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Article

An Eneström–Kakeya Theorem with Monotonicity Conditions on the Even- and Odd-Indexed Coefficients of a Polynomial

Department of Mathematics and Statistics, East Tennessee State University, Johnson City, TN 37614, USA
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Author to whom correspondence should be addressed.
Axioms 2025, 14(9), 651; https://doi.org/10.3390/axioms14090651
Submission received: 21 July 2025 / Revised: 20 August 2025 / Accepted: 20 August 2025 / Published: 22 August 2025

Abstract

The classical Eneström–Kakeya theorem states that an n-degree polynomial p ( z ) = k = 0 n a k z k with real coefficients satisfying 0 a 0 a 1 a n has all of its zeros in | z | 1 in the complex plane. Numerous generalizations of this result exist, many of them weakening the condition on the coefficients in order to be applicable to a larger class of polynomials. In this paper, a monotonicity condition on the real and imaginary parts of the even- and odd-indexed coefficients is imposed and bounds on the location of the zeros are established.
MSC:
12D10; 30C10; 30C15; 30E10

1. Introduction

The study of the zeros of polynomials has a rich history in the conceptual development of the concept of a “number,” as well as a tremendous number of applications. The theory of equations that grew out of the solutions to cubic and quartic equations in the first half of the 16th century led to the wide acceptance of negative numbers and complex numbers. Attempts to solve the general quintic equation ultimately led to the introduction of group theory in the early 19th century with the work of Niels Abel and Évarist Galois, followed by the development of modern algebra in general. A nice history of this is in [1]. Two classic works on the zeros of polynomials with an emphasis on the complex domain are [2,3]. In the absence of an algebraic formula to find the zeros of general n-degree polynomials, it is desirable to find approximations of the zeros. In this paper, we restrict the location of zeros of a polynomial of a complex variable by imposing a monotonicity condition on the coefficients. In Section 2 we state several related results that already exist in the literature and give a brief history. In Section 3 new results are presented, with Section 4 containing the proof. Section 5 describes potential applications and points out a number of directions that future research could take.

2. Previous Results

The study of polynomial zeros’ location with a monotonicity condition imposed on the coefficients dates back to 1893. In that year Gustav Eneström, while considering a problem that arose in the theory of pension funds, published (in Swedish) a bound on the modulus of the zeros of a polynomial with real positive monotone coefficients [4]. He published a verbatim translation into French of this work in 1920 [5]. Independent of Eneström, Sōichi Kakeya published the same result in 1912 (in English) [6]. The result is now known as the Eneström–Kakeya theorem and is as follows.
Theorem 1. 
Eneström –Kakeya Theorem. Let p ( z ) = j = 0 n a j z j be an n-degree polynomial such that 0 a 0 a 1 a 2 a n 1 a n . Then all the zeros of p lie in | z | 1 .
The first generalization of Theorem 1 seems to be the following due to Joyal, Labelle, and Rahman [7]. It eliminates the condition of positivity of the coefficients but still requires them to be real.
Theorem 2. 
Let p ( z ) = j = 0 n a j z j be an n-degree polynomial such that a 0 a 1 a 2 a n 1 a n . Then all the zeros of p lie in
| z | a n a 0 + | a 0 | a n .
Notice that in the event a 0 0 , Theorem 2 reduces to Theorem 1. Aziz and Mohammad considered the zeros of analytic functions of a complex variable with complex coefficients by imposing a monotonicity condition with a reversal on the real and imaginary parts of the coefficients [8]. With modifications, this idea was applied to p ( z ) = j = 0 n a j z j , where a j = α j + i β j with α j , β j R for 0 j n was considered. The result concerning the bound on the location of the zeros is as follows.
Theorem 3. 
Let p ( z ) = j = 0 n a j z j be an n-degree polynomial with coefficients a j = α j + i β j and let t > 0 ,   k ,   r N { 0 } be such that
α 0 t α 1 t 2 α 2 t k α k t k + 1 α k + 1 t n α n   a n d
β 0 t β 1 t 2 β 2 t r β r t r + 1 β r + 1 t n β n .
Then all the zeros of p lie in
min t | a 0 | M 1 , t | z | max M 2 | a n | , 1 t
where
M 1 = ( α 0 + β 0 ) + 2 ( α k t k + β r t r ) ( α n + β n | a n | ) t n
and
M 2 = | a 0 | t n + 1 t n 1 ( α 0 + β 0 ) t ( α n + β n ) + ( t 2 + 1 ) ( t n k 1 α k + t n r 1 β r )
+ ( t 2 1 ) j = 1 k 1 t n j 1 α j + j = 1 r 1 t n j 1 β j + ( 1 t 2 ) j = k + 1 n 1 t n j 1 α j + j = r + 1 n 1 t n j 1 β j
The monotonic conditions of the results in [8,9] were applied separately to the even- and odd-indexed coefficients of a polynomial in [10], which yields the following theorem.
Theorem 4. 
Let p ( z ) = j = 0 n a j z j be an n-degree polynomial with coefficients a j = α j + i β j . Let t > 0 ,   k ,   N { 0 } ,   s ,   q N be such that
α 0 t 2 α 2 t 4 α 4 t 2 k α 2 k t 2 k + 2 α 2 k + 2 t 2 n / 2 α 2 n / 2 ,
α 1 t 2 α 3 t 4 α 5 t 2 2 α 2 1 t 2 α 2 + 1 t 2 n / 2 α 2 ( n + 1 ) / 2 1 ,
β 0 t 2 β 2 t 4 β 4 t 2 s β 2 s t 2 s + 2 β 2 s + 2 t 2 n / 2 β 2 n / 2 ,   a n d
β 1 t 2 β 3 t 4 β 5 t 2 q 2 β 2 q 1 t 2 q α 2 q + 1 t 2 n / 2 β 2 ( n + 1 ) / 2 1 .
Then all the zeros of p lie in
min t | a 0 | M 1 , t | z | max M 2 | a n | , 1 t
where
M 1 = ( α 0 + β 0 ) + ( | α 1 | + | β 1 | ) t ( α 1 + β 1 ) t + 2 ( α 2 k t 2 k + α 2 l 1 t 2 l 1 + β 2 s t 2 s
+ β 2 q 1 t 2 q 1 ) + ( | α n 1 + | β n 1 | ) t n 1 ( α n 1 + β n 1 ) t n 1
+ ( | α n | + | β n | ) t n + ( | α n 1 | + | β n 1 | ) t n ,
and
M 2 = ( | a 0 | α 0 β 0 ) t n + 3 + ( | a 1 | α 1 β 1 ) t n + 2 + ( t 4 + 1 ) ( α 2 k t n 1 2 k + α 2 1 t n 2
+ β 2 s t n 1 2 s + β 2 q 1 t n 2 q ) ( α n 1 + β n 1 ) + | a n 1 | ( α n + β n ) t 1
+ ( t 4 1 ) j = 0 , j   e v e n 2 k 2 α j t n 1 j + j = 1 , j   o d d 2 3 α j t n 1 j + j = 0 , j   e v e n 2 s 2 β j t n 1 j
j = 1 , j   o d d 2 q 3 β j t n 1 j j = 2 k + 2 , j   e v e n 2 n / 2 α j t n 1 j j = 2 + 1 , j   o d d 2 ( n + 1 ) / 2 1 α j t n 1 j
j = 2 s + 2 , j   e v e n 2 n / 2 β j t n 1 j j = 2 q + 1 , j   o d d 2 ( n + 1 ) / 2 1 β j t n 1 j
The purpose of this paper is to impose these types of monotonicity conditions on the complex coefficients of a polynomial, combined with a second reversal in each inequality. In this way, a result is presented that is applicable to a larger class of polynomials than the other results mentioned above.

