How do you simplify #sqrt(-32)#? Precalculus Complex Numbers in Trigonometric Form Complex Number Plane 1 Answer Carl S. Apr 5, 2018 #4isqrt2# Explanation: #i=sqrt{-1}# So #sqrt{-32}=isqrt{32}# #32 = 2 ^5# #sqrt{32} = 2 ^2.5= 2^2 times 2^ 0.5 = 4sqrt2# #sqrt{-32}=4isqrt2# 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 7232 views around the world You can reuse this answer Creative Commons License