectHacker
- Dec 21st 2006, 08:50 AMCaptainBlack
Why. They are complementary, a mistake can occur either way and

their agreement is a sanity check.

It is always important to be able to have two or more independent

techniques available. The ability of the mind to see what it wants is

always with us. An idependent check helps prevent us being carried away

(resolution disagreements also helps develop technique and understanding

of the limitations of our methods).

Finally, it is also instructive to set about computing the integral numericaly

a few things becom clear that otherwise would remain hidden.

RonL - Dec 21st 2006, 09:02 AMTD!