# How do you simplify ((5+i)(4-3i)) / ((2-5i)(1-i))?

Feb 15, 2016

$\frac{4}{29} + \frac{97}{29} i$

#### Explanation:

making use of the following facts related to complex numbers.

• If a + bi , is a complex number then a - bi , is it's conjugate.

•[ i^2 = (sqrt-1)^2 = -1 ]

first step in this question is to distribute the brackets using FOIL ( or any method that you use ).

numerator : $\left(5 + i\right) \left(4 - 3 i\right) = 20 - 15 i + 4 i - 3 {i}^{2} = 20 - 11 i + 3 = 23 - 11 i$

denominator :$\left(2 - 5 i\right) \left(1 - i\right) = 2 - 2 i - 5 i + 5 {i}^{2} = 2 - 7 i - 5$
=-3-7i

hence $\frac{23 - 11 i}{- 3 - 7 i}$
Require the denominator to be real so multiply numerator and denominator by the conjugate of (-3-7i) , which is (-3+7i)

$\Rightarrow \frac{\left(23 - 11 i\right) \left(- 3 + 7 i\right)}{\left(- 3 - 7 i\right) \left(- 3 + 7 i\right)}$

$= \frac{- 69 + 161 i + 33 i - 77 {i}^{2}}{9 - 21 i + 21 i - 49 {i}^{2}} = \frac{- 69 + 194 i + 77}{9 + 49}$

$= \frac{8 + 194 i}{58} = \frac{4}{29} + \frac{97}{29} i$