3. Results

We impose the monotonicity condition given at the end of Section 2, but with a second reversal. For a polynomial with complex coefficients, we have the following.
Theorem 5. 
Let p ( z ) = j = 0 n a j z j be an n-degree polynomial with coefficients a j = α j + i β j . Let t > 0 ,   k ,   s ,   p ,   r N { 0 } ,   l ,   q ,   u ,   v N be such that
α 0 α 2 t 2 α 2 k t 2 k α 2 k + 2 t 2 k + 2 α 2 s t 2 s α 2 s + 2 t 2 s + 2 α 2 n / 2 t 2 n / 2 ,
α 1 α 3 t 2 α 2 l 1 t 2 l 2 α 2 l + 1 t 2 l α 2 q 1 t 2 q 2
α 2 q + 1 t 2 q α 2 ( n + 1 ) / 2 1 t 2 ( n + 1 ) / 2 ,
β 0 β 2 t 2 β 2 p t 2 p β 2 p + 2 t 2 p + 2 β 2 r t 2 r β 2 r + 2 t 2 r + 2
β 2 n / 2 t 2 n / 2 , β 1 β 3 t 2 β 2 u 1 t 2 u 2 β 2 u + 1 t 2 u β 2 v 1 t 2 v 2
β 2 v + 1 t 2 v β 2 ( n + 1 ) / 2 1 t 2 ( n + 1 ) / 2 .
Then all the zeros of p lie in
min t | a 0 | M 1 , t | z | max M 2 | a n | , 1 t
where
M 1 = ( α 0 + β 0 ) + ( | α 1 | + | β 1 | ) t ( α 1 + β 1 ) t + 2 ( α 2 k t 2 k α 2 s t 2 s + α 2 l 1 t 2 l 1 α 2 q 1 t 2 q 1 + β 2 p t 2 p β 2 r t 2 r + β 2 u 1 t 2 u 1 β 2 v 1 t 2 v 1 ) + ( | α n 1 | + | β n 1 | ) t n 1 + ( α n 1 + β n 1 ) t n 1 + ( | α n | + | β n | ) t n + ( α n + β n ) t n
and
M 2 = t n + 3 ( | a 0 | α 0 β 0 ) + t n + 2 ( | a 1 | α 1 β 1 ) + | a n 1 | + ( α n 1 + β n 1 ) + ( α n + β n ) t 1 + ( t 4 + 1 ) ( α 2 k t n 1 2 k α 2 s t n 1 2 s + α 2 l 1 t n 2 l α 2 q 1 t n 2 q + β 2 p t n 1 2 p β 2 r t n 1 2 r + β 2 u 1 t n 2 u β 2 v 1 t n 2 v ) + ( t 4 1 ) ( j = 0 , j   e v e n 2 k 2 α j t n 1 j j = 2 k + 2 , j   e v e n 2 s 2 α j t n 1 j + j = 2 s + 2 , j   e v e n 2 n / 2 α j t n 1 j + j = 1 , j   o d d 2 l 3 α j t n 1 j j = 2 l + 1 , j   o d d 2 q 3 α j t n 1 j + j = 2 q + 1 , j   o d d 2 ( n + 1 ) / 2 1 α j t n 1 j + j = 0 , j   e v e n 2 p 2 β j t n 1 j j = 2 p + 2 , j   e v e n 2 r 2 β j t n 1 j + j = 2 r + 2 , j   e v e n 2 n / 2 β j t n 1 j + j = 1 , j   o d d 2 u 3 β j t n 1 j j = 2 u + 1 , j   o d d 2 v 3 β j t n 1 j + j = 2 v + 1 , j   o d d 2 ( n + 1 ) / 2 1 β j t n 1 j ) .
We give a proof of Theorem 5 in Section 4. With t = 1 we get the following corollary to Theorem 5.
Corollary 1. 
Let p ( z ) = j = 0 n a j z j be an n-degree polynomial with coefficients a j = α j + i β j . Let k ,   s ,   p ,   r N { 0 } ,   l ,   q ,   u ,   v N be such that
α 0 α 2 α 2 k α 2 k + 2 α 2 s α 2 s + 2 α 2 n / 2 ,
α 1 α 3 α 2 l 1 α 2 l + 1 α 2 q 1
α 2 q + 1 t α 2 ( n + 1 ) / 2 1 ,
β 0 β 2 β 2 p β 2 p + 2 β 2 r β 2 r + 2 β 2 n / 2 ,
β 1 β 3 β 2 u 1 β 2 u + 1 β 2 v 1
β 2 v + 1 β 2 ( n + 1 ) / 2 1 .
Then all the zeros of p lie in
min | a 0 | M 1 , 1 | z | max M 2 | a n | , 1
where
M 1 = ( α 0 + β 0 ) + ( | α 1 | + | β 1 | ) ( α 1 + β 1 ) + 2 ( α 2 k α 2 s + α 2 l 1 α 2 q 1 + β 2 p t β 2 r + β 2 u 1 β 2 v 1 ) + ( | α n 1 | + | β n 1 | ) + ( α n 1 + β n 1 ) + ( | α n | + | β n | ) + ( α n + β n )
and
M 2 = ( | a 0 | α 0 β 0 ) + ( | a 1 | α 1 β 1 ) + | a n 1 | + ( α n 1 + β n 1 ) + ( α n + β n ) + 2 ( α 2 k α 2 s + α 2 l 1 α 2 q 1 + β 2 p β 2 r + β 2 u 1 β 2 v 1 ) .
If we apply Theorem 5 to a real polynomial, in which case β j = 0 for all 0 j n , then we get the next corollary.
Corollary 2. 
Let p ( z ) = j = 0 n a j z j be an n-degree polynomial with real coefficients. Let t > 0 , k ,   s N { 0 } , l ,   q N be such that
a 0 a 2 t 2 a 2 k t 2 k a 2 k + 2 t 2 k + 2 a 2 s t 2 s a 2 s + 2 t 2 s + 2 a 2 n / 2 t 2 n / 2
a n d   a 1 a 3 t 2 a 2 l 1 t 2 l 2 a 2 l + 1 t 2 l a 2 q 1 t 2 q 2
a 2 q + 1 t 2 q a 2 ( n + 1 ) / 2 1 t 2 ( n + 1 ) / 2 .
Then all the zeros of p lie in
min t | a 0 | M 1 , t | z | max M 2 | a n | , 1 t
where
M 1 = a 0 + | a 1 | t a 1 t + 2 ( a 2 k t 2 k a 2 s t 2 s + a 2 l 1 t 2 l 1 a 2 q 1 t 2 q 1 ) + | a n 1 | t n 1 + a n 1 t n 1 + | a n | t n + a n t n
