According to a proof of Gauss-Lucas theorem from wikipedia, a logarithmic derivative of polynomial $P(z)$ is
$$\frac{P^{\prime}(z)}{P(z)} = \sum_{i=1}^{n}\frac{1}{z – z_i} = \sum_{i=1}^{n}\frac{\overline{z} – \overline{z}_i}{|z – z_i|^2}.$$
Here $P(z)$ is a non-constant polynomial with $z_i$ to be the zeros of $P(z)$. If $P^{\prime}(z) = 0$ and $P(z) \neq 0$, then
$$z\left(\sum_{i=1}^{n}\frac{1}{|z – z_i|^2}\right) = \sum_{i=1}^{n}\frac{1}{|z – z_i|^2}z_i.$$
Up until here, I do understand what is going on. However, I'm not following what happens next.
According to Wikipedia, "$z$ is a weighted sum with positive coefficients that sum to one, or the barycenter on affine coordinates, of the complex numbers $z_{i}$ (with different mass assigned on each root whose weights collectively sum to $1$)". I do not understand what this sentence means. How come that the weights collectively sum to 1?
Let $$m_i = \frac{1}{|z-z_i|^2}$$ and $$m = \sum_{i=1}^n\frac{1}{m_i}.$$
Why would
$$\sum_{i=1}^{n}\frac{m_i}{m} = 1?$$
If that is the case, then I can see why $z$ lies in the convex hull of the zeros of $P$. I feel like it has something to do with normalization, but I must say that I completely don't see it.
Best Answer
Per comments, we use the second equation above to show that it immediately implies that $z$ is a convex combination of the $z_i$. In the OP we have the following:
"Here $P(z)$ is a non-constant polynomial with $z_i$ to be the zeros of $P(z)$. If $P^{\prime}(z) = 0$ and $P(z) \neq 0$, then $$z\left(\sum_{i=1}^{n}\frac{1}{|z - z_i|^2}\right) = \sum_{i=1}^{n}\frac{1}{|z - z_i|^2}z_i.$$
Looking at the equation above we denote by $m$ the coefficient of $z$ and by $m_i$ the coefficient of $z_i, i=1,..n$, so the equation is:
$mz =\sum_{i=1}^{n}m_iz_i$ or equivalently $z=\sum_{i=1}^{n}\frac{m_i}{m}z_i=\sum_{i=1}^{n}\alpha_iz_i$ where $\alpha_i=m_i/m$
Now we claim that $\alpha_i \ge 0, \sum_{i=1}^{n}\alpha_i=1$ which means precisely that $z$ satisfies the definition of being in the convex hull of the $z_i$.
But notice that $m=\sum_{i=1}^{n}\frac{1}{|z - z_i|^2}$ while $m_i=\frac{1}{|z - z_i|^2}$ so it is self-evident that $m=\sum_{i=1}^{n}m_i$ hence $ \sum_{i=1}^{n}\alpha_i=1$ while it is also immediate that $0 \le \alpha_i \le 1$ by their defintion, so we are done and Gauss-Lucas has been proved