How do you simplify #(15-8i)^2# and write the complex number in standard form? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Gerardina C. Jul 28, 2016 #161-240i# Explanation: Since #(a+b)^2=a^2+2ab+b^2#, then #(15-8i)^2=15^2+2(15)(-8i)+(-8i)^2# that is #225-240i+64i^2# and, since #i^2=-1#, it becomes: #225-240i-64=161-240i# in standard form 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 1734 views around the world You can reuse this answer Creative Commons License