2. The set ${\mathbb {N}}[s]$ for algebraic $s$ without positive conjugates

In this section we assume that $s \in {\mathbb {C}} \setminus {\mathbb {R}}$ is an algebraic number which is not a quadratic algebraic integer and has no conjugates in $(0, \infty )$. We will show how the desired result about the density of ${\mathbb {N}}[s]$ in ${\mathbb {C}}$ can be easily derived from the above-mentioned result [24] about the density of ${\mathbb {Z}}[s]$ in ${\mathbb {C}}$.

Assume that an algebraic number $s \ne 0$ has no conjugates in $(0, \infty )$ (including $s$ itself). Then, $s$ is a root of some nonzero polynomial with nonnegative integer coefficients. (This result is essentially due to Meissner [Reference Meissner19]. It was also proved in [Reference Akiyama3], [Reference Brunotte4], [Reference Correa-Morris and Gotti6], [Reference Curtiss7], [Reference Dubickas9], [Reference Handelman15].) Hence,

\[ \sum_{i=0}^m a_i s^i=0 \]

for some $m, a_0, a_m \in {\mathbb {N}}$ and $a_1, \dots, a_{m-1} \in {\mathbb {N}}_0$. This, by the definition of ${\mathbb {N}}[s]$ in (1.1), yields

\[{-}a_0=\sum_{i=1}^m a_i s^i \in {\mathbb{N}}[s]. \]

Consequently,

\[{-}a_0^2=a_0\cdot ({-}a_0)=\underbrace{({-}a_0)+\dots+({-}a_0)}_{a_0 - \text{times}} \in {\mathbb{N}}[s] \]

by the additivity of ${\mathbb {N}}[s]$ and $a_0^2=(-a_0)^2 \in {\mathbb {N}}[s]$ by its multiplicativity. This implies $\pm a_0^2 s^j \in {\mathbb {N}}[s]$ for every $j \in {\mathbb {N}}_0$. Thus, selecting $a=a_0^2 \in {\mathbb {N}}$ we get the following:

Lemma 2.1 If $s \ne 0$ is an algebraic number whose conjugates over ${\mathbb {Q}}$ (including $s$ itself) do not belong to the interval $(0, \infty )$, then there is a positive integer $a$ which depends on $s$ only such that

\[ a{\mathbb{Z}}[s] \subseteq {\mathbb{N}}[s] \subseteq {\mathbb{Z}}[s]. \]

By [24], we know that, if $s \in {\mathbb {C}} \setminus {\mathbb {R}}$ is an algebraic number that is not a quadratic algebraic integer, then ${\mathbb {Z}}[s]$ is everywhere dense in ${\mathbb {C}}$. This, combined with lemma 2.1, implies the following:

In particular, if $s \in {\mathbb {C}} \setminus {\mathbb {R}}$ is quadratic then it has no real conjugates, since its only conjugate over ${\mathbb {Q}}$, which is not equal to $s$, must be its complex conjugate $\overline {s} \notin {\mathbb {R}}$. Hence, corollary 2.2 implies theorem 1.4.

3. Proof of theorem 1.9

In the proof of theorem 1.9 we shall use Kronecker's approximation theorem; see, for instance, [Reference Aggarwal and Singh1, Theorem 9], [Reference Gonek and Montgomery14, p. 507], [Reference Kueh18].

Lemma 3.1 Let $\lambda _1, \dots, \lambda _N$ be real numbers such that $1, \lambda _1, \dots, \lambda _N$ are linearly independent over ${\mathbb {Q}}$, and let $\omega _1, \omega _2, \dots, \omega _N$ be arbitrary real numbers. Then, for any $\epsilon >0$, there exist $T \in {\mathbb {N}}$ and $T_1, \dots, T_N \in {\mathbb {Z}}$ such that

(3.1)$$ |\lambda_n T-\omega_n-T_n|<\epsilon $$

for $n=1, 2, \dots, N$.

If the point $z=0$ does not belong to the interior of a triangle with vertices at $\alpha, \beta, \gamma \in {\mathbb {C}}$, then there is a line $L$ through the point $z=0$ such that the points $\alpha, \beta, \gamma$ all belong to the same side of the line $L$ (possibly including the line $L$ itself). Then, all the points of the sumset ${\mathbb {N}}\alpha +{\mathbb {N}} \beta +{\mathbb {N}}\gamma$ are also on the same side of $L$ (including $L$ itself). Consequently, this sumset is not dense in ${\mathbb {C}}$. Also, if the imaginary parts $\Im (\alpha \overline {\beta }), \Im (\beta \overline {\gamma }), \Im (\gamma \overline {\alpha })$ are linearly dependent over ${\mathbb {Q}}$, then the sumset ${\mathbb {Z}}\alpha +{\mathbb {Z}}\beta +{\mathbb {Z}}\gamma$ is not dense in ${\mathbb {C}}$ by theorem 1.8. Therefore, ${\mathbb {N}}\alpha +{\mathbb {N}}\beta +{\mathbb {N}}\gamma$, as a subset of ${\mathbb {Z}}\alpha +{\mathbb {Z}}\beta +{\mathbb {Z}}\gamma$, cannot be dense in ${\mathbb {C}}$ either.

From now on we assume that the point $z=0$ belongs to the interior of the triangle with vertices at $\alpha, \beta, \gamma$ and that the three numbers $\Im (\alpha \overline {\beta }), \Im (\beta \overline {\gamma }), \Im (\gamma \overline {\alpha })$ are linearly independent over ${\mathbb {Q}}$. In order to complete the proof of theorem 1.9 it remains to show that the sumset ${\mathbb {N}}\alpha +{\mathbb {N}} \beta +{\mathbb {N}}\gamma$ is everywhere dense in ${\mathbb {C}}$.

Of course, the linear independence of the above imaginary parts implies that $\alpha, \beta, \gamma \ne 0$. Note that by multiplying all three numbers $\alpha, \beta, \gamma$ by the same number $\gamma ^{-1}$ we do not change any of the two conditions. The point $z=0$ still belongs to the interior of a triangle with vertices at $\alpha, \beta, 1$ (where we use the notation $\alpha$ for $\alpha \gamma ^{-1}$ and $\beta$ for $\beta \gamma ^{-1}$) and the three numbers $\Im (\alpha \overline {\beta }), \Im (\beta ), \Im (\overline {\alpha })$ are still linearly independent over ${\mathbb {Q}}$. We will show that the sumset ${\mathbb {N}}+{\mathbb {N}} \alpha +{\mathbb {N}}\beta$ is everywhere dense in ${\mathbb {C}}$.

