Complex Modulus of Sum of Complex Numbers/Proof 2

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:

$\cmod {z_1 + z_2}^2 = \cmod {z_1}^2 + \cmod {z_2}^2 + 2 \cmod {z_1} \cmod {z_2} \, \map \cos {\theta_1 - \theta_2}$

$\ds \cmod {z_1 + z_2}^2$

$=$

$\ds \paren {\cmod {z_1} \cos \theta_1 + \cmod {z_2} \cos \theta_2}^2 + \paren {\cmod {z_1} \sin \theta_1 + \cmod {z_2} \sin \theta_2}^2$

$\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 $

$=$

$\ds 2 \cmod {z_1} \cmod {z_2} \paren {\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2} + \cmod {z_1}^2 + \cmod {z_2}^2$

$\ds $

$=$

$\ds \cmod {z_1}^2 + \cmod {z_2}^2 + 2 \cmod {z_1} \cmod {z_2} \, \map \cos {\theta_1 - \theta_2}$

$\blacksquare$