Complex Modulus of Sum of Complex Numbers
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Theorem
Let $z_1, z_2 \in \C$ be complex numbers.
Let $\theta_1$ and $\theta_2$ be arguments of $z_1$ and $z_2$, respectively.
Then:
- $\cmod {z_1 + z_2}^2 = \cmod {z_1}^2 + \cmod {z_2}^2 + 2 \cmod {z_1} \cmod {z_2} \, \map \cos {\theta_1 - \theta_2}$
Proof 1
We have:
| \(\ds \cmod {z_1 + z_2}^2\) | \(=\) | \(\ds \paren {z_1 + z_2} \paren {\overline {z_1} + \overline {z_2} }\) | Modulus in Terms of Conjugate and Sum of Complex Conjugates | |||||||||||
| \(\ds \) | \(=\) | \(\ds z_1 \overline {z_1} + z_2 \overline {z_2} + z_1\overline {z_2} + \overline {z_1} z_2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \cmod {z_1}^2 + \cmod {z_2}^2 + 2 \, \map \Re {z_1 \overline {z_2} }\) | Modulus in Terms of Conjugate and Sum of Complex Number with Conjugate | |||||||||||
| \(\ds \) | \(=\) | \(\ds \cmod {z_1}^2 + \cmod {z_2}^2 + 2 \, \cmod {z_1} \cmod {z_2} \map \cos {\theta_1 - \theta_2}\) | Product of Complex Numbers in Polar Form and Argument of Complex Conjugate equals Negative of Argument |
$\blacksquare$
Proof 2
| \(\ds \cmod {z_1 + z_2}^2\) | \(=\) | \(\ds \paren {\cmod {z_1} \cos \theta_1 + \cmod {z_2} \cos \theta_2}^2 + \paren {\cmod {z_1} \sin \theta_1 + \cmod {z_2} \sin \theta_2}^2\) | Definition of Complex Modulus | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \cmod {z_1} \cmod {z_2} \cos \theta_1 \cos \theta_2 + \cmod {z_1}^2 \cos^2 \theta_1 + \cmod {z_2}^2 \cos^2 \theta_2\) | ||||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds 2 \cmod {z_1} \cmod {z_2} \sin \theta_1 \sin \theta_2 + \cmod {z_1}^2 \sin^2 \theta_1 + \cmod {z_2}^2 \sin^2 \theta_2\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \cmod {z_1} \cmod {z_2} \paren {\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2} + \cmod {z_1}^2 + \cmod {z_2}^2\) | Sum of Squares of Sine and Cosine | |||||||||||
| \(\ds \) | \(=\) | \(\ds \cmod {z_1}^2 + \cmod {z_2}^2 + 2 \cmod {z_1} \cmod {z_2} \, \map \cos {\theta_1 - \theta_2}\) | Cosine of Difference |
$\blacksquare$