Show that for a polynomial $P\left(z\right)$ the sum $\sum_{\left\{ y\,:\, P\left(y\right)=z\right\} }P^{'}\left(y\right)$ does not depend on $z$ I just came across the following very intriguing question and I'm not sure how to even approach it. Show that for a complex polynomial $P\left(z\right)$ the sum $\sum_{\left\{ y\,:\, P\left(y\right)=z\right\} }P^{'}\left(y\right)
  $ does not depend on $z$.
Help would be appreciated!  
 A: Let $P=(z-a_1)...(z-a_n)$. The logarithmic derivative $$P'(y)/z=\frac{P'}{P}(y)= \sum_{i=1}^{n} \frac{1}{y-a_i}$$
Now, if you sum over all roots of $P(y)-z$ you get (minus) sum of logarithmic derivatives of $P(y)-z$ in the points $a_1, ..., a_n$. So $$\frac{A}{z}=\sum_{i=1}^{n}\frac{P'(a_i)}{z}$$, where $A$ is the desired result.
A: Here's another solution. By a generalization of the argument principle, we have
\begin{align*}
\sum_{y\in P^{-1}(z)} P'(y) = {1\over 2\pi i}\int_C {P'(w)^2\over P(w) - z}\,dw, \tag{1}
\end{align*}
where $C$ is a large circle, say, that contains all the zeroes of $P(w) - z$. Now $C$ will also contain all the zeroes of $P(w) - z'$ for all $z'$ sufficiently close to $z$, so we can differentiate the above equation to obtain
\begin{align*}
{d\over dz} \sum_{y\in P^{-1}(z)} P'(y) & = {d\over dz}{1\over 2\pi i}\int_C {P'(w)^2\over P(w) - z}\,dw = {1\over 2\pi i}\int_C {P'(w)^2\over (P(w) - z)^2}\,dw.\tag{2}
\end{align*}
On the other hand, if $y_1,\dots,y_n$ are the roots of $P(w) - z$, then
\begin{align*}
\left({P'(w)\over P(w) - z}\right)^2 & = \left(\sum_{k=1}^n {1\over w-y_k}\right)^2 = \sum_{j = 1}^n\sum_{k=1}^n {1\over (w-y_j)(w-y_k)}.
\end{align*}
But then $(2)$ is equal to
\begin{align*}
{1\over 2\pi i}\sum_{j = 1}^n\sum_{k=1}^n\int_C {1\over (w-y_j)(w-y_k)}\,dw & = \sum_{j= 1}^n\sum_{y_k\not = y_j} {1\over y_j-y_k},
\end{align*}
and the last sum is equal to its negative (hence equal to $0$) because the indexing is symmetric in $k$ and $j$. It follows that $(2)$ is identically $0$, and therefore that $(1)$ is constant as required.
