Consider a triangle on the complex plane, these three points are $z=1,\omega,\bar{\omega}$, here $\omega=e^{\frac{2}{3}\pi i}$. The boundary of triangle is $\partial T$, whose direction is counterclockwise. Find the Cauchy principal value of the complex integration: \begin{equation} \text{P.V.}\int_{\partial T}\frac{z^3}{z-\omega}\mathrm{d}z \end{equation} I have solved this integral but I'm not sure it is right or not. Here's my solution: I looked up the definition of Cauchy principal value of contour integral, which is defined as \begin{equation} \text{P.V.}\int_{C}f(z)\mathrm{d}z=\lim_{\epsilon\to0+}\int_{C\cup B_{\epsilon}(\omega)}f(z)\mathrm{d}z \end{equation} Here $C$ is the contour including the singularity point and $B_{\epsilon}(\omega)$ is the circle centered at $\omega$ with radius $\epsilon>0$. I made a contour $C$ including $z=\omega$, where at this point its boundary is part of circle $B_{\epsilon}(\omega)$, i.e. the expression is $z=\omega+\epsilon e^{i\theta},\theta\in[-\frac{\pi}{6},\frac{3\pi}{2}]$. Then the rest of $C$ is as same as the rest of $\partial T$.
Applying residue's theorem, the integral \begin{equation} \int_{C}\frac{z^3}{z-\omega}\mathrm{d}z=2\pi i\cdot\text{Res}(f(z),\omega)=2\pi i \end{equation} And the integral of the curve \begin{equation} \int_{\text{part of }B_{\epsilon}(\omega)}\frac{z^3}{z-\omega}\mathrm{d}z=i(\frac{3\pi}{2}-(-\frac{\pi}{6}))\cdot\lim_{z\to\omega}(z-\omega)f(z)=\frac{5\pi}{3}i \end{equation} Let $\epsilon\to0+$, the answer is \begin{equation} \text{P.V.}\int_{\partial T}\frac{z^3}{z-\omega}\mathrm{d}z=2\pi i-\frac{5\pi}{3}i=\frac{\pi}{3}i \end{equation}
Is it feasible to solve this integral like that?