How do you find #abs( 2sqrt3 - πi )#? Precalculus Complex Numbers in Trigonometric Form Complex Number Plane 1 Answer mason m Nov 21, 2016 For a complex number, #abs(a+bi)=sqrt(a^2+b^2)#, or the distance of the segment. So, #abs(2sqrt3-pi i)=sqrt((2sqrt3)^2+(-pi)^2)=sqrt(4(3)+pi^2)=sqrt(12+pi^2)#. 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 1066 views around the world You can reuse this answer Creative Commons License