Complex Modulus of Product of Complex Numbers/Proof 1
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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 $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$, where $x_1, y_1, x_2, y_2 \in \R$.
\(\ds \cmod {z_1 z_2}\) | \(=\) | \(\ds \sqrt {\paren {x_1 x_2 - y_1 y_2}^2 + \paren {x_1 y_2 + x_2 y_1}^2}\) | Definition of Complex Modulus, Definition of Complex Multiplication | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\paren {x_1^2 x_2^2 + y_1^2 y_2^2 - 2 x_1 x_2 y_1 y_2} + \paren {x_1^2 y_2^2 + x_2^2 y_1^2 + 2 x_1 x_2 y_1 y_2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {x_1^2 x_2^2 + y_1^2 y_2^2 + x_1^2 y_2^2 + x_2^2 y_1^2}\) |
\(\ds \cmod {z_1} \cdot \cmod {z_2}\) | \(=\) | \(\ds \sqrt {x_1^2 + y_1^2} \sqrt {x_2^2 + y_2^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\paren {x_1^2 + y_1^2} \paren {x_2^2 + y_2^2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {x_1^2 x_2^2 + y_1^2 y_2^2 + x_1^2 y_2^2 + x_2^2 y_1^2}\) |
$\blacksquare$
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Solved Problems: Fundamental Operations with Complex Numbers: $4 \ \text{(b)}$