Let $0 < \lambda_1 \leq \ldots \leq \lambda_n $ and $k_1, \ldots, k_n> 0$.

Let further $$ \begin{align} P(x)&:=\prod_{i=1}^n (x+\lambda_i) = (x+\lambda_1)\cdot \ldots \cdot (x+\lambda_n) \\ q_i(x) &:= \frac{P(x)}{(x+\lambda_i)}\\ S(x) &:= P(x) + \sum_{i=1}^n k_i q_i(x) \end{align} $$ be polynomials in the complex valued unknown $x$.

E.g. for $n=2$ we have

$$ S(x) = (x + \lambda_1)(x + \lambda_2) + k_1(x+\lambda_2) + k_2(x+\lambda_1). $$

I suspect that S(x) has no zeros with positive real part. I tried it numerically 100 times for $n=5$ with random numbers for $x_i, \lambda_i$ and always got 5 negative real roots. (So, I guess that actually there are no non-real roots, but that is not so important.)

From Descartes rule of signs we know that $S$ has no positive real roots (because all coefficients of $S$ are positive). But how can we exclude complex roots with positive real parts (or as a whole)?


For the special case $\lambda_1 =\ldots = \lambda_n =:\lambda <0$ the assertion is almost trivial. In this case we have

$$ S(x) = (x+ \lambda)^n + \sum_{i=0}^n k_i (x + \lambda)^{n-1} = (x + \lambda)^{n-1} (x + \lambda + \sum_{i=0}^n k_i), $$ i.e., we have $n-1$ zeros unchanged at $\lambda$ and one zero shifted by $\sum_{i=0}^n k_i$ to the left.


1 Answer 1


A former professor of me suggested the following argumentation:

First, we rewrite the polynomial $$ \begin{align} S(x) & = \prod_{j=1}^n (x + \lambda_{j}) + \sum_{i=1}^n k_i \cdot \prod_{j=1, j\neq i}^n (x + \lambda_{j}) \\ &= \sum_{i=1}^n \left( k_i \cdot \hspace{-2mm} \prod_{j=1, j\neq i}^n (x + \lambda_{j}) + \frac{1}{n} \prod_{j=1}^n x + \lambda_{j} \right) \notag \\ &= \sum_{i=1}^n \left( \prod_{j=1, j\neq i}^n (x + \lambda_{j}) \right)\cdot \left( \frac{x + \lambda_{i}}{n} + k_i \right) \\ &= \sum_{i=1}^n \frac{1}{n}\left( \prod_{j=1, j\neq i}^n (x + \lambda_{j}) \right)\cdot \left( x + \lambda_{i} + n\cdot k_i \right). \end{align} $$

Now we let $x = \mu e^{\alpha i}$ with $\mu, \alpha \geq 0$, i.e. $x$ is a complex number in $\mathcal Q_1$, the first quadrant of the complex plain.

Clearly, we have $x + \lambda_j \in \mathcal Q_1$. Further, we remind a) that the angle-arguments of complex numbers add up when multiplying these complex numbers and b) that $\mathrm{arc}(\frac{1}{n} e^{i \alpha}) = \frac{\alpha}{n}$.

Now we see that $$ \frac{1}{n}\left( \prod_{j=1, j\neq i}^n (x + \lambda_{j}) \right)\cdot \left( x + \lambda_{i} + n\cdot k_i \right) \in \mathcal Q_1 $$ holds, because it is the average of $n$ complex numbers from $\mathcal Q_1$.

Finally, $S(x)$ is just the sum of $n$ of such "averages" and thus also in $\mathcal Q_1$. Because $k_i >0$ is a strict inequality, the value $S(x)$ can not be zero. This proves $S(\mathcal Q_1) \neq 0$. For the fourth quadrant an analogous argument holds.


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