The function tangent and polynomials

The function tangent and polynomials

Andrzej Nowicki Version atg-01. May 15, 2015

It is known (see for example [1], [4]) that the numbers tan^π₈, tan^3π₈ , tan^5π₈ , tan^7π₈ are distinct roots of the equation x⁴− 6x²+ 1 = 0, and the numbers tan^π₇, tan^2π₇ , . . . , tan^6π₇ are distinct roots of the equation x⁶− 21x⁴+ 35x²− 7 = 0.

This note contains some proofs of the above facts and their generalizations. Our note is based on [3], [2] and [5].

1 The polynomials f

(x) and g

(x)

If n is a nonnegative integer, then we denote by fn(x) and gn(x) the polynomials in one variable x defined by

(1 + ix)

= f

(x) + ig

(x)

The polynomials fn(x) and gn(x) belong to the polynomial ring Z[x], that is, they are polynomials with integer coefficients.

f₀(x) = 1, f₁(x) = 1,

f₂(x) = −x²+ 1 = −(x − 1)(x + 1), f₃(x) = −3x² + 1,

f₄(x) = x⁴− 6x²+ 1 = (x² + 2x − 1)(x²− 2x − 1), f₅(x) = 5x⁴− 10x²+ 1,

f₆(x) = −x⁶+ 15x⁴− 15x²+ 1 = (x² + 2x − 1)(x²− 2x − 1), f₇(x) = −7x⁶ + 35x⁴− 21x²+ 1,

f8(x) = x⁸− 28x⁶+ 70x⁴− 28x²+ 1

= (x⁴− 4x³− 6x²+ 4x + 1)(x⁴+ 4x³− 6x²− 4x + 1), f₉(x) = 9x⁸− 84x⁶+ 126x⁴− 36x²+ 1

= (3x²− 1)(3x⁶− 27x⁴+ 33x²− 1),

f₁₀(x) = −x¹⁰(x) + 45x⁸− 210x⁶+ 210x⁴− 45x²+ 1

= −(x − 1)(x + 1)(x³− 4x³− 14x²− 4x + 1)(x⁴+ 4x³− 14x²+ 4x + 1), f₁₁(x) = −11x¹⁰(x) + 165x⁸− 462x⁶+ 330x⁴− 55x²+ 1,

g1(x) = x, g₂(x) = 2x,

g₃(x) = −x³ + 3x = −x(x²− 3),

g₄(x) = −4x³+ 4x = −4x(x − 1)(x + 1), g₅(x) = x⁵ − 10x³+ 5x = x(x⁴− 10x² + 5), g₆(x) = 6x⁵− 20x³+ 6x = 2x(3x²− 1)(x²− 3),

g₇(x) = −x⁷ + 21x⁵− 35x³+ 7x = −x(x⁶− 21x⁴+ 35x²− 7),

g₈(x) = −8x⁷+ 56x⁵ − 56x³+ 8x = −8x(x − 1)(x + 1)(x²+ 2x − 1)(x²− 2x − 1), g₉(x) = x⁹ − 36x⁷+ 126x⁵− 84x³+ 9x = x(x²− 3)(x⁶− 33x⁴+ 27x²− 3),

g₁₀(x) = 10x⁹− 120x⁷+ 252x⁵ − 120x³+ 10x = 2x(5x⁴− 10x²+ 1)(x⁴− 10x²+ 5).

Each f_n is an even polynomial, and each g_n is an odd polynomial divisible by x.

Thus we have the equalities

f_n(−x) = f_n(x), g_n(−x) = −g_n(x).

Moreover,

deg f_2s = 2s, deg f_2s+1 = 2s, deg g_2s = 2s − 1, deg g_2s+1 = 2s + 1.

It is clear that we have the following equalities

(1 − ix)ⁿ = f_n(x) − ig_n(x), f_n(x) = ^{(1+ix)n+(1−ix)}₂ ⁿ, g_n(x) = ^(1+ix)n_2i^−(1−ix)n.

If h is a nonzero polynomial in a variable x, then let us denote by w(h) the initial coefficient of h. For example, if h = 5x³ − 4x + 7, then w(h) = 5. It is easy to prove:

Proposition 1.1. w (f_2s) = (−1)^s, w (f_2s+1) = (−1)^s(2s + 1), w (g_2s) = (−1)^s+12s, w (g_2s+1) = (−1)^s.

