Answer
$-i, i,-2i$ and $2i.$
Work Step by Step
First, because all coefficients of f(x) are real numbers, the Conjugate Zeros Theorem applies.
If $-i$ is a zero, so is its conjugate, $+i.$
If k is a zero of f, then, by the factor theorem, (x-k) is a factor of f(x).
Synthetic division gives us a quotient q(x), with degree one less than f(x).
and we know the remainder is zero,
so $f(x)=(x-k)\cdot q(x).$
We use this now
$k=-i$,, note the missing $x^{3}$ and $x$ terms
$\left\{\begin{array}{lllllll}
-i & | & 1 & 0 & 5 & 0 & 4\\
& & & -i & -1 & -4i & -4\\
& & -- & -- & -- & -- & --\\
& & 1 & -i & 4 & -4i & \fbox{$0$}
\end{array}\right.$
$f(x)=(x+i)(x^{3}+ix^{2}+4x+4i)$
Now, we use the fact that $i$ is also a zero.
Apply synthetic division on the quotient.
$\left\{\begin{array}{llllll}
i & | & 1 & -i & 4 & -4i\\
& & & i & 0 & 4i\\
& & -- & -- & -- & --\\
& & 1 & 0 & 4 & \fbox{$0$}
\end{array}\right.$
$f(x)=(x+i)(x-i)(x^{2}+4)$
The zeros of $x^{2}+4:$
$x^{2}+4=0$
$x^{2}=-4$
$x=\pm 2i$
The other zeros (other than the given) are
$i,\quad -2i$ and $2i.$