**Lemma A.1.** Let $(\mathbb{C}^{n \times n}, \|\cdot\|)$ be a normed space and $T \in \mathbb{C}^{n \times n}$. If $\rho(T) < 1$, then

$$ \sum_{k=0}^{\infty} T^k \text{ converges and } \sum_{k=0}^{\infty} T^k = (I - T)^{-1}. $$

Moreover, if $\|T\| < 1$, $\|(I - T)^{-1}\| \le \frac{1}{1 - \|T\|}$.

**Lemma A.2.** Let $T \in \mathbb{C}^{n \times n}$ such that $\rho(T) < 1$. Set

$$ s(T) := \sup_{z \in \mathbb{C}, |z| \ge 1} \|(I - T/z)^{-1}\| \quad (69) $$

then $0 < \|(I - T)^{-1}\| \le s(T) = s(T^*) < +\infty$. Moreover, if $\|T\| < 1$, $0 < s(T) \le \frac{1}{1 - \|T\|}$.

*Proof.* The existence of $s(T)$ (and also $s(T^*)$) is deduced from the fact that the functional $z \mapsto \|(I - T/z)^{-1}\|$, with $z \in \mathbb{C}, |z| \ge 1$, is well-defined and continuous. For every $z \in \mathbb{C}, |z| \ge 1$ we have

$$ \|(I - T/z)^{-1}\| = \|((I - T/z)^{-1})^*\| = \|(I - T^*/z^*)^{-1}\| \le s(T^*) $$

and

$$ \|(I - T^*/z)^{-1}\| = \|((I - T^*/z)^{-1})^*\| = \|(I - T/z^*)^{-1}\| \le s(T), $$

thus $s(T) = s(T^*)$. The second part of conclusion is obtained by Lemma A.1. □

The following lemma says that, for $T \in \mathbb{C}^{n \times n}$ and $\lambda \in \mathbb{C}, |\lambda| \ge 1$, we can decompose $(I - \frac{T}{\lambda})^{-1} = P(\lambda) + iQ(\lambda)$ and $(I - \frac{T^*}{\lambda})^{-1} = P(\lambda)^* + iQ(\lambda)^*$, and gives bounds for $P(\lambda)$ and $Q(\lambda)$.

**Lemma A.3.** Let $T \in \mathbb{C}^{n \times n}$ such that $\rho(T) < 1$ and $\lambda \in \mathbb{C}, |\lambda| \ge 1$. Write $\frac{1}{\lambda} = r(\cos \phi + i \sin \phi)$ in polar form, where $0 < r \le 1$ and $\phi \in [-\pi, \pi]$. Then

$$ \left(I - \frac{T}{\lambda}\right)^{-1} = P(\lambda) + iQ(\lambda) \quad \text{and} \quad \left(I - \frac{T^*}{\lambda}\right)^{-1} = P(\lambda)^* + iQ(\lambda)^* $$

where

$$ P(\lambda) = (I - r \cos \phi T)(I - 2r \cos \phi T + r^2 T^2)^{-1} $$

and

$$ Q(\lambda) = r \sin \phi T (I - 2r \cos \phi T + r^2 T^2)^{-1} $$

are $\mathbb{C}^{n \times n}$-valued functions. We also have the following properties:

$$ (i) \quad \|P(\lambda)\| \le (1 + \|T\|) s(T)^2 \quad \text{and} \quad \|Q(\lambda)\| \le |\sin \phi| \|T\| s(T)^2 \le \|T\| s(T)^2. $$

(ii) Moreover if $\|T\| < 1$ then

$$ \|P(\lambda)\| \le \frac{1}{1 - \|T\|} \quad \text{and} \quad \|Q(\lambda)\| \le \frac{\|T\|}{1 - \|T\|}. $$

