Question

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

Answer 1

`4/29 + 97/29i`

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 : `(5+i)(4-3i) = 20-15i+4i-3i^2 =20-11i +3 = 23-11i`

denominator :`(2-5i)(1-i) = 2-2i-5i+5i^2=2-7i-5`
=-3-7i

hence `(23-11i)/(-3-7i)`
Require the denominator to be real so multiply numerator and denominator by the conjugate of (-3-7i) , which is (-3+7i)

`rArr( (23-11i)(-3+7i))/((-3-7i)(-3+7i))`

`= (-69+161i+33i-77i^2)/(9-21i+21i-49i^2) = (-69+194i+77)/(9+49)`

`=(8+194i)/58 = 4/29 + 97/29i`