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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))

I know the first one is true, but the rest don't have all of the correct input when put over tex. Sorry.