Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.4 De Moivre's Theorem: Powers and Roots of Complex Numbers - 8.4 Exercises - Page 378: 61

If $z$ is an n-th root of 1, then $\frac{1}{z}$ is also an n-th root of 1

Suppose that $z$ is an n-th root of 1. Then $z^n=1$ We can find the value of $(\frac{1}{z})^n$: $(\frac{1}{z})^n = \frac{1^n}{z^n} = \frac{1}{1} = 1$ Since $(\frac{1}{z})^n = 1$, then $\frac{1}{z}$ is also an n-th root of 1