The question

Evaluate $\displaystyle \oint_{|z| = 4} \frac{z^3 \ dz}{(z - 3)^3(z - 2)^2}$ using the Cauchy Integral Formula, where the circle is traversed once anti-clockwise.

My attempt

z = 3, z = 2 are both singularities in the contour, so:

Rewrite the integrand such that one of the singularities is in the denominator:

$\displaystyle \frac{\frac{z^3}{(z - 3)^3}}{(z - 2)^2}$

Thus by CIF, $\displaystyle f(z) = \frac{z^3}{(z - 3)^3}$

This singularity is of pole 2, so we must use $\displaystyle \frac{2\pi i}{1!} f^{1}(2)$:

$\displaystyle 2\pi i (\frac{3z^3(z - 3)^2) - 3z^2(z - 3)^3}{(z - 3)^6})$

= 72\pi i

Now for the other singularity:

$\displaystyle \frac{\frac{z^3}{(z - 2)^2}}{(z - 3)^3}$

Thus by CIF, $\displaystyle f(z) = \frac{z^3}{(z - 2)^2}$

This singularity is of pole 3, so we must use $\displaystyle \frac{2\pi i}{2!} f^{2}(3)$:

$\displaystyle \pi i(\frac{6z^3-6z^2(z - 2) - 6z^2(z - 2) + 6z(z - 2)^2}{(z - 2)^4})$

= $\displaystyle 72\pi i$

Add them together, and we get $\displaystyle 144\pi i$. However, the answer is actually 0. I think I'm supposed to be getting $\displaystyle -72\pi i$ for the second one, but I can't work out where I've gone wrong. :/

Any help would be greatly appreciated!