Since f_n(x) + ig_n(x) = (1 + ix)ⁿ, we have

f_n(x) + ig_n(x) = 1 + i^n₁^x¹−^n₂^x²− i^n₃^x³+^n₄^x⁴+ i^n₅^x⁵− · · · . In the case n = 4k we have

f_4k(x) = x^4k−^_4k−2^{4k }x^4k−2+^_4k−4^{4k }x^4k−4− · · · −^4k₂^x²+ 1, g_4k(x) = −^_4k−1^{4k }x^4k−1+^_4k−3^{4k }x^4k−3− · · · −^4k₃^x³ +^4k₁^x.

Hence:

f_4k(x) =

2k

X

j=0

(−1)^j4k_2j^x^2j, g_4k(x) =

2k−1

X

j=0

(−1)^j_2j+1^{4k }x^2j+1. By the same way we obtain the following 8 equalities:

f4k(x) = ^P2k

j=0

(−1)^j4k_2j^x^2j, g4k(x) = ^2k−1P

j=0

(−1)^j_2j+1^{4k }x^2j+1, f_4k+1(x) = ^P2k

j=0

(−1)^j4k+1_2j ^x^2j, g_4k+1(x) = ^P2k

j=0

(−1)^j4k+1_2j+1^x^2j+1, f_4k+2(x) = ^2k+1P

j=0

(−1)^j4k+2_2j ^x^2j, g_4k+2(x) = ^P2k

j=0

(−1)^j4k+2_2j+1^x^2j+1, f_4k+3(x) = ^2k+1P

j=0

(−1)^j4k+3_2j ^x^2j, g_4k+3(x) = ^2k+1P

j=0

(−1)^j4k+3_2j+1^x^2j+1. Thus, we have the following 4 equalities:

f

(x) =

j=0

(−1)

x

, g

(x) =

j=0

(−1)

x

, f

(x) =

j=0

(−1)

x

, g

(x) =

j=0

(−1)

x

,

.

The polynomials g_n(x) are divisible by x. Let us introduce the notation:

hn(x) = g_n(x) x ,

for all n. Observe that if n is odd, then the polynomials f_n(x) and h_n(x) are of the same degree n − 1.

Proposition 1.2. For every integer k > 0 we have the following two equalities:

x^4kf_4k+1^1_x^= h_4k+1(x), x^4k+2f_4k+3^1_x^= −h_4k+3(x).

Hence

x

f

= (−1)

h

(x).

x^4kf_4k+1^1_x^ = x^{4k P2k}

j=0

(−1)^j4k+1_2j ^{ }_x^12j

= ^P2k

j=0

(−1)^j4k+1_2j ^x^4k−2j = _x^{1 P2k}

j=0

(−1)^j_4k+1−2j^{4k+1 }x^4k+1−2j

= ¹_x ^P2k

j=0

(−1)^j4k+1_2j+1^x^2j+1 = ¹_xg_4k+1(x)

= h4k+1(x).

Now let n = 4k + 3:

x^4k+2f4k+3

1 x

 = x^4k+22k+1P

j=0

(−1)^j4k+3_2j ^{ 1}_x^2j

= ^2k+1P

j=0

(−1)^j4k+3_2j ^x^4k+2−2j = _x^12k+1P

j=0

(−1)^j_4k+3−2j^{4k+3 }x^4k+3−2j

= −¹_x^2k+1P

j=0

(−1)^j4k+3_2j+1^x^2j+1 = −¹_xg_4k+3(x)

= −h_4k+3(x).

This completes the proof. 

2 Divisibility properties of f

(x) and g

(x)

Proposition 2.1. For all nonnegative integers m, n we have the following two equali- ties:

f

(x) = f

(x)f

(x) − g

(x)g

(x) g

(x) = f

(x)g

(x) + f

(x)g

(x)

Proof.

fn+m(x) + ign+m(x) =



1 + ix

n+m

=



1 + ix

n

1 + ix

m

=



f_n(x) + ig_n(x)



f_m(x) + ig_m(x)



=



f_n(x)f_m(x) − g_n(x)g_m(x)



+ i



f_n(x)g_m(x) + f_m(x)g_n(x)



. This completes the proof. 

In particular, f_2n(x) = f_n(x)²− g_n(x)² and

g

(x) = 2f

(x)g

(x)

Proposition 2.2 ([5], [2]). The polynomials fn(x), gn(x) are relatively prime in Q[x].

