I've split the integral around $z_1 = 1 - i$ and $z_2 = 1+ i$ using the contours $C_1$ and $C_2$:
$ \int_{|C|=2} g(z) dz = \int_{C_1} g(z) dz + \int_{C_2} g(z) dz$
In this case, $g(z)$ for $C_1$ is $\displaystyle \frac{\frac{1}{z-z_1}}{z-z_2}$ and $g(z)$ for $C_2$ is $\displaystyle \frac{\frac{1}{z-z_2}}{z-z_1}$. Using Cauchy's Thm I get $\displaystyle \frac{1}{z_2 - z_1}$ for the first one and $\displaystyle \frac{1}{z_1 - z_2}$ for the second. But evaluating
$\displaystyle \int_{|C|=2} g(z) dz = 2\pi i \left(\frac{1}{z_2 - z_1} + \frac{1}{z_1 - z_2} \right) = 0$
I'm not interested in other methods, just this particular version where you split the contour $C$ into $C_1$ and $C_2$ My issue with this answer is that it doesn't make sense. When evaluating an integral like $\displaystyle \int_{\infty}^{-\infty} \frac{dx}{x^2 + 2x + 2}$ you must let $\displaystyle g(z) = \frac{1}{z^2 + 2z + 2}$ and then integrate over the contour $C$. I would've thought that the answer would be give me $\pi$