Question given is: if $h(n)$ is a linear phase causal FIR of order $10$ with real coefficients, find the remaining zeros of this filter if the zeros given are
$$q_{1,2} = -2 \pm 2j$$
$$q_{3} = -\frac12 + \frac{j}{2} $$
$$q_{4} = -1$$
I know for sure that $-\frac {1} 2 - \frac j 2 $ has to be another zero, but there are 5 more zeros unaccounted for. How am I supposed to find them?