Set

\[ \alpha=u+iv \quad\text{and} \quad \beta={-}w-il, \]

where $u, v, w, l \in {\mathbb {R}}$. Since the point $z=0$ lies in the interior of the triangle with vertices at $\alpha, \beta, 1$, the numbers $v, l$ must be either both positive or both negative. Without the restriction of generality (by swapping $\alpha$ and $\beta$ if necessary) we can assume that $v, l>0$. Furthermore, at least one of the numbers $u, -w$ must be negative, since otherwise $z=0$ does not lie in the interior of the triangle $\alpha, \beta, 1$. Again, by swapping $\alpha, \beta$ by $\overline {\beta }, \overline {\alpha }$ if necessary, we can assume that the number $-w$ is negative, i.e. $w>0$. Therefore, without loss of generality, we can assume that $v, w, l>0$. The sign of $u$ can be arbitrary (it is also possible that $u=0$), but the argument of the complex number $\alpha$ plus $\pi$ must be greater than the argument of $\beta$. (The argument $\arg z$ of a complex number $z \ne 0$ is a unique real number in the interval $[0, 2\pi )$ for which $z=|z| e^{i \arg z}$.)

Let $\theta _1, \theta _2 \in (0, 2\pi )$ be the arguments of $\alpha$ and $\beta$, respectively. Then, $\theta _1 \in (0, \pi )$ and $\theta _2 \in (\pi, {3\pi }/{2})$. The condition $\theta _1+\pi >\theta _2$ automatically holds if $u \leq 0$. In the case when $u>0$ the condition $\theta _1+\pi >\theta _2$ is equivalent to $\tan (\theta _1)>\tan (\theta _2)$, namely, $v/u>l/w$. Hence,

(3.2)$$ \frac{vw}{l}-u>0. $$

Of course, by $v, w, l>0$, the inequality (3.2) trivially holds for $u \leq 0$. So, from now on we will assume that $v, w, l>0$ and $u \in {\mathbb {R}}$ satisfy (3.2).

We know that the numbers $\Im (\alpha \overline {\beta })=ul-vw$, $\Im (\beta )=-l$, $\Im (\overline {\alpha })=-v$ are linearly independent over ${\mathbb {Q}}$. Dividing by $-l$ we deduce that the numbers

\[ 1, \quad\frac{v}{l}, \frac{vw}{l}-u \]

are linearly independent over ${\mathbb {Q}}$ too.

Now, the conclusion of the proof is essentially the same as that in [Reference Dubickas11]. Fix two arbitrary real numbers $X$ and $Y$. We need to show that, for any positive number $\varepsilon$, there exist $a, b, c \in {\mathbb {N}}$ for which the sum $a+b\alpha +c\beta$ is close to $X+iY$, namely,

(3.3)$$ |a+bu-cw-X|<\varepsilon \quad \text{and} \quad |bv-cl-Y|<\varepsilon. $$

Set

(3.4)$$ \omega_1=\frac{Y}{l} \quad \text{and} \quad\omega_2={-}X+\frac{Yw}{l}. $$

By lemma 3.1 applied to $\lambda _1=v/l$, $\lambda _2=vw/l-u$ and $\omega _1, \omega _2$ as defined in (3.4), for any $\epsilon >0$, there exist $b \in {\mathbb {N}}$ and $a, c \in {\mathbb {Z}}$ such that

\[ |bv/l-\omega_1-c|<\epsilon \quad \text{and} \quad |b(vw/l-u)-\omega_2-a|<\epsilon. \]

Since the numbers $v/l$ and $vw/l-u$ are irrational (as they both and $1$ are linearly independent over ${\mathbb {Q}}$), the above inequalities have infinitely many solutions in $b \in {\mathbb {N}}$. For $b$ sufficiently large we must have $c \in {\mathbb {N}}$ and $a \in {\mathbb {N}}$ due to $v/l>0$ and (3.2). Hence, we can assume that $a, b, c \in {\mathbb {N}}$.

Next, in view of (3.4), by multiplying the first inequality by $l>0$, we get

(3.5)$$ |bv-cl-Y|<\epsilon l. $$

Similarly, by (3.4), multiplying the second inequality by $l>0$ we deduce

\[ |b(vw-ul)+Xl-Yw-al|<\epsilon l. \]

This inequality can be written in the equivalent form

\[ |-l(a+bu-cw-X)+w(bv-cl-Y)|<\epsilon l. \]

Now, by the triangle inequality, $w>0$ and (3.5), it follows that

\[ |l(a+bu-cw-X)| < w|bv-cl-Y|+\epsilon l \leq \epsilon wl+\epsilon l=\epsilon l(1+w), \]

and hence

(3.6)$$ |a+bu-cw-X|<\epsilon (1+w). $$

It is clear that (3.5) and (3.6) imply (3.3) provided that $\epsilon$ is satisfies

\[ 0<\epsilon \leq \frac{\varepsilon}{\max(l,1+w)}. \]

This completes the proof of the theorem.

4. Auxiliary lemmas

We begin with the following lemma, which will be used in the proof of proposition 5.1 later on.

Proof. Write $s=\varrho e^{i \alpha }$, with $\varrho >0$ and argument $\alpha \in (0, \pi ) \cup (\pi, 2\pi )$. If the numbers $s$, $|s|=\varrho$ and $s/|s|=e^{i \alpha }$ are all three transcendental, then the assertion of the lemma holds with $t=s$. If not, then either $e^{i\alpha }$ or $\varrho$ is algebraic. (They cannot be both algebraic by the assumption of the lemma on $s$.) We will show that then the assertion of the lemma holds for number $t=s+s^2$. Set $t=|t| e^{i\beta }$, where $0 \leq \beta <2\pi$.

From

\[ t=s+s^2=\varrho\cos (\alpha)+\varrho^2 \cos(2\alpha)+ i \big(\varrho\sin (\alpha)+\varrho^2 \sin(2\alpha)\big) \]

we find that

(4.1)$$ |t|^2=\varrho^2+\varrho^4+2\varrho^3 \cos(\alpha) $$

and

(4.2)$$ \big(\cos (\alpha)+\varrho \cos(2\alpha)\big) \tan (\beta)= \sin (\alpha)+\varrho \sin(2\alpha). $$

Assume first that $e^{i\alpha }=\cos (\alpha )+i\sin (\alpha )$ is algebraic and $\varrho$ is transcendental. Then, $\cos (\alpha )={e^{i\alpha }+e^{-i\alpha }}/{2}$ and $\sin (\alpha )={e^{i\alpha }-e^{-i\alpha }}/{2i}$ are both algebraic. If $|t|$ were algebraic, then, as $|t|^2$ and $\cos (\alpha )$ are both algebraic, $\varrho$ were algebraic by (4.1), a contradiction. Hence, $|t|$ is transcendental. Assume that $t/|t|=e^{i\beta }$ is algebraic. If $\beta \in \{{\pi }/{2}, {3\pi }/{2}\}$, then $\Re (t)=0$. Hence, $\varrho \cos (\alpha )+\varrho ^2 \cos (2\alpha )=0$, and so $\cos (\alpha )+\varrho \cos (2\alpha )=0$. This is not the case, because $\varrho \notin \overline {{\mathbb {Q}}}$ (where $\overline {{\mathbb {Q}}}$ stands for the set of algebraic numbers) and $\cos (\alpha ), \cos (2\alpha )=2\cos ^2(\alpha )-1 \in \overline {{\mathbb {Q}}}$ cannot be both zeros. It follows that $\tan (\beta )$ in (4.2) is well defined, namely, $\beta \notin \{{\pi }/{2}, {3\pi }/{2}\}$.

