Evaluate $\oint_C\frac{dz}{z-2}$ around the square with vertices $3 \pm 3i, -3 \pm 3i$. I'm having a tough time figuring out $\gamma(t)$ in $\oint_C f(z) dz = \int_a^b f(\gamma(t)) \gamma '(t) dt$. I think I have to split this into four parts, the four separate lines, but from there I'm trying to get things to cancel out so that I could arrive at the answer $2\pi i$. Clearly, we have to consider $3+3i \to -3+3i \to -3 - 3i \to 3-3i$, but I can't seem to find a good starting point. 
 A: Cauchy's integral formula (and the homotopy invariance of the contour integral) tells you that for holomorphic $f$, $$\oint_C \frac{f(z)}{z-2} = 2\pi i f(2).$$ Apply this to the constant function $f = 1$.
A: We have four curves:
$C_1=x-3\cdot i$ for $-3 \leq x \leq 3$
$C_2=3+y\cdot i$ for $-3 \leq y \leq 3$
$C_3=x+3\cdot i$ for $-3 \leq x \leq 3$
$C_4=-3+y\cdot i$ for $-3 \leq y \leq 3$
Then we just have $\int_C f(z)\,dz=\int_{C_1}f(z)\,dz+\int_{C_2}f(z)\,dz+\int_{C_3}f(z)\,dz+\int_{C_4}f(z)\,dz$.
I can add more to clarify.
A: Just apply the Cauchy Integral formula. And you will get $2\pi(i)f(2)$ = $2\pi(i)$, since $f(z)=1$ and $2$ is an interior point of your contour.
This link should clarify: http://en.wikipedia.org/wiki/Cauchy%27s_integral_formula
By using the Cauchy integral theorem, one can show that the integral over C (or the closed rectifiable curve) is equal to the same integral taken over an arbitrarily small circle around a. Since f(z) is continuous, we can choose a circle small enough on which f(z) is arbitrarily close to f(a). On the other hand, the integral, 
