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
\(\ds \) \(=\) \(\ds \sqrt {\paren {x_1^2 x_2^2 + y_1^2 y_2^2 - 2 x_1 x_2 y_1 y_2} + \paren {x_1^2 y_2^2 + x_2^2 y_1^2 + 2 x_1 x_2 y_1 y_2} }\)
\(\ds \) \(=\) \(\ds \sqrt {x_1^2 x_2^2 + y_1^2 y_2^2 + x_1^2 y_2^2 + x_2^2 y_1^2}\)


\(\ds \cmod {z_1} \cdot \cmod {z_2}\) \(=\) \(\ds \sqrt {x_1^2 + y_1^2} \sqrt {x_2^2 + y_2^2}\)
\(\ds \) \(=\) \(\ds \sqrt {\paren {x_1^2 + y_1^2} \paren {x_2^2 + y_2^2} }\)
\(\ds \) \(=\) \(\ds \sqrt {x_1^2 x_2^2 + y_1^2 y_2^2 + x_1^2 y_2^2 + x_2^2 y_1^2}\)

$\blacksquare$


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