= Solution
Set $w=(z+1)/(z-1)$, with $z\ne1$. For a nonzero integer $n$, all solutions of $w^n=i$ are
$$
w_k=\exp\left(\frac{i(\pi/2+2\pi k)}n\right),\qquad k=0,\ldots,|n|-1.
$$
This covers negative integers too: $w$ is then nonzero and $w^{|n|}=-i$. No $w_k$ equals one. Inverting the <Möbius transformation> and using half-angle identities gives the <cotangent roots of a powered Möbius transformation>:
$$
\boxed{z_k=\frac{w_k+1}{w_k-1}
=-i\cot\left(\frac{\pi/4+\pi k}{n}\right),\quad k=0,\ldots,|n|-1.}
$$
These $|n|$ solutions are distinct and purely imaginary. For $n=0$ there is no solution: every defined zeroth power is one, not $i$. For negative $n$, $z=-1$ must additionally be excluded from the equation's domain; none of the displayed roots equals it.
For the cubic, direct expansion shows
$$
(1-i)(z^3+3iz^2+3z+i)=(z+1)^3-i(z-1)^3.
$$
Since $z=1$ is not a root, the cubic is precisely the $n=3$ case. Taking $k=0,1,2$ yields
$$
\boxed{z=-i(2+\sqrt3),\qquad -i(2-\sqrt3),\qquad i.}
$$
For an algebraic check, the cubic factors as $(z-i)(z^2+4iz-1)$, which gives the same three distinct roots.
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