= Solution
List the <roots> with their <multiplicities>. By the <factor theorem>, $Q(z)=a\prod_{k=1}^n(z-\alpha_k)$ with $a\ne0$. The <product rule>, followed by division where $Q(z)\ne0$, gives its <logarithmic derivative>:
$$
\boxed{\frac{Q'(z)}{Q(z)}=\sum_{k=1}^n\frac1{z-\alpha_k}.}
$$
Let $Q'(\beta)=0$. If $Q(\beta)=0$ too, then $\beta$ is already one of the <roots>, so either desired location conclusion follows immediately. Otherwise the displayed sum vanishes at $\beta$. If all <roots> have positive <imaginary parts> but $\operatorname{Im}\beta\leq0$, then
$$
\operatorname{Im}\frac1{\beta-\alpha_k}
=\frac{\operatorname{Im}\alpha_k-\operatorname{Im}\beta}{|\beta-\alpha_k|^2}>0
$$
for every $k$. Their sum cannot be zero. Thus \b[$\operatorname{Im}\beta>0$].
For the <unit disk> conclusion, the same sum gives an especially transparent <convex combination>. Taking <complex conjugates> and setting $w_k=|\beta-\alpha_k|^{-2}>0$ gives
$$
0=\sum_k w_k(\beta-\alpha_k),\qquad
\boxed{\beta=\frac{\sum_k w_k\alpha_k}{\sum_k w_k}.}
$$
Therefore, if every $|\alpha_k|<1$,
$$
|\beta|\leq\frac{\sum_k w_k|\alpha_k|}{\sum_k w_k}<1.
$$
The strict inequality follows because this is a finite weighted average of numbers strictly below one. This also proves the <Gauss-Lucas theorem>: every <critical point> of a nonconstant <polynomial> lies in the <convex hull> of its <roots>. The common-root case, handled before division, is essential when <multiplicities> occur.
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