Since $\tan (\beta ), \cos (\alpha ), \cos (2\alpha ), \sin (\alpha ), \sin (2\alpha ) \in \overline {{\mathbb {Q}}}$ and $\varrho \notin \overline {{\mathbb {Q}}}$, equality in (4.2) holds only if

(4.3)$$ \cos(\alpha) \tan (\beta)=\sin(\alpha) \quad \text{and} \quad \cos(2\alpha) \tan(\beta)=\sin(2\alpha). $$

We know that $\sin (\alpha ) \ne 0$, so $\cos (\alpha ) \ne 0$ by the first equality in (4.3). Thus, $\sin (2\alpha ) \ne 0$. Hence, by (4.3), we obtain $\tan (\beta ) \ne 0$ and

\[ \frac{\cos(\alpha)}{\sin(\alpha)}=\frac{\cos(2\alpha)}{\sin(2\alpha)}, \]

which implies

\[ \cos(2\alpha)=\sin(2\alpha)\frac{\cos(\alpha)}{\sin(\alpha)}=2\cos^2(\alpha)= \cos(2\alpha)+1, \]

which is absurd. This proves that $e^{i\beta }=t/|t|$ is transcendental.

Now, assume that $\varrho$ is algebraic and $e^{i\alpha }$ is transcendental. Then, $\cos (\alpha )$ must be transcendental too. From (4.1) we see that that $|t|^2$ is transcendental, and hence so is $|t|$. It remains to prove that $e^{i\beta }$ is transcendental. Assume that $e^{i\beta } \in \overline {{\mathbb {Q}}}$. It is clear that then $\tan (\beta ) \in \overline {{\mathbb {Q}}}$ or $\beta \in \{{\pi }/{2}, {3\pi }/{2}\}$. However, if $\beta \in \{{\pi }/{2}, {3\pi }/{2}\}$, then $\Re (t)=0$, which means that $\cos (\alpha )+\varrho \cos (2\alpha )=0$. Then, $x=\cos (\alpha )$ is a root of the nonzero polynomial with algebraic coefficients $2\varrho x^2+x-\varrho$. Hence, $\cos (\alpha ) \in \overline {{\mathbb {Q}}}$, a contradiction. Therefore, $\tan (\beta )$ in (4.2) is well defined. Squaring both sides of (4.2) we obtain

\[ \tan^2(\beta)(2\varrho x^2+x-\varrho)^2=(1-x^2)(1+2\varrho x)^2, \]

where $x=\cos (\alpha )$. Since $x$ is transcendental and $\varrho, \tan (\beta ) \in \overline {{\mathbb {Q}}}$, equality holds only if the resulting quartic polynomials on both sides are identical. However, their coefficients for $x^4$ are $4\varrho ^2 \tan ^2(\beta )$ and $-4\varrho ^2$. Since $\varrho > 0$, they are equal only in the case when $\tan ^2(\beta )=-1$, which is impossible. This completes the proof of the lemma.

The next lemma is similar to lemma 4.1, but deals with algebraic $s$ rather than transcendental.

Proof. Assume that $s=\varrho e^{i \alpha }$. Set $t_m=ms+s^2=|t_m| e^{i\beta _m}$. Then, as in (4.2), we find that

(4.4)$$ \big(m\cos (\alpha)+\varrho \cos(2\alpha)\big) \tan (\beta_m)= m\sin (\alpha)+\varrho \sin(2\alpha). $$

Since $\sin (\alpha ) \ne 0$ and $m+2\varrho \cos (\alpha ) \ne 0$ for $m$ sufficiently large, the right hand side of (4.4) is nonzero. Also, in case $\cos (\alpha )=0$ we have $m\cos (\alpha )+\varrho \cos (2\alpha )=-\varrho \ne 0$. Furthermore, for $\cos (\alpha ) \ne 0$ we have $m\cos (\alpha )+\varrho \cos (2\alpha ) \ne 0$ for $m$ large enough. Consequently, $\tan (\beta _m)$ is (4.4) is well defined, i.e. $\beta _m \notin \{{\pi }/{2}, {3\pi }/{2}\}$. Since $\tan (\beta _m) \ne 0$, we must have

\[ \beta_m \in (0,2\pi) \setminus \left\{\frac{\pi}{2},\pi,\frac{3\pi}{2}\right\}. \]

Also, from (4.4) it follows that

(4.5)$$ \tan(\beta_m) \in F={\mathbb{Q}}(\varrho,\cos(\alpha),\sin(\alpha)) $$

for each sufficiently large $m \in {\mathbb {N}}$.

Now, from (4.4) we deduce

(4.6)$$ m(\cos(\alpha) \tan(\beta_m)-\sin(\alpha))={-}\varrho(\cos(2\alpha) \tan(\beta_m)-\sin(2\alpha)). $$

In case $\cos (\alpha )\tan (\beta _m)-\sin (\alpha )=0$ we have $\cos (\alpha ) \ne 0$, and hence $\tan (\beta _m) =\tan (\alpha )$. By (4.6), this implies

\[ \cos(2\alpha) \tan(\alpha)-\sin(2\alpha)={-}\tan(\alpha)=0, \]

which is not the case in view of $\alpha \notin \{0, \pi \}$. Therefore,

(4.7)$$ \cos(\alpha)\tan(\beta_m)-\sin(\alpha) \ne 0. $$

Assume that $\beta _m=r_m \pi$ with rational number

(4.8)$$ r_m \in (0,2) \setminus \left\{\frac{1}{2},1,\frac{3}{2}\right\}. $$

By (4.5), all values of $\tan (r_m \pi )$ belong to the field $F$. Thus, the degree of the algebraic number $\tan (r_m \pi )$ is bounded from above by a constant, say $c_1=c_1(F)=[F:{\mathbb {Q}}]=c_1(s)$. The exact degree of $\tan (r_m \pi )$ with rational $r_m$ in terms of the denominator $v$ of $r_m$ has been calculated, for instance, in [Reference Tangsupphathawat23, Proposition 4.1]. For $v>8$ it is at least $\varphi (2v)/4$, where $\varphi$ is Euler's totient function. From $\varphi (2v)/4< c_1$ we find that $v < c_2$ for some $c_2$ depending in $s$ only.