and
M 2 = t n + 3 ( | a 0 | a 0 ) + t n + 2 ( | a 1 | a 1 ) + | a n 1 | + a n 1 + a n t 1 + ( t 4 + 1 ) ( a 2 k t n 1 2 k a 2 s t n 1 2 s + a 2 l 1 t n 2 l a 2 q 1 t n 2 q ) + ( t 4 1 ) ( j = 0 , j   e v e n 2 k 2 a j t n 1 j j = 2 k + 2 , j   e v e n 2 s 2 a j t n 1 j + j = 2 s + 2 , j   e v e n 2 n / 2 a j t n 1 j + j = 1 , j   o d d 2 l 3 a j t n 1 j j = 2 l + 1 , j   o d d 2 q 3 a j t n 1 j + j = 2 q + 1 , j   o d d 2 ( n + 1 ) / 2 1 a j t n 1 j ) .
With t = 1 we get the following corollary to Corollary 2.
Corollary 3. 
Let p ( z ) = j = 0 n a j z j be an n-degree polynomial with real coefficients. Let k ,   s N { 0 } ,   l ,   q N be such that
a 0 a 2 a 2 k a 2 k + 2 a 2 s a 2 s + 2 a 2 n / 2   a n d
a 1 a 3 a 2 l 1 a 2 l + 1 a 2 q 1 a 2 q + 1 a 2 ( n + 1 ) / 2 1 .
Then all the zeros of p lie in
min | a 0 | M 1 , 1 | z | max M 2 | a n | , 1
where
M 1 = a 0 + | a 1 | a 1 + 2 ( a 2 k a 2 s + a 2 l 1 a 2 q 1 ) + | a n 1 | + a n 1 + | a n | + a n
and
M 2 = ( | a 0 | a 0 ) + ( | a 1 | a 1 ) + | a n 1 | + a n 1 + a n + 2 ( a 2 k a 2 s + a 2 l 1 a 2 q 1 ) .
The following example illustrates an application of Corollary 3 to a specific polynomial.
Example 1. 
Consider the real polynomial p ( x ) = 1 + x 2 + 99 100 x 4 + x 6 . Here, we have n = 6 , a 0 = a 2 = 1 , a 4 = 0.99 , a 6 = 1 , and a 1 = a 3 = a 5 = 0 . The odd-indexed coefficients satisfy the hypotheses of Corollary 3. For the even-indexed coefficients, we have a 0 a 2 a 4 a 6 , so that with 2 k = 2 and 2 s = 4 we have a 0 a 2 k a 2 s a n , and the even-indexed coefficients satisfy the hypotheses of Corollary 3. Corollary 3 now implies that the zeros of p have a modulus between 50 / 51 0.98039 and 51 / 50 = 1.02 . We can, in fact, algebraically find the zeros of p since we can replace x 2 with y to get the cubic equation 1 + y + 0.99 y 2 + y 3 = 0 . Applying the cubic formula (see page 17 of [11]), we find that the values of y are as follows. With α = 1000 12060822 3338433 3 , we have that the desired values of y are the real number 1 100 33 + α 3 2 / 3 6733 α and the two complex conjugates 33 100 1 ± i 3 α 200 ( 3 2 / 3 ) + 6733 ( 1 i 3 ) 200 α . We then find the six values of x by taking the two square roots of each value of y. Approximating, we find that the values of x are ± 0.99751 i , ± ( 0.70888 0.70710 i ) , and ± ( 0.70888 + 0.70710 i ) . The purely imaginary value is of modulus 0.99751 and the four complex values are of modulus approximately 1.00125 . That is, all zeros of p lie in 50 / 51 | z | 51 / 50 , as implied by Corollary 3.
We now give an example to illustrate the more general Corollary 1.
Example 2. 
Consider the complex polynomial p ( z ) = ( 1 + i ) + ( 1 + i ) z 2 + ( 1 + 99 100 i ) z 4 + ( 1 + i ) z 6 . We have n = 6 , α 0 = α 2 = α 4 = α 6 = β 0 = β 2 = β 6 = 1 , β 4 = 99 / 100 , and α 1 = α 3 = α 5 = β 1 = β 3 = β 5 = 0 . The odd-indexed coefficients satisfy the hypotheses of Corollary 1. For the even-indexed coefficients, we take k = p = 1 and s = r = 2 . We then have α 0 α 2 α 4 α 6 and β 0 β 2 β 4 β 6 so that the even-indexed coefficients satisfy the hypotheses of Corollary 1. Corollary 1 implies that the zeros of p have moduli between 50 2 / 101 0.70011 and 1 + 2 / 100 1.01414 . We can find the exact values of the zeros, similarly to the technique of Example 2. Approximating the zeros we find that the values of z are ± ( 0.00125 + 0.99875 i ) , ± ( 0.70799 + 0.70799 i ) , and ± ( 0.70799 0.70622 i ) . These are of approximate moduli 0.99875 , 1.00125 , and 0.999998 , respectively. That is, all zeros of p lie in 50 2 / 101 | z | 1 + 2 / 100 , as implied by Corollary 1.

