IMC 2013 · Day 2 · P6

Since $z^3 + 1 = (z + 1)(z^2 - z + 1)$, it suffices to prove that $|z^2 - z + 1| \ge \frac{1}{2}$.

Assume that $z + 1 = r e^{\varphi i}$, where $r = |z + 1| > 2$, and $\varphi = \arg(z + 1)$ is some real number. Then $$z^2 - z + 1 = (r e^{\varphi i} - 1)^2 - (r e^{\varphi i} - 1) + 1 = r^2 e^{2\varphi i} - 3 r e^{\varphi i} + 3,$$ and $$\begin{align*} |z^2 - z + 1|^2 &= \bigl( r^2 e^{2\varphi i} - 3 r e^{\varphi i} + 3 \bigr) \bigl( r^2 e^{-2\varphi i} - 3 r e^{-\varphi i} + 3 \bigr) = \\ &= r^4 + 9 r^2 + 9 - (6 r^3 + 18 r) \cos\varphi + 6 r^2 \cos 2\varphi = \\ &= r^4 + 9 r^2 + 9 - (6 r^3 + 18 r) \cos\varphi + 6 r^2 (2 \cos^2 \varphi - 1) = \\ &= 12 \left( r \cos\varphi - \frac{r^2 + 3}{4} \right)^2 + \frac{1}{4} (r^2 - 3)^2 > 0 + \frac{1}{4} = \frac{1}{4}. \end{align*}$$ This finishes the proof.