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

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

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  • $\begingroup$ en.m.wikipedia.org/wiki/Residue_theorem $\endgroup$ – Ant Apr 17 '15 at 12:38
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    $\begingroup$ Isn't $\frac{z^2}{z(z-2)}=\frac{z}{z-2}$? (with $z=0$ being a removal singularity) $\endgroup$ – kennytm Apr 17 '15 at 12:39
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    $\begingroup$ $\frac{z^2}{z\left(z-2\right)}=\frac{z}{z-2}$ so that you can take $f\left(z\right)=z$. $\endgroup$ – Nicolas Apr 17 '15 at 12:40
  • $\begingroup$ Oh, I didn't know we're allowed to cancel out fractions inside an integral. Thanks $\endgroup$ – mr eyeglasses Apr 17 '15 at 12:40
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We have $$\dfrac{z^2}{z(z-2)} = \dfrac{z}{z-2} = 1 + \dfrac2{z-2}$$

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