$z = a + bi$

$\Rightarrow |z| = \sqrt{a^2 + b^2} \quad [1]$

$-z = -a - bi$

$\Rightarrow |-z| = \sqrt{(-a)^2 + (-b)^2}$

$|-z| = \sqrt{a^2 + b^2} \quad [2]$

$\therefore |z| = |-z| \quad \text{from [1] and [2])}$

$z = a + bi$

$\Rightarrow |z| = \sqrt{a^2 + b^2} \quad [1]$

$\overline{z} = a - bi$

$\Rightarrow |\overline{z}| = \sqrt{a^2 + (-b)^2}$

$|\overline{z}|= \sqrt{a^2 + b^2} \quad [2]$

$\therefore |z| = |\overline{z}| \quad \text{from [1] and [2]}$

$z = a + bi$

$\overline{z} = a - bi$

$z \cdot \overline{z} = (a + bi)(a - bi)$

$z \cdot \overline{z}= a^2 - b^2i^2$

$z \cdot \overline{z}= a^2 + b^2 \quad [1]$

$\Rightarrow |z| = \sqrt{a^2 + b^2} \quad [1]$

Squaring both sides

$\Rightarrow |z|^2 = a^2 + b^2 \quad [2]$

$\therefore z \cdot \overline{z} = |z|^2 \quad \text{from [1] and [2]}$

$z_1 = a + bi, \quad z_2 = c + di$

Let's find $\dfrac{z_1}{z_2}$

$\dfrac{z_1}{z_2} = \dfrac{a+bi}{c+di}$

$= \dfrac{a + bi}{c + di} \cdot \dfrac{c - di}{c - di}$

$= \dfrac{(a+bi)(c-di)}{(c+di)(c-di)}$

$= \dfrac{ac - bdi^2 + bci - adi}{c^2 - d^2i^2}$

$= \dfrac{(ac + bd) + (bc - ad)i}{c^2 + d^2}$

$= \dfrac{ac + bd}{c^2+d^2} + i\dfrac{bc - ad}{c^2 + d^2}$

$L.H.S = \overline{\left(\dfrac{z_1}{z_2}\right)}$

$= \overline{\dfrac{ac + bd}{c^2+d^2} + i\dfrac{bc - ad}{c^2 + d^2}}$

$= \dfrac{ac + bd}{c^2+d^2} - i\dfrac{bc - ad}{c^2 + d^2} = L.H.S \quad \quad [1]$

Moving to the right hand side.

$R.H.S = \dfrac{\overline{z_1}}{\overline{z_2}}$

$= \dfrac{a-bi}{c-di}$

$= \dfrac{a-bi}{c-di} \cdot \dfrac{c+di}{c+di}$

$= \dfrac{(a-bi)(c+di)}{(c-di)(c+di)}$

$= \dfrac{ac - bdi^2 +adi - bci}{c^2 - d^2i^2}$

$= \dfrac{(ac + bd) +i(ad - bc)}{c^2 + d^2}$

$= \dfrac{ac + bd}{c^2+d^2} + i\dfrac{ad- bc}{c^2 + d^2}$

$= \dfrac{ac + bd}{c^2+d^2} - i\dfrac{bc - ad}{c^2 + d^2} = R.H.S \quad [2]$

From $[1]$ and $[2]$

$\Rightarrow \overline{\left(\dfrac{z_1}{z_2}\right)}= \dfrac{\overline{z_1}}{\overline{z_2}}$

$z_1 = a + bi, \quad z_2 = c + di$

We want to prove that $|z_1 z_2| = |z_1| |z_2|$

$L.H.S = |z_1 z_2|$

First, let's find $z_1 z_2 = (a + bi)(c + di)$

$= ac + adi + bci + bdi^2$

$= ac + adi + bci - bd$

$= (ac - bd) + (ad + bc)i$

$|z_1 z_2| = |(ac - bd) + (ad + bc)i|$

$= \sqrt{(ac - bd)^2 + (ad + bc)^2}$

$= \sqrt{a^2c^2 - 2abcd + b^2d^2 + a^2d^2 + 2abcd + b^2c^2}$

$= \sqrt{a^2c^2 + b^2d^2 + a^2d^2 + b^2c^2}$

$= \sqrt{a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2}$

$= \sqrt{a^2(c^2 + d^2) + b^2(c^2 + d^2)}$

$= \sqrt{(a^2 + b^2)(c^2 + d^2)}$

$= \sqrt{a^2 + b^2} \cdot \sqrt{c^2 + d^2} = L.H.S \quad [1]$

Now let's evaluate the right hand side.

$R.H.S = |z_1| |z_2|$

$= |a + bi| |c + di|$

$= \sqrt{a^2 + b^2} \cdot \sqrt{c^2 + d^2}$

$= \sqrt{a^2 + b^2} \cdot \sqrt{c^2 + d^2} = R.H.S \quad [2]$

From $[1]$ and $[2]$

$\Rightarrow |z_1 z_2| = |z_1| |z_2|$

This completes the proof that the modulus of a product equals the product of the moduli.