Prove why complex integrals equal zero

Question

I know this is very basic, but I'm having trouble getting started. Am I supposed to use Green's theorem to find Cauchy's integral theorem? Could someone please walk me through it.

Let γ(a,r) represent the circle of radius r centred at a. Explain why each of the following integrals are zero:

$$\oint_{\gamma(1,1)} \frac{1}{z-3}$$

Thank you!

Answer 1

You said the answer yourself: You have a bounded open set $\Omega$ and $f$ holomorphic on $\Omega$ and $C^1$ on a neighborhood of $\overline{\Omega}$. Using Greens theorem and $dz = dx + i\,dy$, you can work out that $$\int_{\partial \Omega}f\,dz = \int_{\Omega}d(f\,dz) = 2i \int_{\Omega}\frac{\partial f}{\partial \overline{z}}\,dx\,dy.$$ Now the Cauchy Riemann equation implies this equals 0.