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 #