How do you find #abs( 4(sqrt3) - 4i)#? Precalculus Complex Numbers in Trigonometric Form Complex Number Plane 1 Answer Konstantinos Michailidis May 26, 2016 The magnitude of a complex number is #abs(a+bi)=sqrt(a^2+b^2)# Hence #abs(4sqrt3-4i)=sqrt((4sqrt3)^2+4^2)=sqrt(4^2(3+1))= 4*sqrt4=4*2=8# Answer link Related questions What is the complex number plane? Which vectors define the complex number plane? What is the modulus of a complex number? How do I graph the complex number #3+4i# in the complex plane? How do I graph the complex number #2-3i# in the complex plane? How do I graph the complex number #-4+2i# in the complex plane? How do I graph the number 3 in the complex number plane? How do I graph the number #4i# in the complex number plane? How do I use graphing in the complex plane to add #2+4i# and #5+3i#? How do I use graphing in the complex plane to subtract #3+4i# from #-2+2i#? See all questions in Complex Number Plane Impact of this question 1306 views around the world You can reuse this answer Creative Commons License