Complex Modulus of Product of Complex Numbers/Proof 3
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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:
- $z_1 = r_1 \paren {\cos \theta_1 + i \sin \theta_1}$
- $z_2 = r_2 \paren {\cos \theta_2 + i \sin \theta_2}$
Then:
\(\ds \cmod {z_1 z_2}\) | \(=\) | \(\ds \cmod {r_1 \paren {\cos \theta_1 + i \sin \theta_1} r_2 \paren {\cos \theta_2 + i \sin \theta_2} }\) | Definition of Polar Form of Complex Number | |||||||||||
\(\ds \) | \(=\) | \(\ds \cmod {r_1 r_2 \paren {\map \cos {\theta_1 + \theta_2} + i \map \sin {\theta_1 + \theta_2} } }\) | Product of Complex Numbers in Polar Form | |||||||||||
\(\ds \) | \(=\) | \(\ds r_1 r_2\) | Definition of Polar Form of Complex Number | |||||||||||
\(\ds \) | \(=\) | \(\ds \cmod {z_1} \cmod {z_2}\) | Definition of Polar Form of Complex Number |
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 2$. Geometrical Representations