Note that, by (4.8), the numerator of $r_m$ is less than $2v$. Thus, there are only finitely many of such rational numbers $r_m$ satisfying (4.8). Hence, as $s=\varrho e^{i \alpha }$ is fixed, there a constant $c_3=c_3(F)=c_3(s)>0$ such that the quotient of

\[ -\varrho(\cos(2\alpha) \tan(\beta_m)-\sin(2\alpha)) \quad \text{and} \quad \cos(\alpha)\tan(\beta_m)-\sin(\alpha), \]

which is nonzero by (4.7), takes at most $c_3$ distinct values. However, by (4.6), this quotient equals $m$. This is clearly impossible for $m$ large enough.

In the next lemma we show the existence of infinitely many triplets of useful complex points such that the point $z=0$ belongs to the interior of a triangle with vertices at each of these triplets.

Proof. For $\theta \in {\mathbb {R}} \setminus {\mathbb {Q}}$ the sequence of the fractional parts $\{\theta k\}$, where $k$ runs over the primes, is everywhere dense in the interval $[0, 1]$. See, e.g., [Reference Strauch and Porubský22, Section 2–70]. (In fact, by an old result of Vinogradov, this sequence is uniformly distributed in $[0, 1]$.)

In particular, as ${\beta }/{2\pi } \notin {\mathbb {Q}}$, for each $\delta >0$, there are infinitely many prime numbers $k$ such that

(4.9)$$ \frac{1}{3}<\left\{\frac{k\beta}{2\pi}\right\}<\frac{1}{3}+\delta. $$

Consequently, for each of those $k$ there is $\ell =\ell (k) \in {\mathbb {Z}}$ such that ${1}/{3}<{k\beta }/{2\pi }-\ell <{1}/{3}+\delta$, which is equivalent to

(4.10)$$ \frac{2\pi}{3} < k\beta-2\pi\ell<\frac{2\pi}{3}+2\pi \delta. $$

This means that the arguments of the complex numbers $t^k=|t|^k e^{ik\beta }$, $t^{2k}$, $t^{3k}$ lie in the intervals

\[ \left(\frac{2\pi}{3}, \frac{2\pi}{3}+2\pi \delta\right), \quad \left(\frac{4\pi}{3}, \frac{4\pi}{3}+4\pi \delta\right), \quad \big(0, 6\pi \delta\big) \]

respectively. It is clear that for $\delta >0$ small enough the point $z=0$ belongs to the interior of a triangle with vertices at $t^k, t^{2k}, t^{3k}$.

Likewise, by (4.10), we see that the arguments of the complex numbers $t^{2k}=|t|^{2k} e^{2ik\beta }$, $t^{4k}$, $t^{6k}$ lie in the intervals

\[ \left(\frac{4\pi}{3}, \frac{4\pi}{3}+4\pi \delta\right), \quad \left(\frac{2\pi}{3}, \frac{2\pi}{3}+8\pi \delta\right), \quad \big(0, 12\pi \delta\big) \]

respectively. Again, for $\delta >0$ small enough, the point $z=0$ belongs to the interior of a triangle with vertices at $t^{2k}, t^{4k}, t^{6k}$.

The following lemma will be useful in treating exceptional numbers in theorem 1.5.

Proof. Set $F={\mathbb {Q}}(\lambda, \psi )$. By the assumption of the lemma, the numbers

\[ \tau_m=\lambda(\lambda+m\psi+m^2), \]

where $m \in {\mathbb {N}}$ is large enough, are all at most quadratic, and all belong to $F$. Since $F$ has only finitely many real quadratic subfields (possibly none), there is a real quadratic field $E$ and an infinite set $M \subset {\mathbb {N}}$ such that $\tau _m \in E$ for each $m \in M$. (In the case when all the numbers $\tau _m$ are rational, we can take any real quadratic field $E$.) Select $m \ne m'$ in $M$. Then,

\[ m'\tau_m-m\tau_{m'}=(m'-m) (\lambda^2-m m' \lambda) \in E, \]

and hence $\lambda ^2-mm'\lambda \in E$. Taking $m' \in M \setminus \{m, m'\}$, by the same argument, we obtain $\lambda ^2-mm' \lambda \in E$. The difference $m(m''-m') \lambda$ of these two numbers is also in $E$, which forces $\lambda \in E$. Since $\lambda \ne 0$ and $\tau _m \in E$, we also obtain

\[ \psi = \frac{\tau_m-\lambda^2-m^2 \lambda}{m \lambda} \in E, \]

which completes the proof of the lemma.

The next lemma gives one more approach to establishing the density of a sumset in ${\mathbb {C}}$. This time we will use four sets instead of three. This lemma will be used in the final part of the proof of theorem 1.5.

Lemma 4.5 Let $s \in {\mathbb {C}} \setminus {\mathbb {R}}$. Assume that the set ${\mathbb {N}}[s]$ contains two elements $s_1, s_2 \ne 0$ whose quotient $\omega ={s_1}/{s_2}$ is a negative real irrational number. Then, the sumset

\[ {\mathbb{N}} s_1+{\mathbb{N}} s_2+{\mathbb{N}} s_1s+{\mathbb{N}} s_2 s \]

is everywhere dense in ${\mathbb {C}}$.

Proof. Since the points $0, s_2, s_2 s$ are not collinear, every complex number can be written in the form $Xs_2+Ys_2 s$ with $X, Y \in {\mathbb {R}}$. The idea is to approximate $Xs_2$ by the sumset ${\mathbb {N}} s_1+{\mathbb {N}} s_2$ and $Ys_2 s$ by the sumset ${\mathbb {N}} s_1s+{\mathbb {N}} s_2 s$.

We will show that for each $\varepsilon >0$ there exist $a_1, a_2, a_3, a_4 \in {\mathbb {N}}$ such that

(4.11)$$ |a_1s_1+a_2s_2-Xs_2|<\varepsilon \quad \text{and} \quad |a_3s_1s+a_4s_2s-Ys_2s|<\varepsilon. $$

Since $s_1=\omega s_2$, setting

\[ \varepsilon_1=\frac{\varepsilon}{|s_2|}, \quad \varepsilon_2=\frac{\varepsilon}{|s_2 s|}, \quad \omega_1=\{{-}X\} \quad \text{and} \quad \omega_2=\{{-}Y\} \]

and using the identities $X=-[-X]-\{-X\}$, $Y=-[-Y]-\{-Y\}$, we can rewrite the inequalities in (4.11) as

\[ |(- \omega) a_1-(a_2+[{-}X])-\omega_1|<\varepsilon_1 \quad \text{and} \quad |(-\omega) a_3 - (a_4+[{-}Y])-\omega_2|<\varepsilon_2. \]

By lemma 3.1 with $N=1$ and $\lambda _1=-\omega$, we see that for each $\varepsilon _1>0$ there is $a_1 \in {\mathbb {N}}$ and $b_1 \in {\mathbb {Z}}$ such that

\[ |(- \omega) a_1-\omega_1-b_1|<\varepsilon_1. \]

Here, for each sufficiently small $\varepsilon _1$, the integer $a_1 \in {\mathbb {N}}$ must be large. Since $-\omega >0$, the integer $b_1$ is positive and large. So the first displayed inequality indeed holds with $a_2=b_1-[-X] \in {\mathbb {N}}$. This proves the first inequality in (4.11). The proof of the second displayed inequality is exactly the same, with some $a_3, a_4 \in {\mathbb {N}}$, which implies the second inequality in (4.11).

