Find all complex numbers $z$ such that $z^6=1$ and $Re(z)>0$. List them.

Step-by-step solution

1. The solutions of $z^6=1$ are the sixth roots of unity: $z_k=e^{,i\frac{2\pi k}{6}}=e^{,i\frac{\pi k}{3}}$ for $k=0,1,\dots ,5$.
2. Compute the six roots: \begin{aligned}k=0&: 1\k=1&: cos\frac{pi}{3}+isin\frac{pi}{3}=\frac12+\frac{sqrt3}{2}i\k=2&: cos\frac{2pi}{3}+isin\frac{2pi}{3}=-\frac12+\frac{sqrt3}{2}i\k=3&: -1\k=4&: -\frac12-\frac{sqrt3}{2}i\k=5&: \frac12-\frac{sqrt3}{2}i\end{aligned}
3. Determine the real part of each root: $1$ has $Re=1$, the roots for $k=1$ and $k=5$ have $Re=\frac{1}{2}>0$, all other roots have non‑positive real part.
4. Hence the roots satisfying $Re(z)>0$ are $1,;\frac{1}{2}+\frac{\sqrt{3}}{2}i,;\frac{1}{2}-\frac{\sqrt{3}}{2}i$.

Concept tested: Roots of unity and real part condition

Published 16 October 2026 · How this solution was prepared and checked