I have $\int_\delta \frac{z}{z^3 -1} dz$, where $\delta(t) = (\frac{1}{2})e^{it }$ with $t \in [0, 2\pi]$. It's clear that the winding number is $Ind_\delta(z_0) = 1$, where $z_0 = 0$. I'm just unsure of how to rewrite the integral so that I can use Cauchy's integral formula.
2026-04-03 19:07:43.1775243263
Contour Integral About A Circle
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By Cauchy's Integral theorem:
$$\int\limits_\delta\frac z{z^3-1}dz=0\;\;\text{since}\;\;\frac z{z^3-1}\;\;\text{analytic on}\;\delta\;\;\text{and within it}$$