How do you find #abs( 12-5i)#? Precalculus Complex Numbers in Trigonometric Form Complex Number Plane 1 Answer Shwetank Mauria Mar 17, 2016 #|12-5i|=13# Explanation: #|12-5i|# is the modulus of complex number #12-5i#. Modulus of complex number #a+bi# is given by #sqrt(a^2+b^2)# hence #|12-5i|=sqrt(12^2+(-5)^2)=sqrt(144+25)=sqrt169=13# 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 8032 views around the world You can reuse this answer Creative Commons License