How do you find the cube root of #81 (cos (pi/12) + isin (pi/12))#?

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)#.

Related questions
See all questions in Roots of Complex Numbers
Impact of this question
2011 views around the world
You can reuse this answer
Creative Commons License