How do you find #abs( i^6 )#? Precalculus Complex Numbers in Trigonometric Form Complex Number Plane 1 Answer Konstantinos Michailidis May 10, 2016 We know that #i^2=-1# hence #i^6=(i^2)^3=(-1)^3=-1# So #abs(i^6)=abs(-1)=1# 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 1062 views around the world You can reuse this answer Creative Commons License