How do you simplify [(3+2i)^ 3 / (-2+3i)^4] ? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Cesareo R. Jun 27, 2016 (3-2i)/13 Explanation: 3+2i=sqrt(3^2+2^2)e^{i phi} -2+3i=sqrt(3^2+2^2)e^{i( phi+pi/2)} with phi = arctan(2/3) [(3+2i)^ 3 / (-2+3i)^4] =((3+2i)/(-2+3i))^3/( (-2+3i))=e^{-i (3pi)/2}/(sqrt(3^2+2^2)e^{i( phi+pi/2)}) =1/sqrt(3^2+2^2)e^{-i(phi+2pi)} = 1/sqrt(3^2+2^2)e^{-i phi} = sqrt(3^2+2^2)/(3^2+2^2)e^{-i phi} = (3-2i)/13 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 2052 views around the world You can reuse this answer Creative Commons License