How do you simplify # ((1 + i)^2 + 2+2i)/ (2+i)#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Konstantinos Michailidis Feb 8, 2016 It is # ((1 + i)^2 + 2+2i)/ (2+i)=(1^2+2i+i^2+2+2i)/(2+i)= (4i+2)/(2+i)=2*(2i+1)/(2+i)=2*[(2i+1)*(2-i)]/[(2+i)*(2-i)]= 8/5+6i/5# Remember that #i^2=-1# Answer link Related questions How do I multiply complex numbers? How do I multiply complex numbers in polar form? What is the formula for multiplying complex numbers in trigonometric form? How do I use the modulus and argument to square #(1+i)#? What is the geometric interpretation of multiplying two complex numbers? What is the product of #3+2i# and #1+7i#? How do I use DeMoivre's theorem to solve #z^3-1=0#? How do I find the product of two imaginary numbers? How do you simplify #(2+4i)(2-4i)#? How do you multiply #(-2-8i)(6+7i)#? See all questions in Multiplication of Complex Numbers Impact of this question 1710 views around the world You can reuse this answer Creative Commons License