Problem:
Evaluate $\int_{\Gamma} \dfrac{z}{(z+2)(z-1)}dz$ , where $\Gamma$ traverses the circle $|z| = 4$ twice in the clockwise direction.
As this is supposed to precede my knowledge of Cauchy's Integral Formula, I proceed using partial fractions and Cauchy's Integral Theorem.
$$\dfrac{z}{(z+2)(z-1)}=\dfrac{2}{3(z+2)} + \dfrac{1}{3(z-1)} $$ by partial fraction decomposition, so the integral can be split into $$\dfrac{2}{3}\int_{\Gamma} \dfrac{1}{z+2}dz + \dfrac{1}{3} \int_{\Gamma} \dfrac{1}{z-1}dz$$
Now each of these integrals would be equal to $2 \pi i$ - if we were traversing once with positive orientation - as the simple poles lie within $\Gamma$ (right?).
But traversing twice gives a factor of $(2)$, and the negative orientation gives a factor of $(-1)$, so we have $$\dfrac{2}{3}(-4 \pi i) + \dfrac{1}{3}(-4 \pi i) = -4\pi i$$
Is this correct?
This has been resolved. Thanks.