How do you simplify #[(1+i)^4 / (2-2i)^3]#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Cesareo R. Aug 2, 2016 #(1-i)/8# Explanation: #(1+i)^4 / (2-2i)^3=1/2^3(1+i)^4/(1-i)^3=1/2^3(1+i)^7/((1-i)^3(1+i)^3) = (1+i)^7/2^6# Now using de Moivre's identity #1+i = sqrt(2)e^{ipi/4}# so #(1+i)^7 = (sqrt(2))^7e^{i7/4pi}# but #e^{i7/4pi} = e^{i 8/4pi} e^{-ipi/4} = e^{-ipi/4}# Finally #(1+i)^4 / (2-2i)^3=(2^3 sqrt(2)e^{-ipi/4})/2^6=sqrt(2)e^{-ipi/4}/2^3 = (1-i)/8# 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 2090 views around the world You can reuse this answer Creative Commons License