Calculate closed curve integral

  • #1
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Homework Statement



Calculate ∫C [itex]\frac{dz}{z(z+a)}[/itex] where C is the unit disk Δ(1) with counterclockwise orientation where a is a complex number with |a| < 1.

Homework Equations



L [itex]\frac{dz}{(z-p)}[/itex]= 2πiν(L,p)

The Attempt at a Solution



Using partial fraction decomposition,

C [itex]\frac{dz}{z(z+a)}[/itex] = ∫C[itex]\frac{-1/a}{z} + \frac{1/a}{(z+a)}[/itex] = (-1/a)2πi + (1/a)2πi = 0
 

Answers and Replies

  • #2
Dick
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Homework Statement



Calculate ∫C [itex]\frac{dz}{z(z+a)}[/itex] where C is the unit disk Δ(1) with counterclockwise orientation where a is a complex number with |a| < 1.

Homework Equations



L [itex]\frac{dz}{(z-p)}[/itex]= 2πiν(L,p)

The Attempt at a Solution



Using partial fraction decomposition,

C [itex]\frac{dz}{z(z+a)}[/itex] = ∫C[itex]\frac{-1/a}{z} + \frac{1/a}{(z+a)}[/itex] = (-1/a)2πi + (1/a)2πi = 0

Well, I think you an overall negative sign wrong in your partial fraction decomposition, but that doesn't matter since the answer is zero anyway. Yes, that's a fine way to do it. Assuming ##\nu## is winding number?
 
  • #3
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Well, I think you an overall negative sign wrong in your partial fraction decomposition, but that doesn't matter since the answer is zero anyway. Yes, that's a fine way to do it. Assuming ##\nu## is winding number?

Yes, it's the winding number. I forgot to mention that. Generally, an integral wouldn't vanish because it contains the origin, correct? This is just a special case that happens to vanish?
 
  • #4
Dick
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Yes, it's the winding number. I forgot to mention that. Generally, an integral wouldn't vanish because it contains the origin, correct? This is just a special case that happens to vanish?

The integral of ##1/z^2## or ##z## vanishes when you integrate around the origin. So that's not a valid general rule. In this case you have two poles inside the contour and they cancel. Isn't that what the calculation is telling you?
 
  • #5
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The integral of ##1/z## doesn't vanish when you integrate around the origin. In this case you have two poles inside the contour and they cancel. Isn't that what the calculation is telling you?

The complex inversion's analyticity is destroyed by the origin so that it doesn't vanish, and the two poles are at z=0 and z=a. Apparently, that's what it's telling me, yes. Heh.
 
  • #6
SteamKing
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The complex inversion's analyticity is destroyed by the origin so that it doesn't vanish, and the two poles are at z=0 and z=a. Apparently, that's what it's telling me, yes. Heh.
Isn't the second pole at z = -a?
 
  • #7
HallsofIvy
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Isn't the second pole at z = -a?
Yes, that was probably a typo. The poles are at 0 and -a.
 
  • #8
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Yes, of course, I apologize for my oversight. Your previous comment about an incorrect negative sign makes sense how. Heh. Thanks again for the help.
 
  • #9
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If C is the curve y = x3-3x2+4x-1 joining points (1,1) and (2,3), find the value of ∫ (12z2-4iz)dz.

(Hint: First show that the integral is independent of path by Cauchy integral theorem, and then choose a special curve to evaluate the integral)

Lemma: Every curve C(φ: [0,1]→D) is homotopic to a piecewise linear curve in D.

Therefore, it is homotopic to the straight line curve connecting (1,1) and (2,3). By the Independence Theorem and Cauchy Integral Theorem, the integrals are equal. I can parametrize the line connecting the points by z(t) = 1 + i + t(1+2i) and insert this into f(z) to get the monstrosity f(z(t)). There has to be a better way of finding the integral.
 
  • #10
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Theorem: Let f be a continuous function on D and let F a primitive function of f. Let C = C(φ:[T0,T1→ℂ) be a curve with finite length with the initial point z0 and the terminal point z1. Then ∫ f(z)dz = Fz1 - Fz0.

So I don't need to go through the rigmarole of finding f(z(t)) and df(z(t))/dt, etc. I can simply integrate with respect to z.
 
Last edited:

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