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

##### 1 Answer

Feb 10, 2016

# 4/5 - 2/5 i #

#### Explanation:

To simplify , require the denominator to be real. To achieve this multiply the numerator and denominator by the 'complex conjugate' of the denominator.

If (a + bi ) is a complex number then (a - bi ) is the conjugate.

Note that : (a + bi )(a - bi ) =

#a^2 + b^2color(black)(" which is real ")# here the conjugate of (4 - 3i ) is (4 + 3i )

hence :

# ((2 - 4i)(4 + 3i))/((4 -3i)(4 + 3i )) # distribute using FOIL

# = (8+6i-16i -12i^2 )/(16 + 12i - 12i - 9i^2 ) # [noting that:

#i^2 =(sqrt-1)^2 = -1 # ]

# = (8 - 10i + 12)/(16 + 9 ) = (20 - 10i )/25 = 20/25 - 10/25 i = 4/5 - 2/5 i #