If you're looking for the cube roots of β(3 + i ), you first have to decide what you mean by the square root β(β¦), since 3 + i is complex and therefore β(3 + i ) is multi-valued. There are 2 choices, but I'll stick with 1 of them.
First write 3 + i in polar form:
3 + i = β(3Β² + 1Β²) exp(i arctan(1/3)) = β10 exp(i arctan(1/3))
Then the 2 possible square roots are
β’Β β(3 + i ) = β10 exp(i arctan(1/3)/2)
β’Β β(3 + i ) = β10 exp(i (arctan(1/3)/2 + Ο))
and I'll take the one with the smaller argument,
β(3 + i ) = β10 exp(i arctan(1/3)/2)
Then the 3 cube roots of β(3 + i ) are
β’Β β(β(3 + i )) = ΒΉΒ²β10 exp(i arctan(1/3)/6)
β’Β β(β(3 + i )) = ΒΉΒ²β10 exp(i (arctan(1/3)/6 + Ο/3))
β’Β β(β(3 + i )) = ΒΉΒ²β10 exp(i (arctan(1/3)/6 + 2Ο/3))
On the off-chance you meant to ask about the cube roots of 3 + i, and not β(3 + i ), then these would be
β’Β β(3 + i ) = βΆβ10 exp(i arctan(1/3)/3)
β’Β β(3 + i ) = βΆβ10 exp(i (arctan(1/3)/3 + 2Ο/3))
β’Β β(3 + i ) = βΆβ10 exp(i (arctan(1/3)/6 + 4Ο/3))
