Location of the Roots of Algebraic Equations

A polynomial of the form

where $p_{1} \geq 0, p_{2} \geq 0, \ldots , p_{n} \geq 0$, $p_{1} + p_{2} + \cdots + p_{n} > 0$, has just one positive zero.

If $z_{0}$ is a zero of the polynomial

then $\left|z_{0}\right|$ is not larger than the only positive zero $\zeta$ of the polynomial $z^{n} - \left|a_{1}\right|z^{n-1} - \left|a_{2}\right|z^{n-2} - \cdots - \left|a_{n}\right|$.

Assume $a_{n} \neq 0$. The absolute value of none of the zeros of the polynomial

is smaller than the only positive zero $\zeta$ of the polynomial

All the zeros of the polynomial $z^{n} + c$ are on the circle centred at $z = 0$ with radius $\left|c\right|^{1/n}$.

Let $c_{1}, c_{2}, \ldots , c_{n}$ be positive numbers and

The absolute values of the zeros of the polynomial

are not larger than

The absolute values of the roots of the equation

are not larger than the largest among the numbers

also they are not larger than the largest of the numbers

and they are certainly smaller than the largest of the numbers

Assume

The polynomial

cannot have a zero in the unit disc $\left|z\right| \leq 1$.

Suppose that all the coefficients $p_{1}, p_{2}, \ldots , p_{n}$ of the polynomial

are positive. Then the zeros of this polynomial lie in the annulus $\alpha \leq \left|z\right| \leq \beta$, where $\alpha$ is the smallest, $\beta$ the largest among the values

Let $a_{0}, a_{1}, a_{2}, \ldots , a_{n}$ be digits (in the ordinary decimal notation, that is integers between $0$ and $9$ inclusively) $n \geq 1$, $a_{n} \geq 1$. Then the zeros of the polynomial

are either in the open left half-plane or in the open disk

The best upper bound that may replace the last number is between $3$ and $4$.

We assume that all the zeros of the polynomial

are in the upper half-plane $\mathrm{Im}\left(z\right) > 0$. Let $a_{\nu } = \alpha_{\nu } + i\beta_{\nu }$, $\alpha_{\nu }, \beta_{\nu }$ real, $\nu = 0, 1, 2, \ldots , n$, and

Then the polynomials $U(z)$ and $V(z)$ have only real zeros.

Let $P(z) = 0$ stand for an algebraic equation of degree $n$ all the zeros of which are in the unit circle $\left|z\right| < 1$. Replacing each coefficient of $P(z)$ by its conjugate we obtain the polynomial $\overline{P}(z)$. We define $P^{*}(z) = z^{n}\overline{P}(z^{-1})$. The roots of the equation $P(z) + P^{*}(z) = 0$ are all on the unit circle $\left|z\right| = 1$.

[M. Fekete.] Define $\gamma = \lambda P(a) + \mu P(b)$, $0 < \lambda < 1$, $\lambda + \mu = 1$. If all the zeros of $P(z) - \gamma = a_{0}(z - z_{1})(z - z_{2})\cdots (z - z_{n})$ were outside the domain in question, the inequality

would hold, thus