How do you find #abs( 2-2i )#? Precalculus Complex Numbers in Trigonometric Form Complex Number Plane 1 Answer Narad T. Dec 6, 2016 The answer is #=2sqrt2# Explanation: Let #z=a +ib# be a complex number. Then the modulus of #z#is #∣z∣=∣a+ib∣=sqrt(a^2+b^2)# so, #∣2-2i∣=sqrt(2^2+(-2)^2)# #=sqrt(4+4)= sqrt8=2sqrt2# 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 2927 views around the world You can reuse this answer Creative Commons License