If you're looking for the cube roots of √(3 + <em>i </em>), you first have to decide what you mean by the square root √(…), since 3 + <em>i</em> is complex and therefore √(3 + <em>i </em>) is multi-valued. There are 2 choices, but I'll stick with 1 of them.
First write 3 + <em>i</em> in polar form:
3 + <em>i</em> = √(3² + 1²) exp(<em>i</em> arctan(1/3)) = √10 exp(<em>i</em> arctan(1/3))
Then the 2 possible square roots are
• √(3 + <em>i</em> ) = ∜10 exp(<em>i</em> arctan(1/3)/2)
• √(3 + <em>i</em> ) = ∜10 exp(<em>i</em> (arctan(1/3)/2 + <em>π</em>))
and I'll take the one with the smaller argument,
√(3 + <em>i</em> ) = ∜10 exp(<em>i</em> arctan(1/3)/2)
Then the 3 cube roots of √(3 + <em>i</em> ) are
• ∛(√(3 + <em>i</em> )) = ¹²√10 exp(<em>i</em> arctan(1/3)/6)
• ∛(√(3 + <em>i</em> )) = ¹²√10 exp(<em>i</em> (arctan(1/3)/6 + <em>π</em>/3))
• ∛(√(3 + <em>i</em> )) = ¹²√10 exp(<em>i</em> (arctan(1/3)/6 + 2<em>π</em>/3))
On the off-chance you meant to ask about the cube roots of 3 + <em>i</em>, and not √(3 + <em>i </em>), then these would be
• ∛(3 + <em>i</em> ) = ⁶√10 exp(<em>i</em> arctan(1/3)/3)
• ∛(3 + <em>i</em> ) = ⁶√10 exp(<em>i</em> (arctan(1/3)/3 + 2<em>π</em>/3))
• ∛(3 + <em>i</em> ) = ⁶√10 exp(<em>i</em> (arctan(1/3)/6 + 4<em>π</em>/3))
is result