Is this line integral correct?

Jamin2112
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Homework Statement



∫C1(0) dz / (z * sin2(z))

Homework Equations



Residue Theorem material

The Attempt at a Solution



z * sin2(z)
= z * (1/2 - cos(2z)/2)
= z * [1/2 - (1/2)∑(-1)n(2z)n/(2n)! ]
= z3 + ...

---> z * sin2(z) has a zero of order 3 at z = 0
---> 1/(z * sin2(z)) has a pole of order 3 at z = 0

The circle C1(0) encloses z = 0 (its only singularity), so we have

2πi * Res[f, 0] = 2πi * 1/(3-1)! * limz-->0 d2/dz2 (z-0)3 1/(z * sin2(z)),

which I don't want to calculate if there's an easier way to do the problem. Is there an easier way? I'm too cool for calculating double derivatives.
 
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Jamin2112 said:

Homework Statement



∫C1(0) dz / (z * sin2(z))

Homework Equations



Residue Theorem material

The Attempt at a Solution



z * sin2(z)
= z * (1/2 - cos(2z)/2)
= z * [1/2 - (1/2)∑(-1)n(2z)n/(2n)! ]
= z3 + ...

---> z * sin2(z) has a zero of order 3 at z = 0
---> 1/(z * sin2(z)) has a pole of order 3 at z = 0

The circle C1(0) encloses z = 0 (its only singularity), so we have

2πi * Res[f, 0] = 2πi * 1/(3-1)! * limz-->0 d2/dz2 (z-0)3 1/(z * sin2(z)),

which I don't want to calculate if there's an easier way to do the problem. Is there an easier way? I'm too cool for calculating double derivatives.

You could expand it in a Laurent series around z=0 and look for the coefficient of the 1/z term if you are so dead set against finding derivatives.
 
Dick said:
You could expand it in a Laurent series around z=0 and look for the coefficient of the 1/z term if you are so dead set against finding derivatives.

Ah, I see! So perform long division using the series from the trig identity I invoked?
 
Jamin2112 said:
Ah, I see! So perform long division using the series from the trig identity I invoked?

Factor out all of the z's like you did so you should be left with something like z^3*(1-g(z)). Then use the expansion 1/(1-g(z))=1+g(z)+g(z)^2+... Remember?
 
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Dick said:
Factor out all of the z's like you did so you should be left with something like z^3*(1-g(z)). Then use the expansion 1/(1-g(z))=1+g(z)+g(z)^2+... Remember?

I got 1/3. That seem legit?
 
Jamin2112 said:
I got 1/3. That seem legit?

That's what I got. Not sure it makes it legit.
 
Dick said:
That's what I got. Not sure it makes it legit.

Awesome, brah. Now let's look at ∫C f(z), where f(z) = dz/(z-1)2(z2+4) and C = C4(0). The integrand clearly has 3 singularities, all of which lie inside C. They are z = 1, which has order 1; z = 2i, which has order 2; and z = -2i, which has order 2. The residue is therefore

2πi * (Res[f, 1] + Res[f, 2i], Res[f, -2i]).​

But hang on a sec, brah. For Res[f, 1] I got limz-->1 (z-1) * 1/(z-1)2(z2+4) = ∞. Wat do?
 
Jamin2112 said:
Awesome, brah. Now let's look at ∫C f(z), where f(z) = dz/(z-1)2(z2+4) and C = C4(0). The integrand clearly has 3 singularities, all of which lie inside C. They are z = 1, which has order 1; z = 2i, which has order 2; and z = -2i, which has order 2. The residue is therefore

2πi * (Res[f, 1] + Res[f, 2i], Res[f, -2i]).​

But hang on a sec, brah. For Res[f, 1] I got limz-->1 (z-1) * 1/(z-1)2(z2+4) = ∞. Wat do?

The pole at z=1 has order 2. The other two have order 1. Why do you think otherwise?
 
Dick said:
The pole at z=1 has order 2. The other two have order 1. Why do you think otherwise?

Let me redeem myself. To find the order of the pole z=1, I need to write (z-1)2(z2+4) in the form ∑an(z-1)n. Is this supposed to be incredibly obvious? I can't figure it out.
 
  • #10
Jamin2112 said:
Let me redeem myself. To find the order of the pole z=1, I need to write (z-1)2(z2+4) in the form ∑an(z-1)n. Is this supposed to be incredibly obvious? I can't figure it out.

No. You are supposed to write f(z)=g(z)/(z-1)^n where is g(z) is analytic in a neighborhood of z=1 and g(1) is not zero. What are n and g(z)? Yes, it's supposed to be pretty obvious.
 
