$$\int_{\gamma} \frac{z^2}{z(z-2)}, \quad \gamma(\theta) = 3e^{i\theta}, 0 \leq \theta \leq 2\pi$$
Cauchy's integral formula is given by:
$$\int\limits_{\gamma} \frac{f(z)}{(z-a)^{n+1}} = \frac{2\pi i}{n!} f^{(n)}(a)$$
And I can choose my holomorphic $f(z) = z^2$. But it doesn't seem like I can get my integral into a form like $(z - a)^n$ in the denominator. Am I missing some algebraic trick to do this?
Also, if $\gamma(\theta)$ was $e^{i\theta}$, then I could choose my holomorphic function to be $\frac{z^2}{z-2}$?
We have $$\dfrac{z^2}{z(z-2)} = \dfrac{z}{z-2} = 1 + \dfrac2{z-2}$$