# How do you divide  (6-i) / (9-4i) ?

Jun 1, 2016

$\frac{58}{97} + \frac{15}{97} i$

#### Explanation:

To divide a complex number by another , we require to $\textcolor{b l u e}{\text{rationalise the denominator}}$ ie. make it a real value.

This is achieved by mutipling the numerator and denominator by the $\textcolor{b l u e}{\text{complex conjugate}}$ of the denominator.

Given a complex number z = x ± iy then the conjugate is x ∓ iy

Note that the values of x and y remain unchanged but the 'sign' changes.

$\Rightarrow \left(x + i y\right) \left(x - i y\right) = {x}^{2} + {y}^{2} \text{ a real number}$

We are also making use of the fact $\left[{i}^{2} = {\left(\sqrt{- 1}\right)}^{2} = - 1\right]$

$\Rightarrow \frac{6 - i}{9 - 4 i} = \frac{6 - i}{9 - 4 i} \times \frac{9 + 4 i}{9 + 4 i} = \frac{\left(6 - i\right) \left(9 + 4 i\right)}{\left(9 - 4 i\right) \left(9 + 4 i\right)}$

now expanding numerator/denominator

$\Rightarrow \frac{54 + 15 i - 4 {i}^{2}}{81 - 16 {i}^{2}} = \frac{58 + 15 i}{81 + 16} = \frac{58}{97} + \frac{15}{97} i$