# Complex Analysis-- Integral

• Dec 20th 2006, 08:16 AM
taypez
Complex 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, 11:27 AM
CaptainBlack
Quote:

Originally Posted by taypez
Evaluate:

∫C e^z (z-1)(z-2)^2 / sin^2 pi z

where C is the circle │z│= 2.5

The poles of this inside the circle are at z=-2, -1, 0, 1, and the residues at
these poles are 0, 0, 0, e/pi^2, so the integral should be:

∫C e^z (z-1)(z-2)^2 / sin^2 pi z = e/pi^2

(but be warned I'm not entirely satisfied with this)

RonL
• Dec 20th 2006, 11:52 AM
TD!

$
\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, 12:50 PM
CaptainBlack
Quote:

Originally Posted by TD!

$
\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 ?

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:

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

is analytic so the residue is (2 pi i times) the coefficient of x in the taylor series
of this about $z=0$, 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, 12:53 PM
TD!
I found the residues to be:

- at z = -2: -8/(e².pi²)
- at z = -1: 3/(e.pi²)
- at z = 0: 4/pi²
- at z = 1: e/pi²
• Dec 20th 2006, 02:01 PM
CaptainBlack
Quote:

Originally Posted by TD!
I found the residues to be:

- at z = -2: -8/(e&#178;.pi&#178;)
- at z = -1: 3/(e.pi&#178;)
- at z = 0: 4/pi&#178;
- at z = 1: e/pi&#178;

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 >
RonL
• Dec 21st 2006, 08:15 AM
TD!
Isn't it wonderful when numerical and analytical results appear to match :)
• Dec 21st 2006, 09:21 AM
CaptainBlack
Quote:

Originally Posted by TD!
Isn't it wonderful when numerical and analytical results appear to match :)

Experience suggests that the opeative word is appear

RonL
• Dec 21st 2006, 09:36 AM
ThePerfectHacker
Quote:

Originally Posted by TD!
Isn't it wonderful when numerical and analytical results appear to match :)

I think it is more wonderful when analytical results do not match, and it turns out the numerical results were wrong.
• Dec 21st 2006, 09:50 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
I think it is more wonderful when analytical results do not match, and it turns out the numerical results were wrong.

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, 10:02 AM
TD!
Quote:

Originally Posted by CaptainBlack
Experience suggests that the opeative word is appear

RonL

My words were carefully chosen ;)