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

1 Answer
George C.
Nov 10, 2015

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

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