Complex Modulus of Product of Complex Numbers/Proof 2
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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: $4 \ \text{(b)}$