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

In summary, Cauchy's theorem states that the integral of all simple closed curves around an analytic region is zero. For curves that are not analytic, the theorem requires the use of a different theorem, Cauchy's integral theorem.
  • #1
mancini0
31
0

Homework Statement



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

Show that [tex]\int[/tex]z^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[tex]\theta[/tex]
then z^n = r^n *e^in[tex]\theta[/tex]
Now pull the r^n in front of the integral, leaving

(r^n)[tex]\oint[/tex]e^in[tex]\theta[/tex] d[tex]\theta[/tex]

Hmmm...Perhaps I can say e^in[tex]\theta[/tex] is analytic since the exponential is never zero. Is this correct? What trouble arises when n=-1?
 
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  • #2
Shucks... the exponential is analytic everywhere, so I can use Cauchy's Integral Theorem. :0
 
  • #3
mancini0 said:
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.
 
  • #4
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.
 
  • #5
mancini0 said:
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.
 
  • #6
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.
 
  • #7
mancini0 said:
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.
 
  • #8
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?
 
  • #9
mancini0 said:
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.
 
  • #10
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.
 
  • #11
By Cauchy's theorem, it is sufficient to integrate around the circle with center at 0 and radius R. On such a circle, [itex]z= Re^{i\theta}[/itex] so that [itex]dz= ri e^{i\theta}d\theta[/itex]. The integral is
[tex]\oint z^n dz= \int_0^{2\pi} (R^ne^{ni\theta})(Re^{i\theta}d\theta[/tex]
[tex]= R^{n+1} \int_0^{2\pi} e^{(n+1)i\theta}d\theta[/tex]

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

What is the formula for the integral of z^n around a smooth curve?

The formula for the integral of z^n around a smooth curve is given by ∫(z^n)dz = 0, where n is any complex number and z is the variable of integration.

Why does the integral of z^n around any smooth curve equal 0?

The integral of z^n around any smooth curve equals 0 because the value of the integral is dependent on the path of integration. Since the curve is smooth, its endpoints can be connected by a straight line, making the integral path-independent. This results in a closed loop, where the starting point is the same as the ending point, and the integral must equal 0.

Can the integral of z^n around a smooth curve ever have a non-zero value?

No, the integral of z^n around a smooth curve cannot have a non-zero value. This is because the integral is path-independent, and any path that connects the starting and ending points must result in a closed loop and an integral value of 0.

Does the value of n affect the result of the integral of z^n around a smooth curve?

Yes, the value of n does affect the result of the integral of z^n around a smooth curve. The only case where the integral equals 0 is when n is any complex number. If n is a real number, the integral will not be equal to 0, and its value will depend on the curve and the limits of integration.

How is the integral of z^n around a smooth curve related to the Cauchy Integral Theorem?

The integral of z^n around a smooth curve is related to the Cauchy Integral Theorem, which states that for a function f(z) that is analytic inside and on a simple closed contour C, the integral of f(z) around C is equal to 0. This is essentially the same as the integral of z^n around a smooth curve, where z^n is an analytic function and the curve is a simple closed contour, resulting in an integral value of 0.

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