*Proof.* The first part of the lemma is verified by direct computation, using

$$ \begin{align*} (I - T/\lambda)^{-1} &= (I - T/\lambda^*) [(I - T/\lambda)(I - T/\lambda^*)]^{-1}, \\ (I - T^*/\lambda)^{-1} &= [(I - T^*/\lambda^*)(I - T^*/\lambda)]^{-1} (I - T^*/\lambda^*) \\ \text{and } (I - T/\lambda)(I - T/\lambda^*) &= I - 2r \cos \phi T + r^2 T^2. \end{align*} $$
After that, with the help of Lemma A.2, it is not difficult to show the inequalities in (i). To prove (ii), first observe that the two series

$$ \sum_{k=0}^{\infty} r^k \cos(k\phi)T^k \quad \text{and} \quad \sum_{k=1}^{\infty} r^k \sin(k\phi)T^k $$

converge. Then, by expanding and simplifying the left-hand sides, we can show that

$$ (\sum_{k=0}^{\infty} r^k \cos(k\phi)T^k)(I - 2r \cos \phi T + r^2 T^2) = I - r \cos \phi T $$

and

$$ (\sum_{k=1}^{\infty} r^k \sin(k\phi)T^k)(I - 2r \cos \phi T + r^2 T^2) = r \sin \phi T, $$

so $P(\lambda)$ and $Q(\lambda)$ can be expressed as the series above, and the inequalities in (ii) follow. $\square$

In Sections 3.2 and 4.2 we identify different cases of $\lambda \in \mathbb{C}$ and we need corresponding estimations, given in the following lemma.

**Lemma A.4.** For $\lambda \in \mathbb{C} \setminus \mathbb{R}, |\lambda| \ge 1$ we write $\lambda = R(\cos \theta + i \sin \theta)$ in polar form where $R \ge 1$, $\theta \in (-\pi, \pi)$, $\theta \ne 0$.

$$ (i) \quad \text{For } \lambda \text{ satisfying } \operatorname{Re}(\lambda^2 - \lambda) \ge 0, \text{ let } \gamma_1 = \gamma_1(\lambda) := \begin{cases} 1 & \text{if } \operatorname{Im}(\lambda^2 - \lambda) \ge 0, \\ -1 & \text{if } \operatorname{Im}(\lambda^2 - \lambda) < 0 \end{cases} \quad \text{then} \\ \operatorname{Re}(\lambda^2 - \lambda) + \gamma_1 \operatorname{Im}(\lambda^2 - \lambda) \ge |\lambda(\lambda - 1)| \ge 2|\sin(\theta/2)|. $$

$$ (ii) \quad \text{Let } 0 < \theta_0 \le \frac{\pi}{4}. \quad \text{For } \lambda \text{ satisfying } \operatorname{Re}(\lambda^2 - \lambda) < 0 \text{ and } \theta \in [\theta_0, \pi - \theta_0] \cup [-\pi + \theta_0, -\theta_0], \text{ let } \\ \gamma_2 = \gamma_2(\lambda) := \begin{cases} -1 & \text{if } \operatorname{Im}(\lambda^2 - \lambda) \ge 0, \\ 1 & \text{if } \operatorname{Im}(\lambda^2 - \lambda) < 0 \end{cases} \quad \text{then} \\ |\operatorname{Re}(\lambda^2 - \lambda) + \gamma_2 \operatorname{Im}(\lambda^2 - \lambda)| \ge |\lambda(\lambda - 1)| \ge 2 \sin(\theta_0/2). $$

$$ (iii) \quad \text{Let } 0 < \theta_0 \le \frac{\pi}{4} \text{ and } \delta_0 > 0. \quad \text{For } \lambda \text{ satisfying } \operatorname{Re}(\lambda^2 - \lambda) < 0 \text{ and } \theta \in (-\theta_0, \theta_0) \setminus \{0\}, \text{ let } \\ \gamma_3 = \gamma_3(\operatorname{sign}(\theta)) := \begin{cases} \left(\delta_0 + \sin \frac{3\theta_0}{2}\right) / \cos \frac{3\theta_0}{2} & \text{if } \theta > 0, \\ -\left(\delta_0 + \sin \frac{3\theta_0}{2}\right) / \cos \frac{3\theta_0}{2} & \text{if } \theta < 0 \end{cases} \quad \text{then} \\ \operatorname{Re}(\lambda^2 - \lambda) + \gamma_3 \operatorname{Im}(\lambda^2 - \lambda) \ge 2\delta_0 |\sin(\theta/2)| \\ \text{and} \\ \frac{|\operatorname{Re}(\lambda - 1) + \gamma_3 \operatorname{Im}(\lambda - 1)|}{\operatorname{Re}(\lambda^2 - \lambda) + \gamma_3 \operatorname{Im}(\lambda^2 - \lambda)} \le \frac{\sqrt{1 + \gamma_3^2}}{\delta_0}. $$

