How do you simplify #2sqrt(-16/7)#? Precalculus Complex Numbers in Trigonometric Form Complex Number Plane 1 Answer mason m Dec 13, 2015 #(8sqrt7i)/7# Explanation: This can be rewritten as #2sqrt(-1xx16/7)# #=2isqrt(16/7)# #=(2isqrt(16))/sqrt7# #=(8i)/sqrt7# #=(8i)/sqrt7(sqrt7/sqrt7)# #=(8sqrt7i)/7# Notice that the #i# is not included in the square root in the final answer. 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 1377 views around the world You can reuse this answer Creative Commons License