4. Proofs of the Result

We now present a proof of Theorem 5.
Proof. 
First, consider the polynomial
g 1 ( z ) = ( t 2 z 2 ) p ( z ) = t 2 a 0 + a 1 t 2 z + j = 2 n ( a j t 2 a j 2 ) z j a n 1 z n + 1 a n z n + 2 = t 2 a 0 + G 1 ( z ) .
On | z | = t we have by the triangle inequality
| G 1 ( z ) | | a 1 | t 3 + j = 2 n | a j t 2 a j 2 | t j + | a n 1 | t n + 1 + | a n | t n + 2 ( | α 1 | + | β 1 | ) t 3 + j = 2 n | α j t 2 α j 2 | t j + j = 2 n | β j t 2 β j 2 | t j + ( | α n 1 | + | β n 1 | ) t n + 1 + ( | α n | + | β n | ) t n + 2 = ( | α 1 | + | β 1 | ) t 3 + j = 2 , j   even 2 k | α j t 2 α j 2 | t j + j = 2 k + 2 , j   even 2 s | α j t 2 α j 2 | t j + j = 2 s + 2 , j   even 2 n / 2 | α j t 2 α j 2 | t j + j = 3 , j   odd 2 l 1 | α j t 2 α j 2 | t j + j = 2 l + 1 , j   odd 2 q 1 | α j t 2 α j 2 | t j + j = 2 q + 1 , j   odd 2 ( n + 1 ) / 2 | α j t 2 α j 2 | t j + j = 2 , j   even 2 p | β j t 2 β j 2 | t j + j = 2 p + 2 , j   even 2 r | β j t 2 β j 2 | t j + j = 2 r + 2 , j   even 2 n / 2 | β j t 2 β j 2 | t j + j = 3 , j   odd 2 u 1 | β j t 2 β j 2 | t j + j = 2 u + 1 , j   odd 2 v 1 | β j t 2 β j 2 | t j + j = 2 v + 1 , j   odd 2 ( n + 1 ) / 2 | β j t 2 β j 2 | t j + ( | α n 1 | + | β n 1 | ) t n + 1 + ( | α n | + | β n | ) t n + 2 = ( | α 1 | + | β 1 | ) t 3 + j = 2 , j   even 2 k ( α j t 2 α j 2 ) t j + j = 2 k + 2 , j   even 2 s ( α j 2 α j t 2 ) t j + j = 2 s + 2 , j   even 2 n / 2 ( α j t 2 α j 2 ) t j + j = 3 , j   odd 2 l 1 ( α j t 2 α j 2 ) t j + j = 2 l + 1 , j   odd 2 q 1 ( α j 2 α j t 2 ) t j + j = 2 q + 1 , j   odd 2 ( n + 1 ) / 2 ( α j t 2 α j 2 ) t j + j = 2 , j   even 2 p ( β j t 2 β j 2 ) t j + j = 2 p + 2 , j   even 2 r ( β j 2 β j t 2 ) t j + j = 2 r + 2 , j   even 2 n / 2 ( β j t 2 β j 2 ) t j + j = 3 , j   odd 2 u 1 ( β j t 2 β j 2 ) t j + j = 2 u + 1 , j   odd 2 v 1 ( β j 2 β j t 2 ) t j + j = 2 v + 1 , j   odd 2 ( n + 1 ) / 2 ( β j t 2 β j 2 ) t j + ( | α n 1 | + | β n 1 | ) t n + 1 + ( | α n | + | β n | ) t n + 2 = ( | α 1 | + | β 1 | ) t 3 + j = 2 , j   even 2 k α j t j + 2 j = 2 , j   even 2 k α j 2 t j + j = 2 k + 2 , j   even 2 s α j 2 t j j = 2 k + 2 , j   even 2 s α j t j + 2 + j = 2 s + 2 , j   even 2 n / 2 α j t j + 2 j = 2 s + 2 , j   even 2 n / 2 α j 2 t j + j = 3 , j   odd 2 l 1 α j t j + 2 j = 3 , j   odd 2 l 1 α j 2 t j + j = 2 l + 1 , j   odd 2 q 1 α j 2 t j j = 2 l + 1 , j   odd 2 q 1 α j t j + 2 + j = 2 q + 1 , j   odd 2 ( n + 1 ) / 2 α j t j + 2 j = 2 q + 1 , j   odd 2 ( n + 1 ) / 2 α j 2 t j + j = 2 , j   even 2 p β j t j + 2 j = 2 , j   even 2 p β j 2 t j + j = 2 p + 2 , j   even 2 r β j 2 t j j = 2 p + 2 , j   even 2 r β j t j + 2 + j = 2 r + 2 , j   even 2 n / 2 β j t j + 2 j = 2 r + 2 , j   even 2 n / 2 β j 2 t j + j = 3 , j   odd 2 u 1 β j t j + 2 j = 3 , j   odd 2 u 1 β j 2 t j + j = 2 u + 1 , j   odd 2 v 1 β j 2 t j j = 2 u + 1 , j   odd 2 v 1 β j t j + 2 + j = 2 v + 1 , j   odd 2 ( n + 1 ) / 2 β j t j + 2 j = 2 v + 1 , j   odd 2 ( n + 1 ) / 2 β j 2 t j + ( | α n 1 | + | β n 1 | ) t n + 1 + ( | α n | + | β n | ) t n + 2 = ( | α 1 | + | β 1 | ) t 3 + j = 2 , j   even 2 k 2 α j t j + 2 + α 2 k t 2 k + 2 α 0 t 2 j = 2 , j   even 2 k 2 α j t j + 2 + α 2 k t 2 k + 2 + j = 2 k + 2 , j   even 2 s 2 α j t j + 2 j = 2 k + 2 , j   even 2 s 2 α j t j + 2 α 2 s t 2 s + 2 + j = 2 s + 2 , j   even 2 n / 2 2 α j t j + 2 + α n / 2 t n / 2 + 2 α 2 s t 2 s + 2 j = 2 s + 2 , j   even 2 n / 2 2 α j t j + 2 + j = 3 , j   odd 2 l 3 α j t j + 2 + α 2 l 1 t 2 l + 1 α 1 t 3 j = 3 , j   odd 2 l 3 α j t j + 2 + α 2 l 1 t 2 l + 2 + j = 2 l + 1 , j   odd 2 q 3 α j t j + 2 j = 2 l + 1 , j   odd 2 q 3 α j t j + 2 α 2 q 1 t 2 q + j = 2 q + 1 , j   odd 2 ( n + 1 ) / 2 2 α j t j + 2 + α 2 ( n + 1 ) / 2 t 2 ( n + 1 ) / 2 + 2 α 2 q 1 t 2 q + 1 j = 2 q + 1 , j   odd 2 ( n + 1 ) / 2 2 α j t j + 2 + j = 2 , j   even 2 p 2 β j t j + 2 + β 2 p 2 p + 2 β 0 t 2 j = 2 , j   even 2 p 2 β j t j + 2 β 2 p t 2 p + 2 + j = 2 p + 2 , j   even 2 r 2 β j t j + 2 j = 2 p + 2 , j   even 2 r 2 β j t j + 2 β 2 r t 2 r + 2 + j = 2 r + 2 , j   even 2 n / 2 2 β j t j + 2 + β 2 n / 2 t 2 n / 2 + 2 β 2 r t 2 r + 2 j = 2 r + 2 , j   even 2 n / 2 2 β j t j + 2 + j = 3 , j   odd 2 u 3 β j t j + 2 + β 2 u 1 t 2 u + 1 β 1 t 3 j = 3 , j   odd 2 u 3 β j t j + 2 + β 2 u 1 t 2 u + 1 + j = 2 u + 1 , j   odd 2 v 3 β j t j + 2 j = 2 u + 1 , j   odd 2 v 3 β j t j + 2 β 2 v 1 t 2 v + 1 + j = 2 v + 1 , j   odd 2 ( n + 1 ) / 2 2 β j t j + 2 + β 2 ( n + 1 ) / 2 t 2 ( n + 1 ) / 2 + 2 β 2 v 1 t 2 v + 1 j = 2 v + 1 , j   odd 2 ( n + 1 ) / 2 2 β j t j + 2 + ( | α n 1 | + | β n 1 | ) t n + 1 + ( | α n | + | β n | ) t n + 2 = ( | α 1 | + | β 1 | ) t 3 ( α 0 + β 0 ) t 2 ( α 1 + β 1 ) t 3 + 2 ( α 2 k t 2 k + 2 α 2 s t 2 s + 2 + α 2 l 1 t 2 l + 1 α 2 q 1 t 2 q + 1 + β 2 p t 2 p + 2 β 2 r t 2 r + 2 + β 2 u 1 t 2 u + 1 β 2 v 1 t 2 v + 1 ) ( α 2 n / 2 + β 2 n / 2 ) t 2 n / 2 + 2 + ( α 2 ( n + 1 ) / 2 + β 2 ( n + 1 ) / 2 ) t 2 ( n + 1 ) / 2 + 2 + ( | α n 1 | + | β n 1 | ) t n + 1 + ( | α n | + | β n | ) t n 2 = ( | α 1 | + | β 1 | ) t 3 ( α 0 + β 0 ) t 2 ( α 1 + β 1 ) t 3 + 2 ( α 2 k t 2 k + 2 α 2 s t 2 s + 2 + α 2 l 1 t 2 l + 1 α 2 q 1 t 2 q + 1 + β 2 p t 2 p + 2 β 2 r t 2 r + 2 + β 2 u 1 t 2 u + 1 β 2 v 1 t 2 v + 1 ) + ( | α n 1 | + | β n 1 | ) t n + 1 + ( α n 1 + β n 1 ) t n + 1 + ( | α n | + | β n | ) t n + 2 + ( α n + β n ) t n + 2 : = t 2 M 1 .
Applying Schwarz’s Lemma [12] (p. 168) to G 1 ( z ) gives
| G 1 ( z ) | t 2 M 1 | z | t = t M 1 | z | f o r | z | t ,
and so
g 1 ( z ) = | t 2 a 0 + G 1 ( z ) | t 2 | a 0 | | G 1 ( z ) | t 2 | a 0 | t M 1 | z | f o r | z | t .
Therefore, if | z | < R 1 = min t | a 0 | M 1 , t , then g 1 ( z ) 0 so that p ( z ) 0 .
Now consider the polynomial
g 2 ( z ) = ( t 2 z 2 ) p ( z ) = t 2 a 0 + a 1 t 2 z + j = 2 n ( a j t 2 a j 2 ) z j a n 1 z n + 1 a n z n + 2 = a n z n + 2 + G 2 ( z ) .
Then
z n + 1 G 2 1 z = t 2 a 0 z n + 1 + a 1 t 2 z n + j = 2 n ( a j t 2 a j 2 ) z n + 1 j a n 1 .
When | z | = t we have by the triangle inequality
z n + 1 G 2 1 z = t 2 a 0 z n + 1 + a 1 t 2 z n + j = 2 n ( a j t 2 a j 2 ) z n + 1 j a n 1 t n + 3 | a 0 | + | a 1 | t n + 2 + j = 2 n | a j t 2 a j 2 | t n + 1 j + | a n 1 | t n + 3 | a 0 | + | a 1 | t n + 2 + | a n 1 | + j = 2 n | α j t 2 α j 2 | t n + 1 j + j = 2 n | β j t 2 β j 2 | t n + 1 j = t n + 3 | a 0 | + | a 1 | t n + 2 + | a n 1 | + j = 0 , j   even 2 k 2 | α j + 2 t 2 α j | t n 1 j + j = 2 k , j   even 2 s 2 | α j + 2 t 2 a j | t n 1 j + j = 2 s , j   even 2 n / 2 2 | α j + 2 t 2 α j | t n 1 j + j = 1 , j   odd 2 l 3 | α j + 2 t 2 α j | t n 1 j + j = 2 l 1 , j   odd 2 q 3 | α j + 2 t 2 α j | t n 1 j + j = 2 q 1 , j   odd 2 ( n + 1 ) / 2 3 | α j + 2 t 2 α j | t n 1 j + j = 0 , j   even 2 p 2 | β j + 2 t 2 β j | t n 1 j + j = 2 p , j   even 2 r 2 | β j + 2 t 2 β j | t n 1 j + j = 2 r , j   even 2 n / 2 2 | β j + 2 t 2 β j | t n 1 j + j = 1 , j   odd 2 u 3 | β j + 2 t 2 β j | t n 1 j + j = 2 u 1 , j   odd 2 v 3 | β j + 2 t 2 β j | t n 1 j + j = 2 v 1 , j   odd 2 ( n + 1 ) / 2 3 | β j + 2 t 2 β j | t n 1 j = t n + 3 | a 0 | + | a 1 | t n + 2 + | a n 1 | + j = 0 , j   even 2 k 2 ( α j + 2 t 2 α j ) t n 1 j j = 2 k , j   even 2 s 2 ( α j + 2 t 2 a j ) t n 1 j + j = 2 s , j   even 2 n / 2 2 ( α j + 2 t 2 α j ) t n 1 j + j = 1 , j   odd 2 l 3 ( α j + 2 t 2 α j ) t n 1 j j = 2 l 1 , j   odd 2 q 3 ( α j + 2 t 2 α j ) t n 1 j + j = 2 q 1 , j   odd 2 ( n + 1 ) / 2 3 ( α j + 2 t 2 α j ) t n 1 j + j = 0 , j   even 2 p 