Square of Complex Modulus equals Complex Modulus of Square

Theorem

Let $z \in \C$ be a complex number.

Let $\cmod z$ be the modulus of $z$.

Then:

$\cmod {z^2} = \cmod z^2$

Proof

From Complex Modulus of Product of Complex Numbers:

$\cmod {z_1 z_2} = \cmod {z_1} \cmod {z_2}$

for $z_1, z_2 \in \C$.

Set $z = z_1 = z_2$ and the result follows.

$\blacksquare$

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 1.2$. The Algebraic Theory