Moreover, if $0 < \theta_0 < \frac{\pi}{4}$ then

$$ \frac{|\gamma_3 \operatorname{Re}(\lambda - 1) - \operatorname{Im}(\lambda - 1)|}{\operatorname{Re}(\lambda^2 - \lambda) + \gamma_3 \operatorname{Im}(\lambda^2 - \lambda)} \le \max \left( \frac{\sqrt{1 + \gamma_3^2}}{\delta_0}, \frac{\sqrt{1 + \gamma_3^2}}{\cos 2\theta_0} \right). $$

$$ (iv) \quad \text{Let } 0 < \theta_0 \le \frac{\pi}{4}. \text{ There exists no } \lambda \text{ satisfying } \operatorname{Re}(\lambda^2 - \lambda) < 0 \text{ and } \theta \in (\pi - \theta_0, \pi) \cup (-\pi, -\pi + \theta_0). $$
After that, with the help of Lemma A.2, it is not difficult to show the inequalities in (i). To prove (ii), first observe that the two series

$$ \sum_{k=0}^{\infty} r^k \cos(k\phi)T^k \quad \text{and} \quad \sum_{k=1}^{\infty} r^k \sin(k\phi)T^k $$

converge. Then, by expanding and simplifying the left-hand sides, we can show that

$$ (\sum_{k=0}^{\infty} r^k \cos(k\phi)T^k)(I - 2r \cos \phi T + r^2 T^2) = I - r \cos \phi T $$

and

$$ (\sum_{k=1}^{\infty} r^k \sin(k\phi)T^k)(I - 2r \cos \phi T + r^2 T^2) = r \sin \phi T, $$

so $P(\lambda)$ and $Q(\lambda)$ can be expressed as the series above, and the inequalities in (ii) follow. $\square$

In Sections 3.2 and 4.2 we identify different cases of $\lambda \in \mathbb{C}$ and we need corresponding estimations, given in the following lemma.

**Lemma A.4.** For $\lambda \in \mathbb{C} \setminus \mathbb{R}, |\lambda| \ge 1$ we write $\lambda = R(\cos \theta + i \sin \theta)$ in polar form where $R \ge 1$, $\theta \in (-\pi, \pi)$, $\theta \ne 0$.

$$ (i) \quad \text{For } \lambda \text{ satisfying } \operatorname{Re}(\lambda^2 - \lambda) \ge 0, \text{ let } \gamma_1 = \gamma_1(\lambda) := \begin{cases} 1 & \text{if } \operatorname{Im}(\lambda^2 - \lambda) \ge 0, \\ -1 & \text{if } \operatorname{Im}(\lambda^2 - \lambda) < 0 \end{cases} \quad \text{then} \\ \operatorname{Re}(\lambda^2 - \lambda) + \gamma_1 \operatorname{Im}(\lambda^2 - \lambda) \ge |\lambda(\lambda - 1)| \ge 2|\sin(\theta/2)|. $$

$$ (ii) \quad \text{Let } 0 < \theta_0 \le \frac{\pi}{4}. \quad \text{For } \lambda \text{ satisfying } \operatorname{Re}(\lambda^2 - \lambda) < 0 \text{ and } \theta \in [\theta_0, \pi - \theta_0] \cup [-\pi + \theta_0, -\theta_0], \text{ let } \\ \gamma_2 = \gamma_2(\lambda) := \begin{cases} -1 & \text{if } \operatorname{Im}(\lambda^2 - \lambda) \ge 0, \\ 1 & \text{if } \operatorname{Im}(\lambda^2 - \lambda) < 0 \end{cases} \quad \text{then} \\ |\operatorname{Re}(\lambda^2 - \lambda) + \gamma_2 \operatorname{Im}(\lambda^2 - \lambda)| \ge |\lambda(\lambda - 1)| \ge 2 \sin(\theta_0/2). $$

