I'm trying to compute $$\int_\gamma \frac{z^4+z^2+1}{z^3-1} \, dz$$ where $\gamma$ is the circle $|z-i|=1$, using Cauchy's integral formula
My solution is as follows:
The integral can be rewritten as $$\int_\gamma \frac{\frac{z^4+z^2+1}{(z-1)\left(z+\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)}}{z+\frac{1}{2}-\frac{\sqrt{3}}{2}i} \, dz = I$$
Then Cauchy's integral formula can be applied, giving $$I=2\pi if\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)=0$$
Is this correct?