5. Final preparation

In this section we will prove the following proposition:

Proposition 5.1 Let $t=|t|e^{i\beta } \in {\mathbb {C}}$ be such that ${\beta }/{\pi } \notin {\mathbb {Q}}$ and $|t|^2$ is not an algebraic number of degree at most $2$. Then, there is a positive integer $n$ such that the sumset

\[ {\mathbb{N}} t^n+{\mathbb{N}} t^{2n} +{\mathbb{N}} t^{3n} \]

is everywhere dense in ${\mathbb {C}}$.

Proof. Set $\xi =|t|^2$. We claim that $\xi ^k$ is not an algebraic number of degree at most $2$ for all sufficiently large prime numbers $k$. This is trivial if $\xi$ is transcendental. Assume that $\xi$ is an algebraic number of degree $d$. By the condition of the proposition, we have $d \geq 3$. Evidently, the degree of $\xi ^k$ is $d$, unless there is a conjugate $\xi ' \ne \xi$ of $\xi$ over ${\mathbb {Q}}$ such that $\xi ^k=\xi '^k$. But then $\zeta =\xi /\xi '$ is a $k$th root of unity, so its degree is $\varphi (k)=k-1$. However, the number $\zeta$ cannot belong to the field ${\mathbb {Q}}(\xi, \xi ')$ of degree at most $d(d-1)$ when $k>d(d-1)$, a contradiction. Thus, $\xi ^k$ is of degree $d$ ($d \geq 3$) for each sufficiently large prime number $k$.

Since the argument $\beta$ of a given number $t$ satisfies the condition of lemma 4.3, there are infinitely many primes $k$ for which $z=0$ belongs to the interior of the triangles with vertices at $t^k, t^{2k}, t^{3k}$ and at $t^{2k}, t^{4k}, t^{6k}$.

Take any $n \in \{k, 2k\}$. By theorem 1.9, the set

\[ {\mathbb{N}} t^n+{\mathbb{N}} t^{2n}+{\mathbb{N}} t^{3n} \]

is everywhere dense in ${\mathbb {C}}$ if the imaginary parts

$$\begin{align*} & \Im(t^n \overline{t^{2n}})={-}|t|^{3n} \sin(n\beta),\\ & \Im(t^{2n} \overline{t^{3n}})={-}|t|^{5n} \sin(n\beta),\\ & \Im(t^{3n} \overline{t^{n}})=|t|^{4n} \sin(2n\beta) \end{align*}$$

are linearly independent over ${\mathbb {Q}}$. We will show that this is the case for either $n=k$ or $n=2k$.

Since the number ${\beta }/{\pi }$ is irrational, we have $\sin (n\beta ) \ne 0$ and $\cos (n\beta ) \ne 0$. Dividing by $-|t|^{3n} \sin (n\beta )$, we see that the above three numbers are linearly dependent over ${\mathbb {Q}}$ if and only if so are $1, |t|^{2n}, -2|t|^n \cos (n\beta )$. This is the case if and only if the numbers

(5.1)$$ |t|^{{-}n}, |t|^n, 2\cos(n \beta) $$

are linearly dependent over ${\mathbb {Q}}$.

We claim that the numbers $|t|^{-n}$ and $|t|^n$ are linearly independent over ${\mathbb {Q}}$. This is trivial if $\xi =|t|^2$ is transcendental. Assume that $\xi$ is algebraic of degree $d \geq 3$. Consider the cases $n=k$ and $n=2k$ separately. If $n=k$ then the numbers $|t|^{-n}$ and $|t|^n$ are linearly dependent if and only if $|t|^{2n}=|t|^{2k}=\xi ^k \in {\mathbb {Q}}$. However, we proved that the degree of $\xi ^k$ is $d \geq 3$. So, the numbers $|t|^{-n}$ and $|t|^n$ are linearly independent for $n=k$. Similarly, for $n=2k$, the numbers $|t|^{-n}$ and $|t|^n$ are linearly dependent if and only if $|t|^{2n}=|t|^{4k}=\xi ^{2k} \in {\mathbb {Q}}$. This is only possible if the degree of $\xi ^k$ over ${\mathbb {Q}}$ is $1$ or $2$, while we know that it is $d \geq 3$. Thus, the numbers $|t|^{-n}$ and $|t|^n$ are linearly independent for $n=2k$ as well.

Now, by the linear dependence of the three numbers (5.1), it follows that in both cases $n=k$ and $n=2k$ for some $\mu _n, \nu _n \in {\mathbb {Q}}$, we must have

(5.2)$$ 2\cos(n\beta)=\mu_n |t|^n+\nu_n |t|^{{-}n}. $$

In order to complete the proof of the proposition it suffices to show that (5.2) cannot hold for both $n=k$ and for $n=2k$. Assume that (5.2) is true for $n=k$ and for $n=2k$. Then, by (5.2), applying the trigonometric identity $\cos (2k\beta )=2\cos ^2(k\beta )-1$ and using the notation $x=|t|^{k}$, $\mu '=\mu _{2k}$, $\nu '=\nu _{2k}$, $\mu =\mu _{k}$, $\nu =\nu _k$, we find that

\[ \frac{\mu' x^2+\nu' x^{{-}2}}{2}=\frac{\big(\mu x+\nu x^{{-}1}\big)^2}{2}-1. \]

This is equivalent to

(5.3)$$ (\mu'-\mu^2)x^4-2(\mu \nu-1) x^2+\nu'-\nu^2=0. $$

Now, since $x^2=|t|^{2k}=\xi ^k$ is either transcendental or an algebraic number of degree $d \geq 3$ and $\mu '-\mu, \nu '-\nu ^2, \mu \nu -1 \in {\mathbb {Q}}$, equality (5.3) must be the identity in $x$. Hence, $\mu '=\mu ^2$, $\nu '=\nu ^2$ and $\mu \nu =1$. In particular, from $\mu \nu =1$ we see that both $\mu =\mu _k$ and $\nu =\nu _k$ have the same sign.

