What does #(1+3i)/(2+2i)# equal in a+bi form?

1 Answer
Oct 21, 2015

Answer:

I found: #1+1/2i#

Explanation:

First multiply and divide by the complex conjugate of the denominator to get a Pure Real denominator:
#(1+3i)/(2+2i)*color(red)((2-2i)/(2-2i))=#
#=((1+3i)(2-2i))/(4+4)=(2-2i+6i-6i^2)/8=#
but #i^2=-1#
#(8+4i)/8=8/8+4/8i=1+1/2i#