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:

\(\displaystyle \cmod {z_1 z_2}\) \(=\) \(\displaystyle \sqrt {\paren {z_1 z_2} \overline {\paren {z_1 z_2} } }\) Modulus in Terms of Conjugate
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {z_1 \overline {z_1} z_2 \overline {z_2} }\) Product of Complex Conjugates, Complex Multiplication is Commutative
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {z_1 \overline {z_1} } \sqrt {z_2 \overline {z_2} }\) Power of Product
\(\displaystyle \) \(=\) \(\displaystyle \cmod {z_1} \cmod {z_2}\)

$\blacksquare$


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