How do you simplify #(3+i)/ (-2+i)#?
1 Answer
Feb 3, 2016
Explanation:
Your denominator is
You should take the complex conjugate of the denominator
# (3+i)/(-2+i) = ((3+i) * color(blue)((-2-i)))/((-2+i) * color(blue)((-2-i)))#
#= (-6 -2i - 3i - i^2) / ((-2)^2 - i^2)#
#= (-6 - 5i - i^2)/(4 - i^2)#
... remember that
#= (-6 - 5i + 1)/(4 + 1)#
#= (-5 - 5i)/5#
#= -5/5 - 5/5 i#
#= - 1 - i#