# How do you simplify (6+i)/(6-i)?

Jan 25, 2016

$z = \frac{6 + i}{6 - i} = \frac{35 + 12 i}{37} = \frac{35}{37} + \frac{12}{37} i$

#### Explanation:

Given:

$z = {z}_{1} / {z}_{2}$

with ${z}_{1} , {z}_{2} \in \mathbb{C}$

to semplify $z$ you have to moltiplicate both numerator and denominator by $\overline{{z}_{2}}$, the complex conjugate of ${z}_{2}$

$\therefore$ if ${z}_{2} = a + i b \implies \overline{{z}_{2}} = a - i b$

$\therefore z = \frac{6 + i}{6 - i} \cdot \frac{6 + i}{6 + i} = {\left(6 + i\right)}^{2} / \left(36 - {i}^{2}\right)$

Remebering that $i = \sqrt{- 1} \implies {i}^{2} = - 1$

${\left(6 + i\right)}^{2} / \left(36 - {i}^{2}\right) = \frac{36 + 12 i + {i}^{2}}{36 - \left(- 1\right)} = \frac{36 - 1 + 12 i}{37} = \frac{35 + 12 i}{37} =$
$= \frac{35}{37} + \frac{12}{37} i$