<s>[INST] How can you find the cubed roots of $i$? I am trying to figure out what the three possibilities of $z$ are such that 
$$
z^3=i
$$
but I am stuck on how to proceed. I tried algebraically but ran into rather tedious polynomials. Could you solve this geometrically? Any help would be greatly appreciated.
 [/INST] 
        $z^3=i$         $⟹ z=\sqrt [3] {i}$

We know,
$\sqrt [3] {i} =a+bi$
$⟹ i=(a+bi)^3 = a^3-3ab^2+3a^2bi-b^3i= a^3-3ab^2+(3a^2b-b^3)i$
From this we can say, 
$a^3-3ab^2=0$________________(i)
$3a^2b-b^3=1$_______________(ii)
From (i) we get,
$a(a^2-3b^2)=0$ 
So,  $a=0$
Or, 
        $a^2-3b^2=0$         $⟹ a=±b\sqrt {3}$

By setting the value of $a$ in (ii) we get,


        If $a=0$,         $ b=-1$



        If $a=±b\sqrt {3}$,         $ b=1/2$


Then, $a=±b\sqrt {3}=±(1/2)\sqrt {3}=±\sqrt {3}/2$
So, $(a,b)=(0,-1),(±\sqrt {3}/2,1/2)$
Now, there are 3 values of $z$. 
(1) $\sqrt [3] {i} =a+bi= 0+(-1)i= -i$
(2) $\sqrt [3] {i} =a+bi= \dfrac {\sqrt {3}}{2}+\dfrac {i}{2}$
(3) $\sqrt [3] {i} =a+bi= -\dfrac {\sqrt {3}}{2}+\dfrac {i}{2}$
</s>