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

1 Answer
Feb 14, 2016

Answer:

Multiply by 1, written as the complex conjugate of the bottom of the fraction divided by itself, in this case #(1-2i)/(1-2i)#:

#(2+3i)/(1+2i)*(1-2i)/(1-2i)=(8-i)/5#

Explanation:

Multiply top and bottom of the fraction by the complex conjugate of the bottom. For a complex expression #a+bi# the complex conjugate is #a-bi#.

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

While simplifying, remember #i^2=-1#

#= (2-i+6)/(1+4)=(8-i)/5#