Original at https://i.stack.imgur.com/Zn8RE.jpg
Lucas' theorem. The zeros of the derivative $P'(z)$ of a polynomial $P(z)$ lie in the convex hull of the zeros of $P(z)$.
Proof. Because any convex domain can be obtained as the intersection of half-planes, it suffices to show that if the zeros of $P(z)$ lie in an open half-plane, then the zeros of $P'(z)$ lie in that half-plane as well. Moreover, by rotating and translating the variable $z$ we can further reduce the problem to the case in which the zeros of $P(z)$ lie in the upper half plane $\text{Im } z > 0$. Here $\text{Im } z$ denotes the imaginary part.
So let $z_1, z_2, \ldots, z_n$ be the (not necessarily distinct) zeros of $P(z)$, which by hypothesis have positive imaginary part. If $\text{Im } w \leq 0$, then $\text{Im } \frac{1}{w-z_k} > 0$, for $k = 1, \ldots, n$, and therefore
$$ \text{Im } \frac{P'(w)}{P(w)} = \sum_{k=1}^n \text{Im } \frac{1}{w-z_k} > 0. $$
This shows that $w$ is not a zero of $P'(z)$ and so all zeros of $P'(z)$ lie in the upper half-plane. The theorem is proved.
We assumed that $ z_1,z_2,...,z_n $ have positive imaginary part. Then if $ Im(w)\leq 0$ how the conclusion $ Im \frac{1}{w-z_k} $ ,for $ k=1,2,...,n $ occurs?
I can't fit this in a comment. We have $$ {1\over \omega-z_k}={\overline{\omega-z_k}\over(\omega-z_k)\overline{(\omega-z_k})} ={\overline{\omega}-\overline{z_k}\over|\omega-z_k|^2}$$
Now the denominator on the right-hand side is real and positive, so we just have to figure out the sign of the imaginary part of the numerator.
I leave that to you.