How do you find the cube root of #81 (cos (pi/12) + isin (pi/12))#? Precalculus Complex Numbers in Trigonometric Form Roots of Complex Numbers 1 Answer A. S. Adikesavan Dec 21, 2016 #3^(4/3)(cos 5^o +i sin 5^o), 3^(4/3)(cos 125 o +i sin 125^o) and 3^(4/3)(cos 245^o +i sin 245^o)#. Explanation: #(81(cos (pi/12)+i sin( pi/12))^(1/3)# #=81^(1/3)(cispi/12)^(1/3)# #3^(4/3)(cis((1/3(2kpi+pi/12)), k = 0, 1, 2# #=3^(4/3)(cis 5^o, cis 125^o, cis 245^o)#, using #pi = 180^o# #=3^(4/3)(cos 5^o + i sin 5^o)#, #3^(4/3)(cos 125^o +i sin 125^o)# and #3^(4/3)(cos 245^o +i sin 245^o)#. Answer link Related questions How do I find the cube root of a complex number? How do I find the fourth root of a complex number? How do I find the fifth root of a complex number? How do I find the nth root of a complex number? How do I find the square root of a complex number? What is the square root of #2i#? What is the cube root of #(sqrt3 -i)#? What are roots of unity? How do I find the square roots of #i#? How do you solve #6x^2-5x+3=0#? See all questions in Roots of Complex Numbers Impact of this question 2010 views around the world You can reuse this answer Creative Commons License