# How to use Cauchy's integral formula with more than one pole?

$$\int_{\gamma} \frac{z^2}{z(z-2)}, \quad \gamma(\theta) = 3e^{i\theta}, 0 \leq \theta \leq 2\pi$$

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 $$\frac{z^2}{z-2}$$?

• en.m.wikipedia.org/wiki/Residue_theorem – Ant Apr 17 '15 at 12:38
• Isn't $\frac{z^2}{z(z-2)}=\frac{z}{z-2}$? (with $z=0$ being a removal singularity) – kennytm Apr 17 '15 at 12:39
• $\frac{z^2}{z\left(z-2\right)}=\frac{z}{z-2}$ so that you can take $f\left(z\right)=z$. – Nicolas Apr 17 '15 at 12:40
• Oh, I didn't know we're allowed to cancel out fractions inside an integral. Thanks – mr eyeglasses Apr 17 '15 at 12:40

We have $$\dfrac{z^2}{z(z-2)} = \dfrac{z}{z-2} = 1 + \dfrac2{z-2}$$