Calculating IA, IB, IC: A Parameterization Approach

[squenshl]

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:
I_A = \int_A 1/z dz

I_B = \int_B 1/z dz

I_C = \int_C 1/z dz

How do I calculate I_A, I_B, I_C? I know to write I_A as an integral by parameterizing A, but how do I parameterize A, I know if I can find I_A then the other two are easy. Could I let z = e^it so dz = ie^it dt. I can't seem to get the limits integration.

 

[jackmell]

You have:

\int_{-3-4i}^{4+3i} \frac{1}{z}dz

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

\int_{-3}^4 \frac{1}{x(t)+iy(t)}(dx(t)+idy(t))

 

[HallsofIvy]

|-3-4i|= |4+ 3i|= 5 so they both lie on the circle |z|= 5 but |-3- i|= \sqrt{10}\ne 5 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 x^2+ y^2= 5, we can use x= 5cos(\theta) 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 beginning and ending values for \theta. At z= 4+ 3i, x= 5 cos(\theta)= 4 and 5 sin(\theta)= 3 so (5 sin(\theta))/(5 cos(\theta))= tan(\theta)= 3/4. For the anti-clockwise circle from 4+ 3i to -3- 4i, \theta goes from arctan(3/4) to \pi+ arctan(3/4) and for the anti-clockwise circle from -3- 4i to 4+ 3i, \theta goes from \pi+ arctan(3/4) to 2\pi+ arctan(3/4).

 

[squenshl]

Sorry C is the arc of a circle |z| = 5 going clockwise from -3-4i to 4+3i.
How do I parameterize z
Is \vartheta the same as for the anticlockwise arc.

 

[squenshl]

Sorry z = 5exp(i\vartheta)

 

[squenshl]

For the anti-clockwise arc I got I_B = ipi
Does that mean the clockwise arc is ipi by symmetry

 

[elibj123]

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.

 

[GRFrones]

1/z is analytic inside the region determined by the curves A and B and on the curves... so, \displaystyle\int_B f(z) dz - \displaystyle\int_A f(z) dz = 0 (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.