Let $a_{1}, a_{2}, \ldots, a_{n}$ be the zeros of a polynomial $p(z)$, what to do next for this problem? For this problem:
Let $a_{1}, a_{2}, \ldots, a_{n}$ be the zeros of a complex polynomial $p(z)$, listed according to multiplicity, and assume that $p(0) \neq 0$. Prove that
$$
\sum_{k=1}^{n} \frac{1}{a_{k}^{2}}=\left(\frac{p^{\prime}(0)}{p(0)}\right)^{2}-\frac{p^{\prime \prime}(0)}{p(0)}
$$

I am thinking:
I know I can write the polynomial as:
$$
p(z)=\sum_{n=0}^{N} c_n z^n
$$
then the derivatives are
$$
p^{'}(z)=\sum_{n=0}^{N} n c_n z^{n-1}
$$
$$
p^{''}(z)=\sum_{n=0}^{N} n (n-1) c_n z^{n-2}
$$
Then the right-side of the equation becomes:
$$
\left(\frac{p^{\prime}(0)}{p(0)}\right)^{2}-\frac{p^{\prime \prime}(0)}{p(0)}
=\frac{n^2 c_1^2 - n (n-1) c_2}{c_0}
$$
because
$$
\frac{p^{\prime}(0)}{p(0)}=\frac{n c_1}{c_0}
$$
and
$$
\frac{p^{\prime\prime}(0)}{p(0)}=\frac{n (n-1) c_2}{c_0}
$$
But I am not sure waht to do next, can anybody help me on this? Thank you very much!
 A: Since $a_{1},a_{2},\ldots,a_{n}$ are zeros for the polynomial $p(z)$,
we have that $p(z)=c(z-a_{1})(z-a_{2})\ldots(z-a_{n})$, where $c\in\mathbb{C}$
is a non-zero constant. By direct calculation,
$$
\frac{p'(z)}{p(z)}=\sum_{k=1}^{n}\frac{1}{z-a_{k}}.
$$
Differentiate again, then we obtain
\begin{eqnarray*}
\frac{p''(z)p(z)-[p'(z)]^{2}}{p^{2}(z)} & = & \sum_{k=1}^{n}\frac{-1}{(z-a_{k})^{2}}.
\end{eqnarray*}
The above is valid as long as $z\notin\{a_{1},\ldots,a_{n}\}$. Note
that it is given that $0\notin\{a_{1},\ldots,a_{n}\}$. Put $z=0$,
then
\begin{eqnarray*}
-\sum_{k=1}^{n}\frac{1}{a_{k}^{2}} & = & \frac{p''(0)}{p(0)}-\left[\frac{p'(0)}{p(0)}\right]^{2}.
\end{eqnarray*}
Hence,
$$
\sum_{k=1}^{n}\frac{1}{a_{k}^{2}}=\left[\frac{p'(0)}{p(0)}\right]^{2}-\frac{p''(0)}{p(0)}.
$$
A: Given differentiable functions $f_1,\dots, f_n$ then if $g(z)=f_1(z)f_2(z)\cdots f_n(z)$ we get the general product rule:
$$g’(z)=\sum_{j=1}^n f_j’(z)\prod_{i\neq j}f_i(z)=g(z)\sum_{j=1}^n \frac{f_j’(z)}{f_j(z)}$$
When $$p(z)=c(z-a_1)(z-a_2)\cdots (z-a_n)$$ this means:
$$\frac{p’(z)}{p(z)}=\sum_{i=1}^n\frac{1}{z-a_k}\tag1$$
You also have:
$$
\begin{align} 
\left(\frac{p’(z)}{p(z)}\right)^2-\frac{p’’(z)}{p(z)}&=\frac{p’(z)p’(z)-p(z)p’’(z)}{p^2(z)}\\&=-\left(\frac{p’(z)}{p(z)}\right)’\\&=\sum_{i=1}^n\frac{1}{(z-a_i)^2}
\end{align} 
$$ The last step by taking the derivative of $(1).$

$T(f)=\frac{f’}f$ is the derivative of $\log f,$ and it called the logorithmic derivative. It has the property $T(fg)=T(f)+T(g),$ because $\log$ does.

Your approach gives: $$p(0)=c_0\\p’(0)=c_{1}\\p’’(0)=2c_2.$$
You want to evaluate $$\frac{c_1^2}{c_0^2}-\frac{2c_2}{c_0}=\frac{c_1^2-2c_0c_2}{c_0^2}$$
Then you need to know that $$c_1=-c_0\sum_{i=1}^n\frac1{a_i}\\
c_2=c_0\sum_{i<j}\frac{1}{a_ia_j}$$
A: Alternative approach:
$$p(z) = \prod_{i=1}^n (z - a_i),$$
with $0$ not an element of $\{a_1, a_2, \cdots, a_n\}.$
You can calculate $p(0), p'(0),$ and $p''(0)$ by calculating the coefficients $c_0, c_1, c_2$ that are associated with $z^0, z^1, z^2$ respectively, in $p(z)$.
$$c_0 = (-1)^n \times \prod_{i=1}^n a_i.$$
$$c_1 = \sum_{i=1}^n \frac{(-a_1) \times (-a_2) \times \cdots \times (-a_n)}{-a_i} = c_0 \times 
\sum_{i=1}^n \frac{1}{-a_i}.$$
Let $S = \sum_{1 \leq i < j \leq n} \frac{1}{a_i a_j}$ [i.e. $S$ is the sum of $\binom{n}{2}$ fractions].
Then, $$c_2 = S \times c_0.$$

$$p(0) = c_0.$$
Since $\frac{d}{dz}c_1z = c_1,$ you have that
$$p'(0) = c_1.$$
Since $\frac{d^2}{dz}c_2z^2 = 2c_2,$ you have that
$$p''(0) = 2c_2.$$
Therefore,
$$\frac{p'(0)}{p(0)} = \sum_{i=1}^n \frac{1}{-a_i},$$
and
$$\frac{p''(0)}{p(0)} = 2S = 2\sum_{1 \leq i < j \leq n} \frac{1}{a_i a_j}.$$
To calculate $\left(\frac{p'(0)}{p(0)}\right)^2$, you have two consider that you will have $n^2$ terms.  $n$ of these terms will be represented by
$$\sum_{i = 1}^n \left(\frac{1}{-a_i}\right)^2.$$
The other $(n^2 - n)$ terms will be represented by
$$\sum_{i \neq j} \frac{1}{a_i a_j} = 2S = 
\frac{p''(0)}{p(0)}.$$
Therefore
$$\left(\frac{p'(0)}{p(0)}\right)^2 = \sum_{i = 1}^n \left(\frac{1}{-a_i}\right)^2 + \frac{p''(0)}{p(0)}.$$
