Compute $$\int_{\gamma} \frac{z}{z^3-1} dz$$ where $\gamma(t)=2e^{it}$, $t\in[0,2\pi]$.
The first part of the problem had me compute the same integral over the path $\gamma(t)=\frac{1}{2} e^{it}$. I used Cauchy's Integral Formula with $z_0 =1$ and $f(z)=\frac{1}{z^2+z+1}$, which is analytic on $\mathbb{C}\smallsetminus \{-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i\}$. However, given that $\gamma$, we could easily find an open convex $U$ such that $\gamma\in U$ and $f(z)$ is analytic on $U$. Thus, I obtained the result $0$.
For this part, we cannot find an open convex $U$ such that $\gamma\in U$ and $f(z)$ is analytic on all of $U$, since the singularities are contained in the path circle. I believe I need to separate the region into multiple parts, but I'm not sure how to go about doing it.
Any help would be greatly appreciated.
$$\int_\gamma \dfrac{z}{z^3-1}dz=\int_\gamma \dfrac{z}{(z-1)(z-z_1)(z-z_2)}dz,$$
with $z_1=-\tfrac{1}{2}+\tfrac{\sqrt 3}{2}i$ and $z_2=-\tfrac{1}{2}-\tfrac{\sqrt 3}{2}i$. Now:
$$\int_\gamma \dfrac{z}{(z-1)(z-z_1)(z-z_2)}dz= \int_{\gamma_1} \dfrac{z/\{(z-z_1)(z-z_2)\}}{(z-1)}dz + \int_{\gamma_2} \dfrac{z/\{(z-1)(z-z_1)\}}{(z-z_2)}dz +\int_{\gamma_3} \dfrac{z/\{(z-1)(z-z_2)\}}{(z-z_1)}dz, $$
with $\gamma_1$ a circle around the pole $z=1$, $\gamma_2$ around $z_2$ and $\gamma_3$ around $z_1$. Using Cauchy's Integral Formula:
$$\int_{\gamma_1} \dfrac{z/\{(z-z_1)(z-z_2)\}}{(z-1)}dz = 2\pi i \dfrac{1}{(1-z_1)(1-z_2)},$$ $$\int_{\gamma_2} \dfrac{z/\{(z-1)(z-z_1)\}}{(z-z_2)}dz = 2\pi i \dfrac{z_2}{(z_2-1)(z_2-z_1)},$$ $$\int_{\gamma_3} \dfrac{z/\{(z-1)(z-z_2)\}}{(z-z_1)}dz= 2\pi i \dfrac{z_1}{(z_1-1)(z_1-z_2)}. $$
I leave the calculations to you.