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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!

Serpahimz
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  • Intterestingly, the sum may or may not count with multiplicities ;) – Hagen von Eitzen Jan 18 '14 at 11:32
  • From an algebraic standpoint, the expression in question is the resultant of the two polynomials $P'(y)$ and $P(y)-z$. (So when $z=0$, it's just the discriminant of $P$.) – Greg Martin Jan 18 '14 at 12:14
  • The solution I'm seeking is definitely something from the field of complex analysis since this is the context in which I encountered the question. – Serpahimz Jan 18 '14 at 13:05

2 Answers2

1

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.

user68061
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  • Can you please provide more detail? I do not follow you after you sum over roots of $P(y)-z$. It seems to me that if you sum over roots of $P(y) -z$ you obtain $\sum_{y\in P^{-1}(z)} \sum_{i=1}^n {1\over y-a_i}$ (assuming that $z\not = 0$), but I don't see why that's equal to $\sum_{i=1}^n{P'(a_i)/z}$. In fact, that's just restating the problem. – Nick Strehlke Jan 18 '14 at 14:20
  • Change the order of summation. Then you get $\sum_{i=1}^n \sum_{y \in P^{-1}(z)}-\frac{1}{a_i-y}$. But the internal sum is minus the value of logarithmic derivative in $a_i$. – user68061 Jan 18 '14 at 14:32
  • Ok I think I get what you're saying. If $a_1,\dots,a_n$ are the roots of $P(w)$ and $b_1,\dots,b_n$ the roots of $P(w)-z$, then $P'(a_j) = -z\sum_i {1\over a_j-b_i}$ and $P'(b_j) = z\sum_i{1\over b_j-a_i}$. So summing $P'(a_j)$ and $P'(b_j)$ over $j$ gives the same result. Nice (+1). – Nick Strehlke Jan 18 '14 at 15:22
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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.

Nick Strehlke
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