Question #2e282 Precalculus Complex Numbers in Trigonometric Form Complex Number Plane 1 Answer Cesareo R. Mar 16, 2017 #1/2(-1+isqrt(3))# Explanation: We have that #z=-1/2(1+isqrt(3))=abs ze^(i "arg"( z))# where #"arg"(z)=arctan((-sqrt(3))/(-1))# but #abs z=1# and #"arg"(z)=-(2pi)/3# so #z^(83)=(e^(-i(2pi)/3))^83 = e^(-i(166pi)/3) = e^(i(2pi)/3) = 1/2(-1+isqrt(3))# 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 1103 views around the world You can reuse this answer Creative Commons License