How do you simplify # (2+2i)/(1+2i) # and write in a+bi form?

Question

3312 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-11T00:49:06

Multiply the numerator and denominator by the complex conjugate of the denominator to find

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

Given a complex number #a+bi# with #a,b in RR# we have

#(a+bi)(a-bi) = a^2 + b^2#

#a-bi# is called the complex conjugate (or conjugate) of #a+bi#. Using this:

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

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

#= (6-2i)/5#

#=6/5 - 2/5i#