Proof. Since Z[i][x] is a unique factorization domain, it is enough to prove that the polynomials f_n(x) and g_n(x) are relatively prime in the domain Q(i)[x].

Let n be a fixed positive integer and suppose that these polynomials are not relatively prime. Then fn(x) and gn(x) are divisible by an irreducible polynomial h(x) ∈ Q(i)[x]. Obviously deg h(x) > 1. Then h(x) divides the polynomials (1 + ix)ⁿ and (1 − ix)ⁿ. As h(x) is irreducible, we deduce that the polynomials (1 + ix) and (1 − ix) are divisible by h(x). This implies that h(x) divides the number 2. But it is a contradiction, because deg h(x) > 0. 

Note next propositions concerning the divisibility of f_n(x) and g_n(x).

Proposition 2.3. If m | n, then g_m(x) | g_n(x) (divisibility in Z[x]).

Proof. We will prove by an induction that for every positive integer k, the poly- nomial g_m divides the polynomial g_km. It is obvious in the case k = 1. Assume that k > 1 and g^m divides gkm in Z[x]. Let g^km = u · gm with u ∈ Z[x]. Then by Proposition 2.1, we have

g_(k+1)m = g_km+m = f_kmg_m+ f_mg_km

= f_kmg_m+ f_mug_m =



f_km+ uf_m



g_m and so, g_m divides g_(k+1)m in Z[x]. 

We may prove also:

Proposition 2.4 ([2]). If m | n, then gn(x) = vm(x)gm(x), where vm(x) ∈ Z[x], and the polynomials v_m(x) and g_m(x) are relatively prime in Q[x].

Proof. Let n = dm, where d is a positive integer. If m = n, then v_m(x) = 1 and we are done. Assume that m < n. Then d> 2 and we have

f_n+ ig_n = (1 + ix)ⁿ = (1 + ix)^dm = (f_m+ ig_m)^d

= f_m^d + idf_m^d−1g_m−^d₂^f_m^d−2g_m² − i^d₃^f_m^d−3g³_m+ · · · ,

and hence g_n = df_m^d−1g_m + u · g_m², where u = u(x) is a polynomial belonging to Z[x].

Put

vm(x) = df_m^d−1+ ugm.

Then g_n(x) = v_m(x)g_m(x). Since gcd(f_m, g_m) = 1 (see Proposition 2.2), we have gcd(g_m, v_m) = gcd^g_m, df_m^d−1+ ug_m^= gcd^g_m, df_m^d−1= 1.

Therefore, the polynomials gm(x) and vm(x) are relatively prime in Q[x]. 

Observe that the polynomials f₃ = −3x² + 1, f₅ = 5x⁴ − 10x² + 1 and f₇ =

−7x⁶+ 35x⁴− 21x²+ 1 are irreducible in Q[x] (and also in Z[x]). Now we will prove that the same is true for all the polynomials f_p, where p is an odd prime number. Let us recall that h_n(x) = ^gn_x^(x).

Proposition 2.5. If p > 3 is a prime number, then the polynomials fp(x) and h_p(x) are irreducible in Z[x].

Proof.

f_p(x) + igp(x) = (1 + ix)^p

= 1 +^p₁^ix −^p₂^x²− i^p₃^x³+ · · · ±^_p−1^{p }x^p−1∓ ix^p.

This implies that f_p(x) = 1 −^p₂^x² +^p₄^x⁴ + · · · ±^_p−1^{p }x^p−1 and g_p(x) = x · h_p(x), where

h_p(x) =^p₁^−^p₃^x²+ · · · ∓^_p−2^{p }x^p−3± x^p−1.

As all the numbers ^p₁^, ^p₂^, . . . , ^_p−1^{p } are divisible by p, we deduce by Eisenstein’s theorem that the polynomials f_p(x) and h_p(x) are irreducible in Z[x]. 

Proposition 2.6 ([5]). If n, k are nonnegative integers, then:

(1) f_n | g_2kn; (2) f_n | f_(2k+1)n;

(3) the polynomials f_kn and g_n are relatively prime;

(4) gcd (gkn+r, gn) = gcd (gr, gn) for integers r > 0.

Proof. (1). Since g_2n = 2f_ng_n, we have f_n| g_2n. But g_2n | g_2kn(see Proposition 2.3).

Hence, f_n| g_2kn.

