How do you find #abs(-8-4i)#? Precalculus Complex Numbers in Trigonometric Form Complex Number Plane 1 Answer Ratnaker Mehta Aug 18, 2016 #|-8-4i|=4sqrt5~=8.9444#. Explanation: For, #z=x+iy in CC, |z|=sqrt(x^2+y^2)#. Hence, #|-8-4i|=sqrt((-8)^2+(-4)^2)=sqrt(64+16)=sqrt80=sqrt(4^2*5)=4sqrt5#. Taking, #sqrt5~=2.2361#, Reqd. Value#~=4*2.2361=8.9444#. 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 1348 views around the world You can reuse this answer Creative Commons License