# How do you simplify 6/ (2+3i)?

Jan 10, 2016

$\frac{12}{13} - \frac{18}{13} i$

#### Explanation:

To divide this expression we attempt to rationalise the denominator ie. We attempt to make it an integer value. To do this we use the following :

Given the complex number a + bi then the complex conjugate is
a - bi

note the following for multiplying

(a + bi )(a - bi )
$= {a}^{2} + a b i - a b i - {b}^{2} {i}^{2}$
= a^2 + b^2 ( i =sqrt(-1 )then(( sqrt(-1))^2 = - 1 )
now ${a}^{2} + {b}^{2}$ is a real number

we now multiply the numerator and denominator by (2 - 3i )

$\Rightarrow \frac{6}{2 + 3 i} . \frac{2 - 3 i}{2 - 3 i}$

$= \frac{6 \left(2 - 3 i\right)}{\left(2 + 3 i\right) \left(2 - 3 i\right)}$

multiply out the brackets gives:

$\frac{12 - 18 i}{4 + 6 i - 6 i - 9 {i}^{2}} = \frac{12 - 18 i}{4 + 9} = \frac{12 - 18 i}{13}$

rewriting in the form a + bi , we get that

$\frac{6}{2 + 3 i} = \frac{12}{13} - \frac{18 i}{13}$