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 1204 views around the world You can reuse this answer Creative Commons License