If you're looking for the cube roots of √(3 + i ), you first have to decide what you mean by the square root √(…), since 3 + i is complex and therefore √(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 = √(3² + 1²) exp(i arctan(1/3)) = √10 exp(i arctan(1/3))
Then the 2 possible square roots are
• √(3 + i ) = ∜10 exp(i arctan(1/3)/2)
• √(3 + i ) = ∜10 exp(i (arctan(1/3)/2 + π))
and I'll take the one with the smaller argument,
√(3 + i ) = ∜10 exp(i arctan(1/3)/2)
Then the 3 cube roots of √(3 + i ) are
• ∛(√(3 + i )) = ¹²√10 exp(i arctan(1/3)/6)
• ∛(√(3 + i )) = ¹²√10 exp(i (arctan(1/3)/6 + π/3))
• ∛(√(3 + i )) = ¹²√10 exp(i (arctan(1/3)/6 + 2π/3))
On the off-chance you meant to ask about the cube roots of 3 + i, and not √(3 + i ), then these would be
• ∛(3 + i ) = ⁶√10 exp(i arctan(1/3)/3)
• ∛(3 + i ) = ⁶√10 exp(i (arctan(1/3)/3 + 2π/3))
• ∛(3 + i ) = ⁶√10 exp(i (arctan(1/3)/6 + 4π/3))