$$ (iii) \quad \text{Let } 0 < \theta_0 \le \frac{\pi}{4} \text{ and } \delta_0 > 0. \quad \text{For } \lambda \text{ satisfying } \operatorname{Re}(\lambda^2 - \lambda) < 0 \text{ and } \theta \in (-\theta_0, \theta_0) \setminus \{0\}, \text{ let } \\ \gamma_3 = \gamma_3(\operatorname{sign}(\theta)) := \begin{cases} \left(\delta_0 + \sin \frac{3\theta_0}{2}\right) / \cos \frac{3\theta_0}{2} & \text{if } \theta > 0, \\ -\left(\delta_0 + \sin \frac{3\theta_0}{2}\right) / \cos \frac{3\theta_0}{2} & \text{if } \theta < 0 \end{cases} \quad \text{then} \\ \operatorname{Re}(\lambda^2 - \lambda) + \gamma_3 \operatorname{Im}(\lambda^2 - \lambda) \ge 2\delta_0 |\sin(\theta/2)| \\ \text{and} \\ \frac{|\operatorname{Re}(\lambda - 1) + \gamma_3 \operatorname{Im}(\lambda - 1)|}{\operatorname{Re}(\lambda^2 - \lambda) + \gamma_3 \operatorname{Im}(\lambda^2 - \lambda)} \le \frac{\sqrt{1 + \gamma_3^2}}{\delta_0}. $$

Moreover, if $0 < \theta_0 < \frac{\pi}{4}$ then

$$ \frac{|\gamma_3 \operatorname{Re}(\lambda - 1) - \operatorname{Im}(\lambda - 1)|}{\operatorname{Re}(\lambda^2 - \lambda) + \gamma_3 \operatorname{Im}(\lambda^2 - \lambda)} \le \max \left( \frac{\sqrt{1 + \gamma_3^2}}{\delta_0}, \frac{\sqrt{1 + \gamma_3^2}}{\cos 2\theta_0} \right). $$

$$ (iv) \quad \text{Let } 0 < \theta_0 \le \frac{\pi}{4}. \text{ There exists no } \lambda \text{ satisfying } \operatorname{Re}(\lambda^2 - \lambda) < 0 \text{ and } \theta \in (\pi - \theta_0, \pi) \cup (-\pi, -\pi + \theta_0). $$
Proof. Notice that $\gamma_1^2 = 1$, $\gamma_1 \operatorname{Im}(\lambda^2 - \lambda) \ge 0$. We have

$$[\operatorname{Re}(\lambda^2 - \lambda) + \gamma_1 \operatorname{Im}(\lambda^2 - \lambda)]^2 = [\operatorname{Re}(\lambda^2 - \lambda)]^2 + [\operatorname{Im}(\lambda^2 - \lambda)]^2 + 2\gamma_1 \operatorname{Re}(\lambda^2 - \lambda) \operatorname{Im}(\lambda^2 - \lambda) \\ \ge [\operatorname{Re}(\lambda^2 - \lambda)]^2 + [\operatorname{Im}(\lambda^2 - \lambda)]^2 = |\lambda^2 - \lambda|^2,$$

which yields $\operatorname{Re}(\lambda^2 - \lambda) + \gamma_1 \operatorname{Im}(\lambda^2 - \lambda) \ge |\lambda(\lambda - 1)|$. Finally,

$$|\lambda - 1| = |R \cos \theta - 1 + iR \sin \theta| = \sqrt{R^2 + 1 - 2R \cos \theta} \ge \sqrt{2 - 2 \cos \theta} = 2 \left| \sin \frac{\theta}{2} \right|$$

since the function $R \mapsto R^2 + 1 - 2R \cos \theta$, for $R \ge 1$, is increasing.

(ii) In this case we have $\frac{\theta}{2} \in [\frac{\theta_0}{2}, \frac{\pi}{2} - \frac{\theta_0}{2}] \cup [-\frac{\pi}{2} + \frac{\theta_0}{2}, -\frac{\theta_0}{2}]$ so $|\sin \frac{\theta}{2}| \ge \sin \frac{\theta_0}{2}$. We also notice that $\gamma_2^2 = 1$ and $\gamma_2 \operatorname{Im}(\lambda^2 - \lambda) \le 0$. Similar to (i), we have $|\operatorname{Re}(\lambda^2 - \lambda) + \gamma_2 \operatorname{Im}(\lambda^2 - \lambda)| = -\operatorname{Re}(\lambda^2 - \lambda) - \gamma_2 \operatorname{Im}(\lambda^2 - \lambda) \ge |\lambda(\lambda - 1)| \ge 2|\sin(\theta/2)|$, that implies the conclusion.

