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

1 Answer
May 21, 2018

Answer:

#(8-i)/(5)#

Explanation:

#(2+3i)/(1+2i)#

You multiply it by the complex conjugate to get rid of the imaginary numbers in the denominator, since the conjugate's value is 1 the value of the expression is not altered:

complex conjugate is the #"denominator"/"denominator"# with the sign changed:

#(1-2i)/(1-2i) =1#

#(2+3i)/(1+2i)*(1-2i)/(1-2i)#

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

#=(2-i-6sqrt(-1)^2)/(1-4sqrt(-1)^2)#

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

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

#=(8-i)/(5)#