2 ( β j + 2 t 2 β j ) t n 1 j j = 2 p , j   even 2 r 2 ( β j + 2 t 2 β j ) t n 1 j + j = 2 r , j   even 2 n / 2 2 ( β j + 2 t 2 β j ) t n 1 j + j = 1 , j   odd 2 u 3 ( β j + 2 t 2 β j ) t n 1 j j = 2 u 1 , j   odd 2 v 3 ( β j + 2 t 2 β j ) t n 1 j + j = 2 v 1 , j   odd 2 ( n + 1 ) / 2 3 ( β j + 2 t 2 β j ) t n 1 j = t n + 3 | a 0 | + | a 1 | t n + 2 + | a n 1 | + j = 0 , j   even 2 k 2 α j + 2 t n + 1 j j = 0 , j   even 2 k 2 α j t n 1 j j = 2 k , j   even 2 s 2 α j + 2 t n + 1 j + j = 2 k , j   even 2 s 2 α j t n 1 j + j = 2 s , j   even 2 n / 2 2 α j + 2 t n + 1 j j = 2 s , j   even 2 n / 2 2 α j t n 1 j + j = 1 , j   odd 2 l 3 α j + 2 t n + 1 j j = 1 , j   odd 2 l 3 α j t n 1 j j = 2 l 1 , j   odd 2 q 3 α j + 2 t n + 1 j + j = 2 l 1 , j   odd 2 q 3 α j t n 1 j + j = 2 q 1 , j   odd 2 ( n + 1 ) / 2 3 α j + 2 t n + 1 j j = 2 q 1 , j   odd 2 ( n + 1 ) / 2 3 α j t n 1 j + j = 0 , j   even 2 p 2 β j + 2 t n + 1 j j = 0 , j   even 2 p 2 β j t n 1 j j = 2 p , j   even 2 r 2 β j + 2 t n + 1 j + j = 2 p , j   even 2 r 2 β j t n 1 j + j = 2 r , j   even 2 n / 2 2 β j + 2 t n + 1 j j = 2 r , j   even 2 n / 2 2 β j t n 1 j + j = 1 , j   odd 2 u 3 β j + 2 t n + 1 j j = 1 , j   odd 2 u 3 β j t n 1 j j = 2 u 1 , j   odd 2 v 3 β j + 2 t n + 1 j + j = 2 u 1 , j   odd 2 v 3 β j t n 1 j + j = 2 v 1 , j   odd 2 ( n + 1 ) / 2 3 β j + 2 t n + 1 j j = 2 v 1 , j   odd 2 ( n + 1 ) / 2 3 β j t n 1 j = t n + 3 | a 0 | + | a 1 | t n + 2 + | a n 1 | + j = 2 , j   even 2 k 2 α j t n + 3 j + α 2 k t n + 3 2 k α 0 t n 1 j = 2 , j   even 2 k 2 α j t n 1 j j = 2 k + 2 , j   even 2 s 2 α j t n + 3 j α 2 s t n + 3 2 s + α 2 k t n 1 2 k + j = 2 k + 2 , j   even 2 s 2 α j t n 1 j + j = 2 s + 2 , j   even 2 n / 2 2 α j t n + 3 j + α 2 n / 2 t n + 3 2 n / 2 α 2 s t n 1 2 s j = 2 s + 2 , j   even 2 n / 2 2 α j t n 1 j + j = 3 , j   odd 2 l 3 α j t n + 3 j + α 2 l 1 t n + 4 2 l α 1 t n 2 j = 3 , j   odd 2 l 3 α j t n 1 j j = 2 l + 1 , j   odd 2 q 3 α j t n + 3 j α 2 q 1 t n + 4 2 q + α 2 l 1 t n 2 l + j = 2 l + 1 , j   odd 2 q 3 α j t n 1 j + j = 2 q + 1 , j   odd 2 ( n + 1 ) / 2 3 α j t n + 3 j + α 2 ( n + 1 ) / 2 1 t n + 4 2 ( n + 1 ) / 2 α 2 q 1 t n 2 q j = 2 q + 1 , j   odd 2 ( n + 1 ) / 2 3 α j t n 1 j + j = 2 , j   even 2 p 2 β j t n + 3 j + β 2 p t n + 3 2 p β 0 t n 1 j = 2 , j   even 2 p 2 β j t n 1 j j = 2 p + 2 , j   even 2 r 2 β j t n + 3 j β 2 r t n + 3 2 r + β 2 p t n 1 2 p + j = 2 p + 2 , j   even 2 r 2 β j t n 1 j + j = 2 r + 2 , j   even 2 n / 2 2 β j t n + 3 j + β 2 n / 2 t n + 3 2 n / 2 β 2 r t n 1 2 r j = 2 r + 2 , j   even 2 n / 2 2 β j t n 1 j + j = 3 , j   odd 2 u 3 β j t n + 3 j + β 2 u t n + 4 2 u β 1 t n 2 j = 3 , j   odd 2 u 3 β j t n 1 j j = 2 u + 1 , j   odd 2 v 3 β j t n + 3 j β 2 v 1 t n + 4 2 v + β 2 u 1 t n 2 u + j = 2 u + 1 , j   odd 2 u 3 β j t n 1 j + j = 2 v + 1 , j   odd 2 ( n + 1 ) / 2 3 β j t n + 3 j + β 2 ( n + 1 ) / 2 1 t n + 4 2 ( n + 1 ) / 2 β 2 v 1 t n 2 v j = 2 v + 1 , j   odd 2 ( n + 1 ) / 2 3 β j t n 1 j = t n + 3 | a 0 | + | a 1 | t n + 2 + | a n 1 | α 0 t n 1 β 0 t n 1 α 1 t n 2 β 1 t n 2 + α 2 n / 2 t n + 3 2 n / 2 + β 2 n / 2 t n + 3 2 n / 2 + α 2 ( n + 1 ) / 2 1 t n + 4 2 ( n + 1 ) / 2 + β 2 ( n + 1 ) / 2 1 t n + 4 2 ( n + 1 ) / 2 + ( t 4 + 1 ) ( α 2 k t n 1 2 k α 2 s t n 1 2 s + α 2 l 1 t n 2 l α 2 q 1 t n 2 q + β 2 p t n 1 2 p β 2 r t n 1 2 r + β 2 u 1 t n 2 u β 2 v 1 t n 2 v ) + ( t 4 1 ) ( j = 2 , j   even 2 k 2 α j t n 1 j j = 2 k + 2 , j   even 2 s 2 α j t n 1 j + j = 2 s + 2 , j   even 2 n / 2 2 α j t n 1 j + j = 3 , j   odd 2 l 3 α j t n 1 j j = 2 l + 1 , j   odd 2 q 3 α j t n 1 j + j = 2 q + 1 , j   odd 2 ( n + 1 ) / 2 3 α j t n 1 j + j = 2 , j   even 2 p 2 β j t n 1 j j = 2 p + 2 , j   even 2 r 2 β j t n 