Line Integral in Complex field

Question

I am trying to compute

$\displaystyle\oint_{|z| = 1/2} \dfrac{z^2+1}{3z^3+z^2} \, dz$ in the positive orientation

I get $14 \pi i $ by the residue theorem, but in my text the answer is $\displaystyle{2\pi i\over 3}$. Thanks for your answers!

Answer 1

I'm not sure how you got $14\pi i$. If you show your work, we might be able to spot the error.

Here is one way to get the answer. Using partial fractions, $\dfrac{z^2+1}{3z^3+z^2} = \dfrac{-3}{z}+\dfrac{1}{z^2}+\dfrac{\tfrac{10}{3}}{z+\tfrac{1}{3}}$. Hence, \begin{align*}\oint_{|z| = 1/2}\dfrac{z^2+1}{3z^3+z^2}\,dz & = \oint_{|z| = 1/2}\left(\dfrac{-3}{z}+\dfrac{1}{z^2}+\dfrac{\tfrac{10}{3}}{z+\tfrac{1}{3}}\right)\,dz \\ &= \oint_{|z| = 1/2}\dfrac{-3}{z}\,dz + \oint_{|z| = 1/2}\dfrac{1}{z^2}\,dz + \oint_{|z| = 1/2}\dfrac{\tfrac{10}{3}}{z+\tfrac{1}{3}}\,dz \\ &= -3 \cdot 2\pi i + 0 + \dfrac{10}{3} \cdot 2\pi i \\ &= \dfrac{2\pi i}{3}.\end{align*}