  • #11
Dick said:
[...] where is g(z) is analytic in a neighborhood of z=1 [...]

A punctured neighborhood of z=1?
 
  • #12
Jamin2112 said:
A punctured neighborhood of z=1?

Ok. Punctured. It might have a removable singularity at z=1.
 
  • #13
Dick said:
No. You are supposed to write f(z)=g(z)/(z-1)^n where is g(z) is analytic in a neighborhood of z=1 and g(1) is not zero. What are n and g(z)? Yes, it's supposed to be pretty obvious.

You must be referring to g(z) = 1/(z2+4), n = 2.
 
  • #14
Jamin2112 said:
You must be referring to g(z) = 1/(z2+4), n = 2.

Sure. Pole of order 2.
 
  • #15
Dick said:
Sure. Pole of order 2.

Ah, I understand the reasoning now.

Since g(z) = 1/(z2+4) is analytic in a neighborhood about z=1, it will equal some power series centered at z=1:

g(z) = ∑an(z-1)n.​

Thus
g(z)/(z-1)2
= (z-1)-2 (a0 + a1(z-1) + a2(z-2)2 + ...)
= a0/(z-1)2 + ...

Thus f(z) = g(z)/(z-1)2 has a finite number of terms of the form a-k/(z-1)k (with k>0), the first of them being being k = 2. Is that basically right?
 
  • #16
Jamin2112 said:
Ah, I understand the reasoning now.

Since g(z) = 1/(z2+4) is analytic in a neighborhood about z=1, it will equal some power series centered at z=1:

g(z) = ∑an(z-1)n.​

Thus
g(z)/(z-1)2
= (z-1)-2 (a0 + a1(z-1) + a2(z-2)2 + ...)
= a0/(z-1)2 + ...

Thus f(z) = g(z)/(z-1)2 has a finite number of terms of the form a-k/(z-1)k (with k>0), the first of them being being k = 2. Is that basically right?

Sure. The point is near z=1 f(z) behaves like a0/(z-1)^2 plus small corrections.
 
  • #17
Dick said:
Sure. The point is near z=1 f(z) behaves like a0/(z-1)^2 plus small corrections.

Sweet deal. (Maybe I should actually buy the textbook and stop sleeping in class.)

I have another question, though. (See this as an opportunity to become an even higher contributor to the homework help sub-forum.)

Am I correct that z-2csc(z) has a pole of order 3 at z=0?

For z2sin(z)
= z2(z - z3/3! + z5/5! - z7/7! + z9/9! + ...)
= z3 - z5/3! + ...
undeniably has a zero of order 3 at z=0.


And in that case,
Res[z-2csc(z)] = 1/(2-1)! limz-->0 ((z3 * z-2csc(z))'' = limz-->0 (z*sin2(z) - 2cos(z)sin(z) + 2z*cos2(z))/sin3(z) ??
 
  • #18
Jamin2112 said:
Sweet deal. (Maybe I should actually buy the textbook and stop sleeping in class.)

I have another question, though. (See this as an opportunity to become an even higher contributor to the homework help sub-forum.)

Am I correct that z-2csc(z) has a pole of order 3 at z=0?

For z2sin(z)
= z2(z - z3/3! + z5/5! - z7/7! + z9/9! + ...)
= z3 - z5/3! + ...
undeniably has a zero of order 3 at z=0.And in that case,
Res[z-2csc(z)] = 1/(2-1)! limz-->0 ((z3 * z-2csc(z))'' = limz-->0 (z*sin2(z) - 2cos(z)sin(z) + 2z*cos2(z))/sin3(z) ??

Sure again. Now you can either use l'Hopital on the trig stuff to find the limit, or you can use the Laurent series trick like in the first problem.
 
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  • #19
Dick said:
Sure again. Now you can either use l'Hopital on the trig stuff to find the limit, or you can use the Laurent series trick like in the first problem.

Sweet deal. Again, how do I solve z6+1=0?

I tried this:

z6+1=0 <---> (z3+i)(z3-i)=0


z3+i
= (x+iy)3+i = 0
<---> (x2+2xyi - y2)(x+iy) + i = (x3 - 3xy2) + i * (3x2y - y3 + 1) = 0
<---> x2 = 3y2; 3x2y - y3 = -1
<---> y = (1/8)-1/3 = -1/2

So two of our zeroes are √3/2 -1/2 * i and -√3/2 - 1/2 * i

Ditto for (z3-i)=0? amidoinitrite?
 
  • #20
Jamin2112 said:
Sweet deal. Again, how do I solve z6+1=0?