1 j + j = 2 r + 2 , j   even 2 n / 2 2 β j t n 1 j + j = 3 , j   odd 2 u 3 β j t n 1 j j = 2 u + 1 , j   odd 2 v 3 β j t n 1 j + j = 2 v + 1 , j   odd 2 ( n + 1 ) / 2 3 β j t n 1 j ) = t n + 3 | a 0 | α 0 t n + 3 + α 0 t n + 3 α 0 t n 1 β 0 t n + 3 + β 0 t n + 3 β 0 t n 1 + | a 1 | t n + 2 α 1 t n + 2 + α 1 t n + 2 α 1 t n 2 β 1 t n + 2 + β 1 t n + 2 β 1 t n 2 + α 2 n / 2 t n + 3 2 n / 2 α 2 n / 2 t n 1 2 n / 2 + α 2 n / 2 t n 1 2 n / 2 + β 2 n / 2 t n + 3 2 n / 2 β 2 n / 2 t n 1 2 n / 2 + β 2 n / 2 t n 1 2 n / 2 + α 2 ( n + 1 ) / 2 1 t n + 4 2 ( n + 1 ) / 2 α 2 ( n + 1 ) / 2 1 t n 2 ( n + 1 ) / 2 + α 2 ( n + 1 ) / 2 1 t n 2 ( n + 1 ) / 2 + β 2 ( n + 1 ) / 2 1 t n + 4 2 ( n + 1 ) / 2 β 2 ( n + 1 ) / 2 1 t n 2 ( n + 1 ) / 2 + β 2 ( n + 1 ) / 2 1 t n 2 ( n + 1 ) / 2 + | a n 1 | ( t 4 + 1 ) ( α 2 k t n 1 2 k α 2 s t n 1 2 s + α 2 l 1 t n 2 l α 2 q 1 t n 2 q + β 2 p t n 1 2 p β 2 r t n 1 2 r + β 2 u 1 t n 2 u β 2 v 1 t n 2 v ) + ( t 4 1 ) ( j = 2 , j   even 2 k 2 α j t n 1 j j = 2 k + 2 , j   even 2 s 2 α j t n 1 j + j = 2 s + 2 , j   even 2 n / 2 2 α j t n 1 j + j = 3 , j   odd 2 l 3 α j t n 1 j j = 2 l + 1 , j   odd 2 q 3 α j t n 1 j + j = 2 q + 1 , j   odd 2 ( n + 1 ) / 2 3 α j t n 1 j + j = 2 , j   even 2 p 2 β j t n 1 j j = 2 p + 2 , j   even 2 r 2 β j t n 1 j + j = 2 r + 2 , j   even 2 n / 2 2 β j t n 1 j + j = 3 , j   odd 2 u 3 β j t n 1 j j = 2 u + 1 , j   odd 2 v 3 β j t n 1 j + j = 2 v + 1 , j   odd 2 ( n + 1 ) / 2 3 β j t n 1 j ) = t n + 3 ( | a 0 | α 0 β 0 ) + ( t 4 1 ) ( α 0 + β 0 ) t n 1 + t n + 2 ( | a 1 | α 1 β 1 ) ( t 4 1 ) ( α 1 + β 1 ) t n 2 + ( t 4 1 ) ( α 2 n / 2 + β 2 n / 2 ) t n 1 2 n / 2 + ( α 2 n / 2 + β 2 n / 2 ) t n 1 2 n / 2 + ( t 4 1 ) ( α 2 ( n + 1 ) / 2 1 + β 2 ( n + 1 ) / 2 1 ) t n 2 ( n + 1 ) / 2 + ( α 2 ( n + 1 ) / 2 1 + β 2 ( n + 1 ) / 2 1 ) t n 2 ( n + 1 ) / 2 + | a n 1 | ( t 4 + 1 ) ( α 2 k t n 1 2 k α 2 s t n 1 2 s + α 2 l 1 t n 2 l α 2 q 1 t n 2 q + β 2 p t n 1 2 p β 2 r t n 1 2 r + β 2 u 1 t n 2 u β 2 v 1 t n 2 v ) + ( t 4 1 ) ( j = 2 , j   even 2 k 2 α j t n 1 j j = 2 k + 2 , j   even 2 s 2 α j t n 1 j + j = 2 s + 2 , j   even 2 n / 2 2 α j t n 1 j + j = 3 , j   odd 2 l 3 α j t n 1 j j = 2 l + 1 , j   odd 2 q 3 α j t n 1 j + j = 2 q + 1 , j   odd 2 ( n + 1 ) / 2 3 α j t n 1 j + j = 2 , j   even 2 p 2 β j t n 1 j j = 2 p + 2 , j   even 2 r 2 β j t n 1 j + j = 2 r + 2 , j   even 2 n / 2 2 β j t n 1 j + j = 3 , j   odd 2 u 3 β j t n 1 j j = 2 u + 1 , j   odd 2 v 3 β j t n 1 j + j = 2 v + 1 , j   odd 2 ( n + 1 ) / 2 3 β j t n 1 j ) = t n + 3 ( | a 0 | α 0 β 0 ) + t n + 2 ( | a 1 | α 1 β 1 ) + ( α n 1 + β n 1 ) + ( α n + β n ) t 1 + | a n 1 | + ( t 4 + 1 ) ( α 2 k t n 1 2 k α 2 s t n 1 2 s + α 2 l 1 t n 2 l α 2 q 1 t n 2 q + β 2 p t n 1 2 p β 2 r t n 1 2 r + β 2 u 1 t n 2 u β 2 v 1 t n 2 v ) + ( t 4 1 ) ( j = 0 , j   even 2 k 2 α j t n 1 j j = 2 k + 2 , j   even 2 s 2 α j t n 1 j + j = 2 s + 2 , j   even 2 n / 2 α j t n 1 j + j = 1 , j   odd 2 l 3 α j t n 1 j j = 2 l + 1 , j   odd 2 q 3 α j t n 1 j + j = 2 q + 1 , j   odd 2 ( n + 1 ) / 2 1 α j t n 1 j + j = 0 , j   even 2 p 2 β j t n 1 j j = 2 p + 2 , j   even 2 r 2 β j t n 1 j + j = 2 r + 2 , j   even 2 n / 2 β j t n 1 j + j = 1 , j   odd 2 u 3 β j t n 1 j j = 2 u + 1 , j   odd 2 v 3 β j t n 1 j + j = 2 v + 1 , j   odd 2 ( n + 1 ) / 2 1 β j t n 1 j ) : = M 2 .
By the Maximum Modulus Theorem, | z n + 1 G 2 ( 1 z ) | M 2 when | z | t and hence | G 2 ( z ) | M 2 | z | n + 1 for | z | 1 / t . So for | z | 1 / t ,
| g 2 ( z ) | = | a n z n + 2 + G 2 ( z ) | | a n | | z | n + 2 M 2 | z | n + 1 = | z | n + 1 ( | a n | | z | M 2 ) .
We have g 2 ( z ) 0 (and consequently p ( z ) 0 ) as long as | z | > max M 1 | a n | , 1 t , which proves the theorem. □