(2). It is obvious for k = 0. Let k > 0 and assume that fⁿ| f_(2k+1)n. Since f_(2k+3)n = f_(2k+1)n+2n= f_(2k+1)nf_2n− g_(2k+1)ng_2n

and f_n| g_2n, we have f_n | f_(2k+3)n.

(3). It is obvious for k = 0, because f₀ = 1. If k = 1, then it follows from Proposi- tion 2.2. Let k > 1 and assume that gcd (fkn, g_n) = 1. Then we have: gcd^f_(k+1)n, g_n^= gcd (f_kn+n, g_n) = gcd (f_knf_n− g_kng_n, g_n) = gcd (f_knf_n, g_n) = gcd (f_kn, g_n) = 1.

(4). We use the previous properties:

gcd (gkn+r, gn) = gcd (fkngr+ gknfr, gn) = gcd (fkngr, gn) . But gcd (f_kn, g_n) = 1, so gcd (g_kn+r, g_n) = gcd (g_r, g_n).

Now, by Proposition 2.6 and the Euclid algorithm, we have:

Theorem 2.7. If n, m are positive integers, then

gcd (g

, g

) = g

3 Zeros of the polynomials f

(x) and g

(x)

Let n> 1 be a fixed integer. Consider the numbers of the form

t

= tan jπ n

!

,

where {0, 1, 2, . . . , n − 1} and n 6= 2j.

If n jest odd, then we have n distinct real numbers t₀ = 0, t₁, . . . , t_n−1. If n = 2m is even, then we have n − 1 distinct real numbers

t₀ = 0, t₁, . . . , t_m−2, t_m−1, t_m+1, t_m+2, . . . , t_n−1.

Theorem 3.1. All zeros of each polynomial g_n(x) are real numbers.

More precisely, if n is odd, then all the real numbers t₀, t₁, . . . , t_n−1 are distinct zeros of g_n. If n = 2m is even then all the real numbers t₀, t₁, . . . , t_m−1, t_m+1, t_m+2, . . . , t_n−1 are distinct zeros of g_n.

Proof. Assume that j ∈ {0, 1, . . . , n − 1}, n 6= 2j. Then we have:

f_n(t_j) + ig_n(t_j) = (1 + it_j)ⁿ=^1 + i tan^jπ_n^n =



1 + i^sin

jπ n

cos^jπ_n

n

= ^cos^jπ_n^−ncos^jπ_n + i sin^jπ_n^n

= (−1)^jcos^jπ_n^−n and this implies that g_n(t_j) = 0. 

Theorem 3.2. All zeros of each polynomial fn(x) are real numbers.

Proof. It follows from Theorem 3.1 and the equality g_2n= 2f_ng_n.  As a consequence of the above theorems we obtain

Proposition 3.3. For every n > 1, the polynomials fn(x) and g_n(x) are square-free.

The next proposition is also a consequence of the above theorems.

Proposition 3.4. If q 6= ¹₂ is a rational number, then the number tan (qπ) is algebraic.

Proof. Put r = tan(qπ). Then ±r = tan^_n^jπ^, where j, n are nonnegative integers with n > 1 and j < n, and of course n 6= 2j. We know, by Theorem 3.1, that ±r is a zero of the polynomial g_n(x). All the coefficients of g_n(x) are integers, so ±r is algebraic, and so r is algebraic. 

Theorem 3.5. If q is a rational number, then all the numbers sin(qπ), cos(qπ), tan (qπ), cot(qπ) (except when undefined) are algebraic.

Proof. It is a consequence of Proposition 3.4 and the following known equalities:

sin α = 2t

1 + t², cos α = 1 − t²

1 + t², cot α = 1 tan α, where t = tan^α₂. 

Theorem 3.6. If q is a rational number, then all the numbers sin(q^o), cos(q^o), tan (q^o), cot(q^o) are algebraic.

Proof. Since the angle 180^o (180 degrees) equals π radians, we have q^o = ₁₈₀^q π.

Therefore, this theorem is a consequence of the previous theorem. 

4 Equalities with tangents

Look again at Theorems 3.1, 3.2 and the initial coefficients described in Proposition 1.1.

Observe that as a consequence of the above facts we obtain the following equalities.

Proposition 4.1.

f

(x) = (−1)

k=1



x − tan



, f

(x) = (−1)

(2m + 1)

k∈A



x − tan



,

where m> 1 and A = {0, 1, 2, . . . , 2m} r {m}.

