A New Description of the Complex Roots Using Zeros of Some Polynomials

Inspired by a problem found in a textbook of Faddeev and Sominsky we propose a new form of roots of an arbitrary complex number. Our idea uses the zeros of some polynomials connected with the modified Chebyshev polynomials.


Results
In the known textbook [1] authored by Faddeev and Sominsky one can encounter the following task No 171.

Task 1. Show that
for every n ∈ N 0 . Now let us introduce two new families of polynomials associated with Ω n (x). Let a, b ∈ R with a 2 + b 2 = 0 and let us put λ := √ a 2 + b 2 . Definition 1. We call the following complex polynomial: λ the first polynomial associated with the modified Chebyshev polynomial, whereas the complex polynomial sin πn 2 is called the second polynomial associated with the modified Chebyshev polynomial.
The polynomials defined above admit the following factorization.
Vol. 78 (2023) A new description of the complex roots Page 3 of 6 74 and W (a,b) where ϕ ∈ [0, 2π) is an angle such that Proof. In [4] one can find the following identity that holds for any t ∈ R and n ∈ 2N + 1: To show that (1) holds, let us replace the angle t from (3) by ϕ satisfying the relation t = ϕ n : what immediately proves (1). In order to show that (2) holds, let us now replace the angle t from (3) by ϕ satisfying the relation t = π 2 − ϕ n . Using trigonometric identities and fact that n ∈ 2N + 1, we obtain Using the substitution of indices j = n − k, we get Moreover, from a trigonometric identity, it follows: Clearly, latter yields (2).
If v, w are zeros of V Using the trigonometric identity we obtain that it is also equivalent to: From (not only) trigonometric identities we get If ϕ ∈ [0, 2π)\ 0, 1 2 π, π, 3 2 π , then one can check that Thus, Equation (4) is satisfied if and only if sin 2π(k − j) n = 0, so one of the following cases holds: Vol. 78 (2023) A new description of the complex roots Page 5 of 6 74 However, as n is odd, only the case j = k is possible. Thus, we have v = 2 cos ϕ n + 2πk n and w = 2 sin ϕ n + 2πk n .
Whereas in the case where the equality k = n − j is consistent. However, one has v = 2 cos 2π(n − j) n = 2 cos 2πj n , so w = 2 sin 2πj n .
The other three cases can be checked similarly.
Using Lemma 2 we can give a full description of the set n √ a + ib for odd n that, in fact, is a generalization of the correct form of Task 1. Data availability statement Data sharing not applicable to this article as no datasets were generated or analysed during the current study.