How do you find #abs( sqrt 5 + 2i sqrt2)#? Precalculus Complex Numbers in Trigonometric Form Complex Number Plane 1 Answer Shwetank Mauria Apr 29, 2016 #|sqrt5+2sqrt2i|=sqrt13# Explanation: #|a+bi|=sqrt(a^2+b^2)# Hence, #|sqrt5+2sqrt2i|# = #sqrt((sqrt5)^2+(2sqrt2)^2)# = #sqrt(5+8)# = #sqrt13# 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 1839 views around the world You can reuse this answer Creative Commons License