What does #(3+i)^(1/3)# equal in a+bi form?
1 Answer
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
The two other cube roots of
#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#