How do you simplify #((-1+2i)(5-4i)) / ((1-2i)(5+4i))#? Precalculus Complex Numbers in Trigonometric Form Division of Complex Numbers 1 Answer Ratnaker Mehta Jul 20, 2016 #-9/41+40/41i# Explanation: The Given Exp.#=(-cancel((1-2i))(5-4i))/(cancel((1-2i))(5+4i))# #=-(5-4i)/(5+4i)xx(5-4i)/(5-4i)# #=-(5-4i)^2/{5^2-(4i)^2}# #=-(25-40i+16i^2)/{25-16i^2}# #=-(25-40i-16)/(25+16)# #=-(9-40i)/41# #=-9/41+40/41i# Answer link Related questions How do I graphically divide complex numbers? How do I divide complex numbers in standard form? How do I find the quotient of two complex numbers in polar form? How do I find the quotient #(-5+i)/(-7+i)#? How do I find the quotient of two complex numbers in standard form? What is the complex conjugate of a complex number? How do I find the complex conjugate of #12/(5i)#? How do I rationalize the denominator of a complex quotient? How do I divide #6(cos^circ 60+i\ sin60^circ)# by #3(cos^circ 90+i\ sin90^circ)#? How do you write #(-2i) / (4-2i)# in the "a+bi" form? See all questions in Division of Complex Numbers Impact of this question 1393 views around the world You can reuse this answer Creative Commons License