Cauchy Integral Formula Problem

[Joshk80k]

Homework Statement

[tex] \oint \frac{dz}{z^2 + z} = 0, C: abs(z) > 1 [/tex]

Homework Equations

[tex] \oint \frac{f(z)}{z-z_0} dz = 2i\pi * f(z_0) [/tex]

The Attempt at a Solution

Under normal circumstances, I usually deal with these in the following way.

I say that F(z) = 1 (the value in the numerator) and I find the value that will make the denominator go to zero (in this case, 1 or 0).

Evaluating for F(z) = 1 and z_0 = 1, and plugging these values into the equation I provided, the answer comes out to be

[tex] 2i\pi [/tex]

Which is definitely not what the question was asking me to prove.

I am pretty sure that since we have the condition that abs(z) > 1, I am not allowed to pick the value of "1 or 0" for this problem. I'm at a loss as to what to do next.

Additionally, I was considering factoring out a z in the denominator and letting f(z) = 1/z, but I'm not sure that would help anything, or if that's even the correct way to go about doing this. If anyone could show me what to do when my values of z are restricted, I'd be very appreciative!

 

[JSuarez]

Well, I will assume that you are being asked to prove that the integral is zero for any closed contour that satisfies |z| > 1.

If that is the case, then you may consider f(z) analytic in the outside of the contour, because [itex]lim_{z \rightarrow \infty}f(z)=0[/itex] and apply Cauchy's integral theorem directly.

Another is to perform the substitution w = 1/z and then apply the said theorem.