How do you simplify #(1+2i)/(2-3i)#?

1 Answer
Jan 12, 2016

Answer:

#(1+2i)/(2-3i)=(-4+7i)/13=-4/13+7/13i#

Explanation:

  1. Find the complex coniugate of denominator

denominator: #z=2-3i#

denominator complex coniugate: #bar(z)=2color(red)+3i#

  1. Multiply both numerator and denominator for the complex coniugate

#(1+2i)/(2-3i)*(2+3i)/(2+3i)=(2+3i+4i+6i^2)/(2^2-(3i)^2)=#

Remembering that: #i^2=-1#

#=(2+6*(-1)+7i)/(4-9i^2)=(2-6+7i)/(4-9*(-1))=(-4+7i)/(4+9)=#
#=(-4+7i)/13=-4/13+7/13i#