Inequality contour integral

Question

Given a curve $\gamma:[- \frac{\pi}{6}, \frac{\pi}{6}] \rightarrow \mathbb{C}$ and $\mathop{\gamma}\,(t):=2\exp(it)$, I want to show that $$\left\vert\,\int_{\gamma} \frac{1}{z^3+1} \mathrm{d}z\, \right| \le \frac{\pi}{12}$$

Answer 1

Since

$$z=2e^{it}\;,\;\;-\frac\pi 6\le t\le\frac \pi 6\implies |z^3+1|=|8e^{3it}+1|=|(8\cos 3t+1)+8i\sin3t|=$$

$$\sqrt{64\cos^23t+16\cos 3t+1+64\sin^23t}=\sqrt{65+16\cos3t}\ge\sqrt{65+16\cos 3\frac\pi 6}=\sqrt{65}\ge 8$$

we get from Cauchy's Estimates:

$$\left|\;\int\limits_\gamma\frac{dz}{z^3+1}\;\right|\le\frac18\cdot\frac{4\pi}6=\frac{\pi}{12}$$