5. Discussion

We now discuss applications of Eneström–Kakeya theorems, and the possible direction of future research, based on the results of this paper.

5.1. Applications

Applications involving zeros of polynomials are seen early in an undergraduate education in mathematics or science. The study of extrema of functions in a calculus class is initially illustrated with polynomial functions, or at least algebraic functions. Finding critical values often involves finding zeros of polynomials. The characteristic equation of a linear differential equation with constant coefficients is a polynomial, the zeros of which (both real and complex) yield the general solution of the differential equation. Similarly, in a system of linear homogeneous differential equations with constant coefficients, the solutions are determined from the eigenvalues of a matrix, and (as with any computation involving real or complex matrices) the eigenvalues are the zeros of the characteristic polynomial of the matrix. In a nonlinear differential equation, the eigenvalues of the linearization of the equation at critical points allow for the classification of the stability of the critical points. These applications of the zeros of polynomials are presented in most introductory-level differential equations textbooks. An example of such a book, which also addresses nonlinear equations and stability, is [13].

5.2. Future Research

Future research could focus on more general monotonicity conditions on the coefficients of a real or complex polynomial. For example, monotonicity conditions such as those discussed above could be imposed on the moduli of the coefficients instead of on the real and imaginary parts. The first such result is due to Govil and Rahman and is as follows [14].
Theorem 6. 
Let p ( z ) = j = 0 n a j z j be an n-degree polynomial with coefficients satisfying | a r g ( a j ) β | α π / 2 for some α and β and for j = 0 , 1 , , n and | a 0 | | a 1 | | a n | . Then all the zeros of p lie in
| z | cos α + sin α + 2 sin α | a n | j = 0 n 1 | a j | .
Additional reversals in the inequalities could be introduced. This has been performed for an arbitrary number of reversals in the monotonicity condition on real and imaginary parts in [15] as follows.
Theorem 7. 
Let p ( z ) = j = 0 n a j z j be an n-degree polynomial with coefficients a j = α j + i β j . Let t > 0 , and k 1 , k 2 , , k p , r 1 , r 2 , , r q N { 0 } be such that
α t α 1 t k 1 α k 1 t k 1 + 1 α k 1 + 1 t k 2 α k 2 t k 2 + 1 α k 2 + 1   a n d
β 0 t β 1 t r 1 β r 1 t r 1 + 1 β r 1 + 1 t r 1 β r 2 t r 2 + 1 β r 2 + 1
with inequalities reversed at p indices k 1 , k 2 , , k p for the real parts and inequalities reversed at q indices r 1 , r 2 , , r q for the imaginary parts. Then all the zeros of p lie in
min t | a 0 | M 1 , t | z | max M 2 | a n | , 1 t
where
M 1 = α 0 + ( 1 ) p + 1 α n t n + u = 1 p ( 1 ) u α k u t k u β 0 + ( 1 ) q + 1 β n t n + s = 1 q ( 1 ) s β r s t r s + | a n | t n
and
M 2 = | a 0 | t n + 1 ( α 0 + β 0 ) t n 1 + ( 1 ) p + 1 α n t ( 1 ) q + 1 β n t
+ ( t 2 + 1 ) u = 1 p ( 1 ) u α k u t n k u 1 ( t 2 + 1 ) s = 1 q ( 1 ) s β r s t n r s 1
+ ( t 2 1 ) u = 0 p ( 1 ) u + 1 m = k u + 1 k u + 1 1 α m t n m 1 ( t 2 1 ) s = 0 q ( 1 ) s + 1 v = r s + 1 r s + 1 1 β v t n v 1 ,
where we take k 0 = r 0 = 0 and k p + 1 = r q + 1 = n .
The arbitrary number of reversals could be imposed on the even-indexed and odd-indexed parts of the coefficients, or on the moduli of the coefficients, similarly to Theorem 6.
Conditions could be imposed on different congruence classes modulo m of coefficient subscripts. The results of this paper could be interpreted as the imposition of a monotonicity condition on the coefficients with the subscripts in congruence classes modulo m = 2 . This is performed for an arbitrary number of equivalence classes, but with a single reversal in [16].
Theorem 8. 
Let p ( z ) = j = 0 n a j z j be an n-degree polynomial with coefficients a j = α j + i β j . Let t > 0 , and let m N satisfy m < n such that
α k α k + m t m α k + 2 m t 2 m α r k t m r k / m α r k + m r m r k / m + m
α m ( n m k + 1 ) / m + k t m ( n m k + 1 ) / m
and
β k β k + m t m β k + 2 m t 2 m β s k t m s k / m β s k + m t m s k / m + m
β m ( n m k + 1 ) / m + k t m ( n m k + 1 ) / m
for k { 0 , 1 , , m 1 } , r k k (mod m ) , 0 r k n , s k k (mod m ) , and 0 s k n . Then all the zeros of p lie in
min t | a 0 | M 1 , t | z | max M | a n | , 1 t
where
M 1 = ( α 0 + β 0 ) + k = 1 m 1 ( | α k | ( α k + β k ) ) t k + 2 k = 0 m 1 ( α r k t r k + β s k t s k ) + k = n m + 1 n ( | α k | ( α k + β k ) ) t k
and
M 2 = k = 0 m 1 ( | a k | t 2 m ( α k + β k ) ) t n 1 k + ( t 2 m + 1 ) k = 0 m 1 ( α r k t n 1 r k + β s k t n 1 s k )
+ k = n m + 1 n 1 ( | a k | ( α k + β k ) t 2 m ) t n 1 k ( α n + β n ) t 2 m 1
+ ( t 2 m 1 ) k = 0 m 1 = 1 ( r k k ) / m 1 α k + m t n 1 k m + = 1 ( s k k ) / m 1 β k + m t n 1 k m
+ ( 1 t 2 m ) k = 0 m 1 = ( r k k ) / m + 1 ( n m k + 1 ) / m 1 α k + m t n 1 k m + = ( s k k ) / m + 1 ( n m k + 1 ) / m 1 β k + m t n 1 k m .
Additional reversals could be added to the hypotheses of Theorem 8, or this could be applied to the moduli of the coefficients.
The quaternions are the canonical example of a noncommutative division ring. With the recent introduction of an analytic theory of functions of a quaternionic variable [17], monotonicity conditions can be imposed on coefficients of a quaternionic polynomial or on its real and imaginary parts. The first such result in this setting was a proof of the Eneström–Kakeya theorem [18].
Theorem 9. 
If p ( q ) = = 0 n q a is an n-degree polynomial, where q is a quaternionic variable, with real coefficients satisfying 0 a 0 a 1 a n , then all the zeros of p lie in | a | 1 .
The paper presenting Theorem 9 also addresses polynomials with monotone real and imaginary parts and monotone moduli. Since the first appearance of Theorem 9, a number of generalizations have been presented. Monotonicity of the real and imaginary parts with the t condition and reversals could be imposed and/or the monotonicity of even- and odd-indexed coefficients could be considered separately (or indexes modulo m, as in Theorem 8).
The bicomplex numbers are a commutative algebra, but they contain zero divisors and so do not form a field. They were introduced by Carrado Segre in 1892 (see [19] for more details). A version of the Eneström–Kakeya theorem exists in the bicomplex arena [20] as follows.
Theorem 10. 
If p ( z ) = = 0 n a z is a bicomplex n-degree polynomial with real coefficients satisfying 0 a 1 a 2 a n , then all the zeros of p lie in the closed discus D ( 0 ; 1 , 1 ) .
The set D ( 0 ; 1 , 1 ) corresponds to the unit ball in the bicomplex numbers; the center is 0 and there are two radii of 1 based on the idempotent representation of bicomplex numbers. Reference [20] gives a brief description of this terminology. This new result (2023) will likely soon be the focus of additional research, and generalizations based on the idea of modifying the monotonicity conditions undoubtedly will be forthcoming.

Author Contributions

Conceptualization, R.G. and L.H.; formal analysis, R.G. and L.H.; writing—original draft preparation, R.G. and L.H.; writing—review and editing, R.G. and L.H. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

This research was completed as part of the ETSU Honors College’s Undergraduate Research Honors Program. The ETSU Honors College provided support for one of the authors (Herrell) through their scholarship program. The authors also acknowledge the useful comments of the reviewers, which led to a much more readable paper.

Conflicts of Interest

The authors declare no conflicts of interest.

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Gardner, R.; Herrell, L. An Eneström–Kakeya Theorem with Monotonicity Conditions on the Even- and Odd-Indexed Coefficients of a Polynomial. Axioms 2025, 14, 651. https://doi.org/10.3390/axioms14090651

AMA Style

Gardner R, Herrell L. An Eneström–Kakeya Theorem with Monotonicity Conditions on the Even- and Odd-Indexed Coefficients of a Polynomial. Axioms. 2025; 14(9):651. https://doi.org/10.3390/axioms14090651

Chicago/Turabian Style

Gardner, Robert, and Luke Herrell. 2025. "An Eneström–Kakeya Theorem with Monotonicity Conditions on the Even- and Odd-Indexed Coefficients of a Polynomial" Axioms 14, no. 9: 651. https://doi.org/10.3390/axioms14090651

APA Style

Gardner, R., & Herrell, L. (2025). An Eneström–Kakeya Theorem with Monotonicity Conditions on the Even- and Odd-Indexed Coefficients of a Polynomial. Axioms, 14(9), 651. https://doi.org/10.3390/axioms14090651

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