If they are both positive then, by (5.2) with $n=k$ and $\mu _k\nu _k=1$, we derive that

\[ 2\cos(k\beta)=\mu_k |t|^k+\nu_k |t|^{{-}k} \geq 2\sqrt{\mu_k |t|^k \nu_k |t|^{{-}k}} =2\sqrt{\mu_k \nu_k}=2. \]

Therefore, $\cos (k \beta )=1$, which implies that ${\beta }/{\pi } \in {\mathbb {Q}}$, a contradiction. Similarly, if $\mu _k$ and $\nu _k$ are both negative, then from $-\mu _k>0$, $-\nu _k>0$ and $\mu _k \nu _k=1$ we deduce

\[{-}2\cos(k\beta)={-}\mu_k |t|^k-\nu_k |t|^{{-}k} \geq 2\sqrt{(-\mu_k) |t|^k (-\nu_k) |t|^{{-}k}}=2, \]

and hence $\cos (k\beta )=-1$. Hence, ${\beta }/{\pi } \in {\mathbb {Q}}$, which is again a contradiction. This completes the proof of the proposition.

6. Proof of theorem 1.3

Let $s \in {\mathbb {C}} \setminus {\mathbb {R}}$ be a transcendental number. By lemma 4.1, for some $t \in \{s, s+s^2\}$ the three numbers $t=|t|e^{i\beta }$, $|t|$ and $t/|t|=e^{i\beta }$ are all transcendental. In particular, this implies that the quotient ${\beta }/{\pi }$ is irrational and that $|t|^2$ is not an algebraic number of degree at most $2$. Thus, proposition 5.1 yields theorem 1.3 for transcendental $s$.

Assume now that $s \in {\mathbb {C}} \setminus {\mathbb {R}}$ is algebraic and has degree $d$ over ${\mathbb {Q}}$. Clearly, $d \ne 1$. Also, $d \ne 2$ by the condition of the theorem. By lemma 4.2, we can take a sufficiently large $m \in {\mathbb {N}}$ such that the argument $\beta _m$ of $t=ms+s^2$ satisfies ${\beta _m}/{\pi } \notin {\mathbb {Q}}$. Now, proposition 5.1 implies theorem 1.3 in case for at least one sufficiently large $m \in {\mathbb {N}}$ the algebraic number $|ms+s^2|^2$ has degree greater than $2$.

For a contradiction, assume that all numbers

\[ |ms+s^2|^2=(ms+s^2)(m \overline{s}+\overline{s}^2)=s \overline{s}(s \overline{s}+(s+\overline{s})m+m^2), \]

$m \geq m_0$, are of degree at most $2$. Setting $\lambda =s \overline {s}$ and $\psi =s+\overline {s}$, by lemma 4.4, we see that there is a real quadratic field $E$ such that $\lambda, \psi \in E$. Since $s$ is the root of

\[ x^2-\psi x+\lambda =(x-s)(x-\overline{s}) \in E[x] \]

and $s$ is of degree $d>2$, at least one of the numbers $\lambda, \psi$ must be quadratic (otherwise $x^2-\psi x+\lambda \in {\mathbb {Q}}[x]$ and $d \leq 2$).

Now, we will show that such $s$ must be exceptional. For this, by (1.5), it suffices to show that $s$ is quartic. Assume that the conjugates of $\lambda$ and $\psi$ over ${\mathbb {Q}}$ are $\lambda '$ and $\psi '$ respectively. Here, $\lambda '=\lambda$ if $\lambda \in {\mathbb {Q}}$ and $\psi '=\psi$ if $\psi \in {\mathbb {Q}}$. By the above, we must have either $\lambda ' \ne \lambda$ or $\psi ' \ne \psi$ (or both). With this notation, it follows that $s$ is a root of the polynomial

(6.1)$$ Q(x)=(x^2-\psi x+ \lambda)(x^2-\psi' x +\lambda') \in {\mathbb{Q}}[x]. $$

If the polynomial $Q$ were reducible over ${\mathbb {Q}}$ then its irreducible factor with the root $s$ must be cubic. So $Q$ must have a rational root. However, if $r \in {\mathbb {Q}}$ is a root of $Q$, then, by (6.1),

\[ r^2-\psi r +\lambda=0 \quad \text{or} \quad r^2-\psi' r +\lambda'=0. \]

Taking an automorphism of the Galois group ${\rm Gal}(E/{\mathbb {Q}})$ that maps $\lambda$ to $\lambda '$ and so $\psi$ to $\psi '$ (or vice versa) we see that both displayed equalities must hold. Hence, $r$ is at least a double root of $Q$, so $s$ cannot be cubic. This proves that $Q$ is irreducible over ${\mathbb {Q}}$, and hence $s$ is a quartic number ($d=4$), which is not allowed by the condition of the theorem. This completes the proof of theorem 1.3.

7. Proof of theorem 1.5

Note that in the previous section we have proved the required result for the first part of the theorem for all quartic numbers $s$ as well except for those with minimal polynomial $Q$ defined in (6.1). By (1.5) and the irreducibility of $Q$, these are exactly exceptional numbers.

To prove the second part of the theorem we assume that for some exceptional $s$ and some $s_1, s_2, s_3 \in {\mathbb {N}}[s]$ the sumset (1.6) is everywhere dense in ${\mathbb {C}}$. Then, by theorem 1.9, the imaginary parts

\[ \Im(s_1 \overline{s_2}), \quad \Im(s_2 \overline{s_3}), \quad \Im(s_3 \overline{s_1}), \]

must be linearly independent over ${\mathbb {Q}}$. Since $s=\varrho e^{i\alpha } \in {\mathbb {C}} \setminus {\mathbb {R}}$, we have $\varrho \sin (\alpha ) \ne 0$, so that the three numbers

(7.1)$$ \frac{\Im(s_1 \overline{s_2})}{\varrho \sin (\alpha)}, \quad \frac{\Im(s_2 \overline{s_3})}{\varrho \sin(\alpha)}, \quad \frac{\Im(s_3 \overline{s_1})}{\varrho \sin(\alpha)}, $$

must be linearly independent over ${\mathbb {Q}}$. As $s$ is exceptional, by (1.5), there is a real quadratic number field $E$ such that

(7.2)$$ \varrho^2=s \overline{s} \in E \quad \text{and} \quad \varrho \cos(\alpha) = \frac{s+\overline{s}}{2} \in E. $$

Below, we will show that the numbers in (7.1) all belong to $E$. Since $E$ is quadratic, any three (not necessarily distinct) numbers in $E$ must be linearly dependent. This contradicts the linear independence of the three numbers (7.1).

