Let \(z_1,\ldots,z_n\) be distinct complex numbers, and let \(a_1,\ldots,a_n\) be nonzero complex numbers. Suppose \[ S_p=\sum_{j=1}^n a_jz_j^p=0 \qquad(p=0,1,\ldots,m-1), \] but \(S_m\ne0\), where \(1\le m\le n-1\).
How many zeros does \[ f(z)=\sum_{j=1}^n\frac{a_j}{z-z_j} \] have in \(\mathbb C\)? Why is \(m\ge n\) impossible?
Solution
Define \[ Q(z)\coloneqq\prod_{j=1}^n(z-z_j). \]
1. The function \(f\) can be written\[ f(z)=\frac{P(z)}{Q(z)}, \]where\[ P(z) \coloneqq \sum_{j=1}^n a_j \prod_{k\neq j}(z-z_k). \]
Put the summands \[ \frac{a_j}{z-z_j} \] over the common denominator \(Q(z)\).
2. For every \(j\),\[ P(z_j)\neq0. \]
At \(z=z_j\), every term in the sum defining \(P\) vanishes except the \(j\)th term. Thus \[ P(z_j) = a_j \prod_{k\neq j}(z_j-z_k). \] The coefficient \(a_j\) is nonzero and the numbers \(z_1,\ldots,z_n\) are distinct, so every factor in the product is nonzero.
3. For all sufficiently large \(\abs z\),\[ f(z) = \sum_{p=0}^{\infty} S_pz^{-p-1}. \]
Choose \(\abs z>\max_j\abs{z_j}\). Then \[ \frac1{z-z_j} = \frac1z \frac1{1-z_j/z} = \sum_{p=0}^{\infty} z_j^p z^{-p-1}. \] Summing over \(j\) gives \[ \begin{aligned} f(z) &= \sum_{j=1}^n a_j \sum_{p=0}^{\infty} z_j^p z^{-p-1}\\ &= \sum_{p=0}^{\infty} \left( \sum_{j=1}^n a_jz_j^p \right) z^{-p-1}\\ &= \sum_{p=0}^{\infty} S_pz^{-p-1}. \end{aligned} \] The finite sum over \(j\) permits termwise rearrangement.
4. The hypotheses on the moments imply\[ f(z) = S_mz^{-m-1} +O(z^{-m-2}) \]as \(z\to\infty\).
By assumption, \[ S_0=\cdots=S_{m-1}=0 \] and \[ S_m\neq0. \] Apply these facts to the Laurent expansion from step 3.
5. The polynomial \(P\) has degree\[ \deg P=n-m-1. \]
The polynomial \(Q\) is monic of degree \(n\). Let \[ d=\deg P. \] Since \(P\neq0\) by step 4, the rational function \[ \frac{P(z)}{Q(z)} \] has leading behavior \[ c z^{d-n} \] at infinity for some nonzero constant \(c\). Step 4 says that the leading power is instead \[ z^{-m-1}. \] Therefore \[ d-n=-m-1, \] so \[ d=n-m-1. \]
6. The function \(f\) has exactly\[ \boxed{n-m-1} \]zeros in \(\CC\), counted with multiplicity.
By step 2, \(P\) and \(Q\) have no common zero, so no factor of the denominator cancels. Thus the finite zeros of \(f=P/Q\) are exactly the zeros of \(P\). By step 5, \(P\) has degree \(n-m-1\), and the fundamental theorem of algebra gives exactly that many zeros counted with multiplicity.
7. If\[ m\geq n, \]then the moment conditions are impossible under the assumption that all\(a_j\) are nonzero.
If \(m\geq n\), then in particular \[ S_p = \sum_{j=1}^n a_jz_j^p =0 \] for \[ p=0,1,\ldots,n-1. \] This is the homogeneous linear system \[ \begin{pmatrix} 1&1&\cdots&1\\ z_1&z_2&\cdots&z_n\\ \vdots&\vdots&&\vdots\\ z_1^{n-1}&z_2^{n-1}&\cdots&z_n^{n-1} \end{pmatrix} \begin{pmatrix} a_1\\ a_2\\ \vdots\\ a_n \end{pmatrix} =0. \] The coefficient matrix is Vandermonde, with determinant \[ \prod_{1\leq i<j\leq n}(z_j-z_i)\neq0 \] because the \(z_j\) are distinct. Hence \[ a_1=\cdots=a_n=0, \] contrary to the hypotheses.
8. Q.E.D.
Step 6 gives the zero count, and step 7 explains the impossible range.