What does #(3+i)^(1/3)# equal in a+bi form?

1 Answer
Nov 10, 2015

Answer:

#root(6)(10)cos(1/3 arctan(1/3)) + root(6)(10)sin(1/3 arctan(1/3)) i#

Explanation:

#3+i = sqrt(10)(cos(alpha)+i sin(alpha))# where #alpha = arctan(1/3)#

So

#root(3)(3+i) = root(3)(sqrt(10))(cos(alpha/3)+i sin(alpha/3))#

#=root(6)(10)(cos(1/3 arctan(1/3)) + i sin(1/3 arctan(1/3)))#

#=root(6)(10)cos(1/3 arctan(1/3)) + root(6)(10)sin(1/3 arctan(1/3)) i#

Since #3+i# is in Q1, this principal cube root of #3+i# is also in Q1.

The two other cube roots of #3+i# are expressible using the primitive Complex cube root of unity #omega = -1/2+sqrt(3)/2 i#:

#omega (root(6)(10)cos(1/3 arctan(1/3)) + root(6)(10)sin(1/3 arctan(1/3)) i)#

#=root(6)(10)cos(1/3 arctan(1/3) + (2pi)/3) + root(6)(10)sin(1/3 arctan(1/3)+(2pi)/3) i#

#omega^2 (root(6)(10)cos(1/3 arctan(1/3)) + root(6)(10)sin(1/3 arctan(1/3)) i)#

#=root(6)(10)cos(1/3 arctan(1/3) + (4pi)/3) + root(6)(10)sin(1/3 arctan(1/3)+(4pi)/3) i#