Complex Modulus of Product of Complex Numbers/Proof 2

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 $\overline z$ denote the complex conjugate of $z$.

Then:

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

\(=\)

\(\ds \sqrt {\paren {z_1 z_2} \overline {\paren {z_1 z_2} } }\)

Modulus in Terms of Conjugate

\(\ds \)

\(=\)

\(\ds \sqrt {z_1 \overline {z_1} z_2 \overline {z_2} }\)

Product of Complex Conjugates, Complex Multiplication is Commutative

\(\ds \)

\(=\)

\(\ds \sqrt {z_1 \overline {z_1} } \sqrt {z_2 \overline {z_2} }\)

Power of Product

\(\ds \)

\(=\)

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

$\blacksquare$

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 1.2$. The Algebraic Theory: $(1.10)$
• 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Subgroups
• 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Solved Problems: Fundamental Operations with Complex Numbers: $4 \ \text{(b)}$