I tried this:

z6+1=0 <---> (z3+i)(z3-i)=0


z3+i
= (x+iy)3+i = 0
<---> (x2+2xyi - y2)(x+iy) + i = (x3 - 3xy2) + i * (3x2y - y3 + 1) = 0
<---> x2 = 3y2; 3x2y - y3 = -1
<---> y = (1/8)-1/3 = -1/2

So two of our zeroes are √3/2 -1/2 * i and -√3/2 - 1/2 * i

Ditto for (z3-i)=0? amidoinitrite?

Now can't you think of a simpler way to do it than that? You are trying to find the six sixth roots of -1. Try using polar form r*exp(i*theta).
 
  • #21
Dick said:
Now can't you think of a simpler way to do it than that? You are trying to find the six sixth roots of -1. Try using polar form r*exp(i*theta).

(reiø)6 = -1
<---> r6cos(6ø) + r6isin(6ø) = -1.
<---> r6cos(6ø) = -1; r6isin(6ø) = 0

That right so far?
 
  • #22
Jamin2112 said:
(reiø)6 = -1
<---> r6cos(6ø) + r6isin(6ø) = -1.
<---> r6cos(6ø) = -1; r6isin(6ø) = 0

That right so far?

Still too hard. (r*e^(iø))^6=r^6*e^(6iø). r must be 1. Now express (-1) in polar form.
 
  • #23
Dick said:
Still too hard. (r*e^(iø))^6=r^6*e^(6iø). r must be 1. Now express (-1) in polar form.

cos(6ø)+isin(6ø)= -1
---> cos(6ø) = -1
---> 6ø = (2k+1)π
---> ø = (2k+1)π/6 =π/6, π/2, -π/6, -π/2, 7π/6, -7π/6, ... (everything after the "..." is redundant)
 
  • #24
Jamin2112 said:
cos(6ø)+isin(6ø)= -1
---> cos(6ø) = -1
---> 6ø = (2k+1)π
---> ø = (2k+1)π/6 =π/6, π/2, -π/6, -π/2, 7π/6, -7π/6, ... (everything after the "..." is redundant)

Yes. This is review for you, isn't it?
 
  • #25
Dick said:
Yes. This is review for you, isn't it?

This is me catching up after slacking last quarter.
 
  • #26
Jamin2112 said:
This is me catching up after slacking last quarter.

I think buying the textbook and not sleeping in class are really good ideas.
 
  • #27
Dick said:
I think buying the textbook and not sleeping in class are really good ideas.

Question: Do you think it's a good idea to drink a few beers to get a good buzz going before an exam? It helps one not second-guess himself when writing down answers.
 
  • #28
Jamin2112 said:
Question: Do you think it's a good idea to drink a few beers to get a good buzz going before an exam? It helps one not second-guess himself when writing down answers.

If you've done all your homework, bought the textbook and not snoozed through all you classes, it MIGHT not hurt you that much. If you haven't, well, you are probably dead in the water anyway and hoping everybody else sucks more than you do. No. I don't think it's a good strategy. Do the beers after the exam. You'll enjoy the buzz more. So we're done with complex analysis then?
 
  • #29
Dick said:
So we're done with complex analysis then?

Almost. This problem has been ticking me off. I've checked my work several times.

Using residues, I'm supposed to find the partial fractional representation of (z2-7z+4)/(z2(z+4)).

I have it on good word that

(z2-7z+4)/(z2(z+4)) = A/z2 + B/z + C/(z+4),

where

A = Res[(z-0)*f, 0],
B = Res[f, 0],
C = Res[f, -4].

Thus

A= limz-->0 d/dz (z-0)*(z-0)2(z2-7z+4)/(z2(z+4),

B = limz-->0 d/dz (z-0)2(z2-7z+4)/(z2(z+4)),

C = limz-->4 (z+4)(z2-7z+4)/(z2(z+4)).Right?
 
  • #30
Jamin2112 said:
Almost. This problem has been ticking me off. I've checked my work several times.

Using residues, I'm supposed to find the partial fractional representation of (z2-7z+4)/(z2(z+4)).

I have it on good word that

(z2-7z+4)/(z2(z+4)) = A/z2 + B/z + C/(z+4),

where

A = Res[(z-0)*f, 0],
B = Res[f, 0],
C = Res[f, -4].

Thus

A= limz-->0 d/dz (z-0)*(z-0)2(z2-7z+4)/(z2(z+4),

B = limz-->0 d/dz (z-0)2(z2-7z+4)/(z2(z+4)),

C = limz-->4 (z+4)(z2-7z+4)/(z2(z+4)).Right?

I'm getting tired. What part of that are you having a problem with? Can you be specific? Since you haven't shown your work, what part isn't checking?
 
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