Proposition 4.2.

g

(x) = (−1)

x

k=1



x − tan

, g

(x) = (−1)

2mx

k∈A



x − tan

π

,

where m> 1 and A = {0, 1, 2, . . . , 2m − 1} r {m}.

Since tan (π − α) = −tan (α), we may rewrite the above two propositions in the following way.

Proposition 4.3.

f

(x) = (−1)

k=1



x

− tan

(

)



f

(x) = (−1)

(2m + 1)

k=0



x

− tan

(

)



g

(x) = (−1)

2mx

k=1



x

− tan

(

)

g

(x) = (−1)

x

k=1



x

− tan

(

)

.

Let us recall (see Proposition 1.2) that x^2mf_2m+1

1 x



= (−1)^mh_2m+1(x).

where h_2m+1(x) = ^g2m+1_x ^(x). The zeros of the polynomial h_2m+1(x) are real numbers of the form tan_2m+1^kπ , where k = 1, . . . , 2m. Thus, we have:

Proposition 4.4. The zeros of the polynomial f_2m+1(x) are real numbers of the form cot kπ

2m + 1, k = 1, 2, . . . , 2m.

Thus, we have the equalities

f

(x) = u

k=1



x − cot

= u

k=1



x

− cot

,

where u_m = (−1)^m(2m + 1).

Recall that hn(x) = ^gn_x^(x). Look again at the polynomial hn(x) in the case n = 4k+1:

h_4k+1(x) =

2k

X

j=0

(−1)^j4k+1_2j+1^x^2j =

2k

Y

j=1

x²− tan²(_4k+1^jπ )^.

Comparing the constant terms, we obtain the equality ^4k+1₁ ^ = ^Q2k

j=1

tan²(_4k+1^jπ ). Com- paring the coefficients of x^4k−2, we obtain the equality ^4k+1₂ ^= ^P2k

j=1

tan²(_4k+1^jπ ). Hence, we have the following two identities:

2k

Y

j=1

tan

 jπ 4k + 1



=√

4k + 1,

2k

X

j=1

tan²

 jπ 4k + 1



= 2k(4k + 1).

Using the same in the case n = 4k + 3, we obtain:

2k+1

Y

j=1

tan

 jπ 4k + 3



=√

4k + 3,

2k+1

X

j=1

tan²

 jπ 4k + 3



= (2k + 1)(4k + 3).

Assume now that n is even. For n = 4k we have:

h_4k(x) = −4k

2k−1

Y

j=1

x²− tan²(^jπ_4k)^=

2k−1

X

j=0

(−1)^j_2j+1^{4k }x^2j,

Comparing the constant terms, we obtain the equality ^4k₁^ = 4k^2k−1Q

j=1

tan²(^jπ_4k). Com- paring the coefficients of x^4k−4, we obtain the equality ^4k₃^ = 4k^2k−1P

j=1

tan²(^jπ_4k). Hence, we have the following two identities:

2k−1

Y

j=1

tan

jπ 4k



= 1,

2k−1

X

j=1

tan²

jπ 4k



= 1

3(2k − 1)(4k − 1).

Using the same in the case n = 4k + 2, we obtain:

2k

Y

j=1

tan

 jπ 4k + 2



= 1,

2k

X

j=1

tan²

 jπ 4k + 2



= 1

32k(4k + 1).

Therefore, we proved the following proposition.

Proposition 4.5.

m−1 Q

j=1

tan

= 1,

j=1

tan

=

(m − 1)(2m − 1),

m Q

j=1

tan

= √

2m + 1,

j=1

tan

= m(2m + 1).

.

By the same method, using Proposition 4.4, we obtain the following equality with cotangents.

Proposition 4.6 ([1] 123).

m X

j=1

cot

jπ

2m + 1 = m(2m − 1)

3

5 Examples and applications

5.1. Multiples of ^π₅ = 36^o. (1) tan^π₅ · tan ^2π₅ =√

5.

(2) tan^2π₅^+ tan^22π₅ ^= 10.

(3) cot^2π₅^+ cot^22π₅ ^= 2.

(4) tan^π₅ =

q

5 − 2√

5, tan^2π₅ =

q

5 + 2√ 5.

(5) The numbers tan^π₅, tan^2π₅ , tan^3π₅ , tan^4π₅ are zeros of the polynomial h₅(x) = x⁴− 10x²+ 5.

