Complex Modulus of Product of Complex Numbers/Proof 3

Theorem

Let $z_1, z_2 \in \C$ be complex numbers.

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

Then:

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

Proof

Let:

$z_1 = r_1 \paren {\cos \theta_1 + i \sin \theta_1}$

$z_2 = r_2 \paren {\cos \theta_2 + i \sin \theta_2}$

Then:

\(\ds \cmod {z_1 z_2}\)

\(=\)

\(\ds \cmod {r_1 \paren {\cos \theta_1 + i \sin \theta_1} r_2 \paren {\cos \theta_2 + i \sin \theta_2} }\)

Definition of Polar Form of Complex Number

\(\ds \)

\(=\)

\(\ds \cmod {r_1 r_2 \paren {\map \cos {\theta_1 + \theta_2} + i \map \sin {\theta_1 + \theta_2} } }\)

Product of Complex Numbers in Polar Form

\(\ds \)

\(=\)

\(\ds r_1 r_2\)

Definition of Polar Form of Complex Number

\(\ds \)

\(=\)

\(\ds \cmod {z_1} \cmod {z_2}\)

Definition of Polar Form of Complex Number

$\blacksquare$

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 2$. Geometrical Representations