How do you divide #-2-2i div 5-2i#?
1 Answer
Feb 1, 2016
# -6/29 - 14/29 i#
Explanation:
When dividing complex numbers ,multiply numerator and
denominator by the complex conjugate of the denominator.
If a + bi is a complex number then
#color(red)(" a - bi is conjugate")# This ensures the denominator is real.
as
# (a+bi)(a-bi) = a^2 + b^2 color(black)( " which is real")# [Note:
# i^2 =( sqrt-1 )^2 = -1 ]# the conjugate of 5 - 2i is 5 + 2i .
# rArr( (-2-2i)(5+2i))/((5-2i)(5+2i)) # ( distribute to obtain)
#(-10-14i-4i^2)/(25-4i^2) #
# =( -6-14i)/29 = -6/29 -14/29 i#
