Showing the integral of z^n around any smooth curve = 0.

[mancini0]

Homework Statement

Hello everyone. I'm trying to finish the following problem:

Show that \intz^n dz = 0 for any closed smooth path and any integer not equal to -1. [If n is negative, assume that γ does not pass through the origin, since otherwise the integral is not defined.]

Homework Equations

Cauchy's Integral Theorem states that the integral of all simple closed curves around an analytic region is zero.

Cauchy's Theorem.

The Attempt at a Solution

First, I recognize that if n>0, z^n is analytic everywhere (entire) and Cauchy's Integral Theorem yields the desired result.

My trouble arises when n is negative. By the problem statement, I must only consider n not equal to -1. I plan on using a circle (call it "C1" )as my path of integration, somehow arrive at the conclusion that the integral of z^n with (n<0, n does not = 1) is always zero, then apply Cauchy's Theorem to extend that result to any smooth curve C2. But when n is negative, z^n is not analytic, and Cauchy's Integral Theorem does not apply.

Can I proceed as follows?

Let z =re^i\theta
then z^n = r^n *e^in\theta
Now pull the r^n in front of the integral, leaving

(r^n)\ointe^in\theta d\theta

Hmmm...Perhaps I can say e^in\theta is analytic since the exponential is never zero. Is this correct? What trouble arises when n=-1?

 

[mancini0]

Shucks... the exponential is analytic everywhere, so I can use Cauchy's Integral Theorem. :0

 

[Dick]

Shucks... the exponential is analytic everywhere, so I can use Cauchy's Integral Theorem. :0

It's not that simple. r isn't analytic and neither is theta. You should really figure out what's wrong with the n=(-1) case. Integrate around the unit circle for general n. Then use that the only place 1/z^n isn't analytic is the origin.

 

[mancini0]

Thank you. The fact that theta and r are not analytic follow from the fact that theta and r are functions of z = x+iy. That thought was bothering me. When n = -1, my problem is that I calculate the integral of z^-1 dz to in fact be zero, when z = re^i*theta. But a previous problem showed me that the integral of (z-w)^-1 dz around a unit circle centered on w is 2(pi)i.

 

[Dick]

Thank you. The fact that theta and r are not analytic follow from the fact that theta and r are functions of z = x+iy. That thought was bothering me. When n = -1, my problem is that I calculate the integral of z^-1 dz to in fact be zero, when z = re^i*theta. But a previous problem showed me that the integral of (z-w)^-1 dz around a unit circle centered on w is 2(pi)i.

I don't get z^(-1)*dz around the unit circle to be zero. Use z=e^(i*theta). Find dz in terms of d(theta). Integrate for theta from 0 to 2*pi.

 

[mancini0]

Yikes. I made an elementary mistake in taking the derivative of e^i*theta. Then the integral reduces to the integral of i d(theta) from theta = 0 to theta = 2pi, which is 2(pi)i.

 

[Dick]

Yikes. I made an elementary mistake in taking the derivative of e^i*theta. Then the integral reduces to the integral of i d(theta) from theta = 0 to theta = 2pi, which is 2(pi)i.

Ok. Now you're getting it. Try z^n where n<1.

 

[mancini0]

Okay, for example suppose n = -2. Then integrating (e^i*theta)^-2 from 0 to 2pi gives me zero, as desired. I suppose I could use mathematical induction from here on out?

 

[Dick]

Okay, for example suppose n = -2. Then integrating (e^i*theta)^-2 from 0 to 2pi gives me zero, as desired. I suppose I could use mathematical induction from here on out?

How did you get n=(-2) to be zero? You really don't need induction. You just need that the integral of cos(k*theta) and sin(k*theta) from 0 to 2*pi is zero with k an integer unless k=0, don't you? It's elementary integration. And using the trig functions are periodic.

 

[mancini0]

I meant the integral of z^n dz, which should be 0 by the problem statement. Somebody get this man a Field's Medal! Thanks for your help.

 

[HallsofIvy]

By Cauchy's theorem, it is sufficient to integrate around the circle with center at 0 and radius R. On such a circle, z= Re^{i\theta} so that dz= ri e^{i\theta}d\theta. The integral is
\oint z^n dz= \int_0^{2\pi} (R^ne^{ni\theta})(Re^{i\theta}d\theta
= R^{n+1} \int_0^{2\pi} e^{(n+1)i\theta}d\theta

Now, just do that integral. Note that you will have to do the cases n= -1 and n\ne -1 separately.