Integrate 1/z^2 Over a Closed Curve

[Poopsilon]

\int_\alpha\frac{1}{z^2}dz

I can't figure out how to integrate this over a closed circle which contains the origin on its interior. I'm assuming it is equal to 2πi; is there a way to apply Cauchy's Integral Theorem? If I set f(z)=1/z then that is not analytic on the interior, so I don't see how I can apply the theorem.

 

[micromass]

You can either solve it directly by applying the definition or you can use Cauchy's integral formula ( http://en.wikipedia.org/wiki/Cauchy's_integral_formula ).

 

[Poopsilon]

oops I mean't cauchy's integral theorem. But that theorem would require that f(z)=1/z by analytic at zero.

 

[micromass]

oops I mean't cauchy's integral theorem. But that theorem would require that f(z)=1/z by analytic at zero.

Yeah, you can't use that theorem here...

 

[Poopsilon]

oops shoot I meant cauchys integral formula! sorry about that lol. How can I use that? Would f(z) just equal 1?

Edit: Ah wait so I could use the generalized formula with f(z)=1 and get 2πi, is that right?

Super Duper Edit: But wait I'm sorry that would give me 0 because the derivative of a constant function is 0, but I'm quite sure it should be 2πi, so something's wrong.

 

[Dick]

It's not 2pi*i. It's 0. Do what you did in the other thread and integrate around the unit circle and put z=exp(it). Change it to an integral dt. Use the cauchy integral theorem on the difference between the unit circle and the contour alpha.