How do you write #(1-9i)(1-4i)(4-3i)# in standard form? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Binayaka C. Oct 18, 2017 In standard form #(1-9i)(1-4i)(4-3i) = -179 +53 i# Explanation: #(1-9i)(1-4i)(4-3i) = ( 1 -13i - 36)(4-3i) [i^2=-1]# #( -35 -13i )(4-3i) = -140 +53i -39 = -179 +53 i# In standard form #(1-9i)(1-4i)(4-3i) = -179 +53 i# [Ans] 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 2531 views around the world You can reuse this answer Creative Commons License