Cube Root of Unity if Modulus is 1 and Real Part is Minus Half

Theorem

Let $z \in \C$ be a complex number such that:

$\cmod z = 1$

$\Re \paren z = -\dfrac 1 2$

where:

$\cmod z$ denotes the complex modulus of $z$

$\Re \paren z$ denotes the real part of $z$.

Then:

$z^3 = 1$

Proof

Let $z = x + i y$.

From $\Re \paren z = -\dfrac 1 2$:

$x = -\dfrac 1 2$

by definition of the real part of $z$.

Then:

\(\ds \cmod z\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds x^2 + y^2\)

\(=\)

\(\ds 1\)

Definition of Complex Modulus

\(\ds \leadsto \ \ \)

\(\ds \paren {-\dfrac 1 2}^2 + y^2\)

\(=\)

\(\ds 1\)

substituting for $x$

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds \sqrt {1 - \dfrac 1 4}\)

simplifying

\(\ds \)

\(=\)

\(\ds \pm \dfrac {\sqrt 3} 2\)

Thus:

$z = -\dfrac 1 2 \pm \dfrac {\sqrt 3} 2$

and the result follows from Cube Roots of Unity.

$\blacksquare$

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 3$. Roots of Unity: Exercise $4$.