Line integral

  • Thread starter squenshl
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  • #1
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I'm studying for a test.
The question is:
Let A be the straight line segment from -3-4i to 4+3i. Let B be the arc of the circle |z| = 5 going anti-clockwise from -3-4i to 4+3i. Let C be the arc of the circle |z| = 5 going anti-clockwise from -3-i to 4+3i. Define:
IA = [tex]\int_A[/tex] 1/z dz

IB = [tex]\int_B[/tex] 1/z dz

IC = [tex]\int_C[/tex] 1/z dz

How do I calculate IA, IB, IC? I know to write IA as an integral by parameterizing A, but how do I parameterize A, I know if I can find IA then the other two are easy. Could I let z = eit so dz = ieit dt. I can't seem to get the limits integration.
 
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Answers and Replies

  • #2
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You have:

[itex]\int_{-3-4i}^{4+3i} \frac{1}{z}dz[/itex]

Just figure out the equation of the straight line from start to end. I get simply [itex]y=x-1[/itex]. So let [itex]x(t)=t[/itex] and [itex]y(t)=t-1[/itex] and solve:

[itex]\int_{-3}^4 \frac{1}{x(t)+iy(t)}(dx(t)+idy(t))[/itex]
 
  • #3
HallsofIvy
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|-3-4i|= |4+ 3i|= 5 so they both lie on the circle |z|= 5 but [itex] |-3- i|= \sqrt{10}\ne 5[/itex] so the circle |z|= 5 does NOT go "from -3- i to 4+ 3i". Did you mean "from -3- 4i to 4+ 3i"? But then B and C are the same!? Perhaps you mean "from 4+ 3i to -3- 4i"? That is, B and C are two halves of the same circle.

Since z= x+ iy is on the circle |z|= 5 if and only if [itex]x^2+ y^2= 5[/itex], we can use [itex]x= 5cos(\theta)[/itex] and [/itex]y= 5 sin(\theta)[/itex] (which is the same as z= 5e^{i\theta}[/itex] for the circle. The only "difficulty" is getting the begining and ending values for [itex]\theta[/itex]. At z= 4+ 3i, [itex]x= 5 cos(\theta)= 4[/itex] and [itex]5 sin(\theta)= 3[/itex] so [itex](5 sin(\theta))/(5 cos(\theta))= tan(\theta)= 3/4[/itex]. For the anti-clockwise circle from 4+ 3i to -3- 4i, [itex]\theta[/itex] goes from [itex]arctan(3/4)[/itex] to [itex]\pi+ arctan(3/4)[/itex] and for the anti-clockwise circle from -3- 4i to 4+ 3i, [itex]\theta[/itex] goes from [itex]\pi+ arctan(3/4)[/itex] to [itex]2\pi+ arctan(3/4)[/itex].
 
  • #4
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Sorry C is the arc of a circle |z| = 5 going clockwise from -3-4i to 4+3i.
How do I parameterize z
Is [tex]\vartheta[/tex] the same as for the anticlockwise arc.
 
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  • #5
479
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Sorry z = 5exp(i[tex]\vartheta[/tex])
 
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  • #6
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For the anti-clockwise arc I got IB = ipi
Does that mean the clockwise arc is ipi by symmetry
 
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  • #7
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No.. 1/z is not a symmetric function.
It is also not an analytic function (in the origin), and I guess this exercise comes to teach you what it means about its line integrals.
 
  • #8
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1/z is analytic inside the region determined by the curves A and B and on the curves... so, [tex]\displaystyle\int_B f(z) dz - \displaystyle\int_A f(z) dz = 0[/tex] (the minus sign is to correct orientation of the curves). So, if you found B, you have A.

Joining C with B (or A, it doesn't really matter), correcting the orientations, you can find the integral using residue theorem (which is simple for f(z) = 1/z, as it's already a Laurent series, and the pole is simple (order 1)). So, if you have B, and the residue, you also have C, and the problem is solved.
 

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