Indeed, write $s_1=\sum _{i \in I} a_i s^i$ and $s_2=\sum _{j \in J} b_j s^j$, where $I, J$ are some finite subsets of ${\mathbb {N}}$ and $a_i, b_j \in {\mathbb {N}}$ for $i \in I$, $j \in J$. Note that

\[ \Im(s_1 \overline{s_2})=\sum_{i \in I, \> j \in J} a_i b_j \varrho^{i+j} \sin((i-j)\alpha). \]

The terms with $i=j$ are all equal to zero. So the first number on the list (7.1) is the sum of several terms of the form

\[ c_{i,j} \varrho^{i+j-1} \frac{\sin((i-j)\alpha)}{\sin(\alpha)}, \]

where $i>j$ and $c_{i,j} \in {\mathbb {Z}}$. It is well known that the quotient ${\sin ((i-j)\alpha )}/{\sin (\alpha )}$ is $U_{i-j-1}(\cos (\alpha ))$, where $U_0(x)=1$ and

\[ U_{n}(x)=\sum_{k=0}^{[n/2]}=({-}1)^k {n-k \choose k} (2x)^{n-2k} \]

is the Chebyshev polynomial of the second kind for $n \in {\mathbb {N}}$. Therefore, ${\Im (s_1 \overline {s_2})}/{\varrho \sin (\alpha )}$ is the sum of integral multiples of several terms of the form

\[ \varrho^{i+j-1} \cos(\alpha)^{i-j-2k+1}, \]

where $i, j, k$ are positive integers satisfying $i-j-2k+1 \geq 0$. Note that

\[ \varrho^{i+j-1} \cos(\alpha)^{i-j-2k+1}=\varrho^{2j+2k-2} \cdot (\varrho \cos(\alpha))^{i-j-2k+1}, \]

where the factors $\varrho ^{2j+2k-2}$ and $(\varrho \cos (\alpha ))^{i-j-2k+1}$ are both in $E$ by (7.2). Consequently, the first number in (7.1) belongs to $E$. By exactly the same argument, the second and the third numbers in (7.1) are also in $E$. This completes the proof of the second part of theorem 1.5.

In all that follows we will prove that for a quartic exceptional number $s \in {\mathbb {C}} \setminus {\mathbb {R}}$, whose minimal polynomial (6.1) has two nonreal roots $s$, $\overline {s}$, and two other roots $s', s''$, there exist $t_1, t_2, t_3, t_4 \in {\mathbb {N}}[s]$ such that the sumset (1.7) is everywhere dense in ${\mathbb {C}}$. We remark that the density of ${\mathbb {N}}[s]$ in ${\mathbb {C}}$ for some quartic numbers $s$ with minimal polynomial (6.1) can be established by corollary 2.2. Indeed, for a quartic $s$ with minimal polynomial (6.1), corollary 2.2 implies the density of ${\mathbb {N}}[s]$ in ${\mathbb {C}}$ in the case when $s$ has no real positive conjugates over ${\mathbb {Q}}$. However, the result that we are going to prove is stronger, since in (1.7) we only use the sumset of four sets.

Firstly, by lemma 4.2, we can replace $s$ by $ms+s^2$ with appropriate sufficiently large $m \in {\mathbb {N}}$ such that the argument $\beta$ of $ms+s^2$ satisfies ${\beta }/{\pi } \notin {\mathbb {Q}}$. Then, $ms+s^2 \ne m \overline {s}+\overline {s}^2$, so $ms+s^2$ is a quartic exceptional number by (1.5). Since ${\mathbb {N}}[ms+s^2] \subseteq {\mathbb {N}}[s]$, we can further argue with the number $ms+s^2$, which we denote by $s$. Its conjugates are $\overline {s}, s', s''$, its minimal polynomial is (6.1), and its argument $\beta$ satisfies ${\beta }/{\pi } \notin {\mathbb {Q}}$.

Assume first $\lambda =s \overline {s}$ is a rational number. Since the argument $\beta$ of $s$ satisfies ${\beta }/{\pi } \notin {\mathbb {Q}}$, as in lemma 4.3 (see (4.9)), we can select $l \in {\mathbb {N}}$ such that

(7.3)$$ \frac{1}{2}< \left\{\frac{l\beta}{2\pi}\right\} < \frac{2}{3}. $$

Then, for some $k \in {\mathbb {Z}}$ we get ${1}/{2}< {l\beta }/{2\pi }-k < {2}/{3}$. Hence, $\pi < l\beta -2\pi k <{4\pi }/{3}$, which implies

(7.4)$$ \cos(l\beta)<{-}\frac{1}{2}. $$

Note that $s^l \ne \overline {s}^l$ by the above-mentioned property of $\beta$, so $s^l$ is exceptional. Since ${\mathbb {N}}[s^l] \subseteq {\mathbb {N}}[s]$, it suffices to prove the density of (1.7) in ${\mathbb {C}}$ for $s$ replaced by $s^l$. We will write $s$ for $s^l$. For this $s$, we have $\lambda \in {\mathbb {Q}}$, $\lambda >0$ and $\psi =2|s|^l \cos (l\beta )<0$ by (7.4). From $s^2-\psi s +\lambda =0$ we thus obtain $\psi ={s^2+\lambda }/{s}$. Choosing $L \in {\mathbb {N}}$ for which $L \lambda \in {\mathbb {N}}$ we deduce that the negative irrational number $\psi$ is a quotient of $s_1=Ls^3+L\lambda s \in {\mathbb {N}}[s]$ and $s_2=Ls^2 \in {\mathbb {N}}[s]$. By lemma 4.5, this proves (1.7) with the choice $t_1=s_1$, $t_2=s_2$, $t_3=s_1 s$ and $t_4=s_2 s$.

Assume now that $\lambda =s \overline {s}>0$ is irrational. Then, $\lambda \ne \lambda '=s' s''$. We now reduce the problem to the case when the minimal polynomial $Q$ of $s$ (as in (6.1)) has two nonreal roots $s, \overline {s}$ and two other roots $s', s''$ satisfying

(7.5)$$ \psi=s+\overline{s}<0, \quad \psi<\psi'=s'+s' \quad \text{and} \quad \lambda=s \overline{s} > \lambda' =s's' >0. $$

Consider the number $s^2$, which we denote by $s$. It is exceptional, with conjugates $\overline {s}, s', s''$. Note that for this number we have $\lambda '=s' s''>0$ (which was possibly negative in case the original $s$ had a positive and a negative conjugate). Since $\lambda \ne \lambda '$, the third inequality in (7.5) is either true, or we have $\lambda <\lambda '$. But in that case we can replace $s$ by $s^{-1}$ (which is also exceptional). Then, the third inequality in (7.5) becomes true. Furthermore, if a real negative number $\omega$ is expressible by a quotient of two polynomials in the variable $s^{-1}$ with coefficients in ${\mathbb {N}}$, then, by multiplying these two polynomials by an appropriate power of $s$, we see that the same number $\omega$ is also a quotient of two polynomials in $s$ with coefficients in ${\mathbb {N}}$. So, we can always assume that the third inequality in (7.5) is true.

