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

1 Answer
Lotusbluete
Feb 3, 2016

#- 1 - i#

Explanation:

Your denominator is #-2 + i#.

You should take the complex conjugate of the denominator #(-2-i)# and expand your fraction with it:

# (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 #i^2 = -1#...

#= (-6 - 5i + 1)/(4 + 1)#

#= (-5 - 5i)/5#

#= -5/5 - 5/5 i#

#= - 1 - i#

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