(6) The numbers cot^π₅, cot^2π₅ , cot^3π₅ , cot^4π₅ są are zeros of the polynomial f₅(x) = x⁴− 10x²+ 1.

5.2. Multiplies of ^π₇.

(1) tan^π₇ · tan ^2π₇ · tan ^3π₇ =√ 7.

(2) tan^2π₇^+ tan^22π₇ ^+ tan^23π₇ ^= 21.

(3) cot^2π₇^+ cot^22π₇ ^+ cot^23π₇ ^= 5.

(4) The numbers tan^π₇, tan^2π₇ , . . . , tan^6π₇ are zeros of the polynomial h₇(x) = x⁶− 21x⁴+ 35x²− 7.

(5) The numbers cot^π₇, cot^2π₇ , . . . , cot^6π₇ are zeros of the polynomial f7(x) = −7x⁶+ 35x⁴− 21x²+ 1.

5.3. Multiplies of ^π₈ = 22, 5^o. (1) tan^π₈ · tan ^2π₈ · tan ^3π₈ = 1.

(2) tan^2π₈^+ tan^22π₈ ^+ tan^23π₈ ^= 7.

(3) tan^π₈ =

q

3 − 2√

2, tan^2π₈ = 1, tan^3π₈ =

q

3 + 2√ 2.

(4) The numbers tan^π₈, tan^3π₈ , tan^5π₈ , tan^7π₈ are zeros of the polynomial f₄(x) = x⁴− 6x²+ 1.

5.4. Multiplies of ^π₉ = 20^o.

(1) tan^π₉ · tan ^2π₉ · tan ^3π₉ · tan ^4π₉ = 3.

(2) tan^2π₉^+ tan^22π₉ ^+ tan^23π₉ ^+ tan^24π₉ ^= 36.

(3) cot^2π₉^+ cot^22π₉ ^+ cot^23π₉ ^+ cot^24π₉ ^= ²⁸₃.

(4) The numbers tan^π₉, tan^2π₉ , . . . , tan^8π₉ are zeros of the polynomial h₉(x) = x⁸− 36x⁶+ 126x⁴− 84x² + 9 = (x²− 3)(x⁶− 33x⁴+ 27x²− 3) (5) The numbers cot^π₉, cot^2π₉ , . . . , cot^8π₉ are zeros of the polynomial

f₉(x) = 9x⁸− 84x⁶+ 126x⁴− 36x²+ 1 = (3x²− 1)(3x⁶− 27x⁴+ 33x²− 1).

5.5. Multiplies of ₁₀^π = 18^o.

(1) tan₁₀^π · tan ^2π₁₀ · tan ^3π₁₀ · tan^4π₁₀ = 1.

(2) tan^2₁₀^π+ tan^22π₁₀^+ tan^23π₁₀^+ tan^24π₁₀^= 12.

(3) tan₁₀^π =^q5−2

√ 5

5 , tan^3π₁₀ =^q5+2

√ 5 5 .

(4) All numbers of the form tan^jπ₁₀, where j ∈ {1, 2, 3, 4, 6, 7, 8, 9}, are zeros of the polynomial

h₁₀(x) = 10x⁸− 120x⁶+ 252x⁴− 120x²+ 10 = 2(5x⁴− 10x²+ 1)(x⁴− 10x²+ 5).

(5) The numbers tan₁₀^π, tan^3π₁₀, tan^7π₁₀, tan^9π₁₀ are zeros of the polynomial f₅(x) = 5x⁴− 10x²+ 1.

Literatura

[1] T. Andreescu, Z. Feng, 103 Trigonometry Problems. From the training of the USA IMO team, Birkh¨auser, Boston - Basel - Berlin, 2005.

[2] X. Lin, Infinitely many primes in the arithmetic progression kn − 1, The American Mathematical Monthly, 122(1)(2015), 48-50.

[3] T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964.

[4] A. Nowicki, Real Numbers and Functions (in Polish), Podróże po Imperium Liczb, part 10, Second Edition, OWSIiZ, Toruń, Olsztyn, 2013.

[5] A. Nowicki, S. Spodzieja, Polynomial imaginary decomposition for finite extensions of fields of characteristic zero, Bull. Pol. Sci. Acad., 51(2)(2003), 157-168.

Nicolaus Copernicus University, Faculty of Mathematics and Computer Science, 87-100 Toruń, Poland, (e-mail: anow@mat.uni.torun.pl).