Now, we will show that without the restriction of generality we may also assume the first two inequalities in (7.5). Indeed, if the conjugates $s', s''$ of $s$ are both real, then we can choose an even integer $l$ for which (7.3) holds. Then, (7.4) is also true, and we can replace $s$ by $s^l$. Thus, replacing $s$ by $s^l$, which is exceptional, we find that the new $\psi$ and $\psi '$ satisfy

\[ \psi=s^l+\overline{s}^l=2|s|^l \cos(l\beta) < 0 < (s')^l+(s'')^l = \psi'. \]

The new $\lambda$, namely $\lambda ^l$, is still a positive rational number, satisfying $\lambda ^l>(\lambda ')^l$. So all three inequalities in (7.5) hold.

Alternatively, if $s'$ and $s''$ are both nonreal, then they are complex conjugate numbers. Thus, they have the same modulus, say $\varrho '$, satisfying $\varrho '<|s|$ in view of the third inequality in (7.5). In that case we choose $l \in {\mathbb {N}}$ satisfying (7.3) (and so (7.4)) and in addition so large that

\[ \left(\frac{|s|}{\varrho'}\right)^l>3. \]

Now, replacing $s$ by $s^l$, we find that

\[ \psi=s^l+\overline{s}^l=2|s|^l \cos(l\beta) < 2(\varrho')^l \cos(l \arg s')=(s')^l+(s'')^l=\psi', \]

where the (only) inequality above holds due to

\[ \left(\frac{|s|}{\varrho'}\right)^l \cos(l \beta) <{-}\frac{1}{2} \left(\frac{|s|}{\varrho'}\right)^l <{-} \frac{3}{2}<{-}1 \leq \cos(l \arg s'). \]

This proves the second inequality in (7.5). The first inequality there, i.e. $\psi <0$, is also true, since $\cos (l\beta )<0$. This, replacing $s$ by $s^l$, reduces the problem to the case of exceptional numbers satisfying (7.5).

In order to apply lemma 4.5 we will show that there is a negative real irrational number expressible by two nonzero elements $s_1, s_2$ of ${\mathbb {N}}[s]$ for $s$ with its conjugates $\overline {s}$ and $s', s''$ satisfying (7.5). Of course, then lemma 4.5 implies the density of the sumset (1.7) in ${\mathbb {C}}$ with

(7.6)$$ t_1=s_1, \quad t_2=s_2, \quad t_3=s_1 s \quad \text{and} \quad t_4=s_2 s. $$

By (1.5), we have $\psi, \lambda \in E$ and so their conjugates $\psi ', \lambda '$ are also in the same real quadratic field $E$. Take a square-free integer $D>1$ for which $E={\mathbb {Q}}(\sqrt {D})$. Then, there are some rational numbers $\psi _1, \psi _2, \lambda _1, \lambda _2$ such that

(7.7)$$ \psi=\psi_1-\psi_2 \sqrt{D} \quad \text{and} \quad \lambda=\lambda_1+\lambda_2 \sqrt{D}. $$

This implies

\[ \psi'=\psi_1+\psi_2 \sqrt{D} \quad \text{and} \quad \lambda'=\lambda_1-\lambda_2 \sqrt{D}. \]

From the second inequality in (7.5), namely $\psi <\psi '$, we obtain $\psi _2>0$. Likewise, from the third inequality in (7.5), $\lambda > \lambda '$, it follows that $\lambda _2 > 0$. Now, from $\lambda '>0$ we obtain $\lambda _1>\lambda _2 \sqrt {D}>0$. Consequently,

(7.8)$$ \lambda_1>0, \quad \lambda_2 > 0, \quad \psi_2>0. $$

Let $L$ be the least positive integer for which

(7.9)$$ L\psi_1, L\psi_2, L\lambda_1, L\lambda_2 \in {\mathbb{Z}}. $$

Consider two cases, $\psi _1 \leq 0$ and $\psi _1 > 0$. In the first case, $\psi _1 \leq 0$, from

\[ s^2-\psi s+ \lambda=s^2-(\psi_1-\psi_2\sqrt{D})s+\lambda_1+\lambda_2\sqrt{D}=0 \]

it follows that

\[ -\sqrt{D}=\frac{s^2-\psi_1 s+\lambda_1}{\psi_2s+ \lambda_2}=\frac{Ls^3+L(-\psi_1)s^2+L\lambda_1 s}{L\psi_2 s^2+L \lambda_2 s}. \]

Hence, by (7.8), (7.9) and $-\psi _1 \geq 0$, the negative irrational number $-\sqrt {D}$ is a quotient of two elements of ${\mathbb {N}}[s]$, i.e. $s_1=Ls^3+L(-\psi _1)s^2+L\lambda _1 s$ and $s_2=L\psi _2 s^2+L \lambda _2 s$. Now, lemma 4.5 implies that the corresponding sumset of four sets (1.7) with the choice (7.6) is everywhere dense in ${\mathbb {C}}$.

In the second case, $\psi _1>0$, a key observation is the following identity:

\[ \psi=\frac{s^2+\lambda}{s}=\frac{s^2+\lambda_1+\frac{\psi_1 \lambda_2}{\psi_2}}{s+\frac{\lambda_2}{\psi_2}}. \]

Here, the second equality can be verified directly using $s^2+\lambda =\psi s$ and (7.7):

$$\begin{align*} (s^2+\lambda)\left(s+\frac{\lambda_2}{\psi_2}\right) & =s^3+\frac{\lambda_2}{\psi_2}s^2+\lambda s+ \frac{\lambda \lambda_2}{\psi_2}=s^3+\lambda s+ \frac{\lambda_2}{\psi_2}(s^2+\lambda)\\ & =s^3+\lambda s+ \frac{\lambda_2}{\psi_2} \psi s=s^3+ \left(\lambda+\frac{\lambda_2 \psi}{\psi_2}\right)s \\& =s^3+ \left(\lambda_1+\lambda_2 \sqrt{D}+\frac{\lambda_2(\psi_1-\psi_2 \sqrt{D})}{\psi_2}\right)s \\ & =s^3+\left(\lambda_1+\frac{\psi_1 \lambda_2}{\psi_2}\right)s=s \left(s^2+\lambda_1+\frac{\psi_1 \lambda_2}{\psi_2}\right). \end{align*}$$

Now, multiplying numerator and denominator by the factor $L^2\psi _2 s$, we derive that

\[ \psi=\frac{L^2 \psi_2 s^3+(L^2\psi_2\lambda_1+L^2\psi_1 \lambda_2)s}{L^2\psi_2 s^2+L^2\lambda_2s}. \]

By (7.8), (7.9) and $\psi _1>0$, we conclude that $\psi$ is a quotient of two nonzero elements of ${\mathbb {N}}[s]$. Note that, by the first inequality in (7.5), $\psi$ is negative, whereas, by $\psi _2>0$, it is irrational. Therefore, lemma 4.5 again implies that the sumset of four sets (1.7) with the choice (7.6) is everywhere dense in ${\mathbb {C}}$. This completes the proof of the last part of theorem 1.5.