(iii) Note that $\cos 2\theta > 0$, $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$, and $\sin 2\theta$ has the same sign as $\theta$ and $\gamma_3$, so we have

$$\begin{aligned} \operatorname{Re}(\lambda^2 - \lambda) + \gamma_3 \operatorname{Im}(\lambda^2 - \lambda) &= R(R \cos 2\theta - \cos \theta + \gamma_3 R \sin 2\theta - \gamma_3 \sin \theta) \\ &\ge \cos 2\theta - \cos \theta + \gamma_3 \sin 2\theta - \gamma_3 \sin \theta \\ &= -2 \sin \frac{3\theta}{2} \sin \frac{\theta}{2} + 2\gamma_3 \cos \frac{3\theta}{2} \sin \frac{\theta}{2} \\ &= 2 \sin \frac{\theta}{2} \left( \gamma_3 \cos \frac{3\theta}{2} - \sin \frac{3\theta}{2} \right). \end{aligned}$$

Then we consider two cases: if $0 < \theta < \theta_0$ then $\gamma_3 > 0$, $|\sin \frac{\theta}{2}| = \sin \frac{\theta}{2} > 0$, $0 < \frac{3\theta}{2} < \frac{3\theta_0}{2} < \frac{\pi}{2}$ and $\gamma_3 \cos \frac{3\theta}{2} - \sin \frac{3\theta}{2} > \gamma_3 \cos \frac{3\theta_0}{2} - \sin \frac{3\theta_0}{2} = \delta_0$; if $-\theta_0 < \theta < 0$ then $-\gamma_3 > 0$, $|\sin \frac{\theta}{2}| = -\sin \frac{\theta}{2} > 0$, $-\frac{\pi}{2} < -\frac{3\theta_0}{2} < \frac{3\theta}{2} < 0$ and $-\gamma_3 \cos \frac{3\theta}{2} + \sin \frac{3\theta}{2} > -\gamma_3 \cos \frac{3\theta_0}{2} - \sin \frac{3\theta_0}{2} = \delta_0$.

Next, we will show that $\frac{|\operatorname{Re}(\lambda-1)+\gamma_3 \operatorname{Im}(\lambda-1)|}{\operatorname{Re}(\lambda^2-\lambda)+\gamma_3 \operatorname{Im}(\lambda^2-\lambda)}$ is bounded. First,

$$\frac{|\operatorname{Re}(\lambda - 1) + \gamma_3 \operatorname{Im}(\lambda - 1)|}{\operatorname{Re}(\lambda^2 - \lambda) + \gamma_3 \operatorname{Im}(\lambda^2 - \lambda)} = \frac{|(\cos \theta + \gamma_3 \sin \theta)R - 1|}{R[(\cos 2\theta + \gamma_3 \sin 2\theta)R - (\cos \theta + \gamma_3 \sin \theta)]} \\ \le \frac{|(\cos \theta + \gamma_3 \sin \theta)R - 1|}{(\cos 2\theta + \gamma_3 \sin 2\theta)R - (\cos \theta + \gamma_3 \sin \theta)}.$$

Since $\gamma_3$ does not depend on $R$, let us study $f_1(R) := \left(\frac{aR-1}{bR-a}\right)^2$ where $a := \cos \theta + \gamma_3 \sin \theta$ and $b := \cos 2\theta + \gamma_3 \sin 2\theta$. We observe that:

* $a > 0$ and $b > 0$. Indeed, $\cos \theta > 0$, $\cos 2\theta > 0$, and $\theta$ and $\gamma_3$ have the same sign.

* $bR - a > 0$ since $\operatorname{Re}(\lambda^2 - \lambda) + \gamma_3 \operatorname{Im}(\lambda^2 - \lambda) > 0$, thus $R > \frac{a}{b}$.

* $a^2 > b$ (equivalently $\frac{a}{b} > \frac{1}{a}$), since $a^2 = \cos^2 \theta + \gamma_3^2 \sin^2 \theta + \gamma_3 \sin 2\theta > \cos^2 \theta - \sin^2 \theta + \gamma_3 \sin 2\theta = b$.

Now, $f'_1(R) = 2 \cdot \frac{aR-1}{bR-a} \cdot \frac{b-a^2}{(bR-a)^2} < 0$ for $R > \frac{a}{b} > \frac{1}{a}$ and we would like to have $\frac{a}{b} < 1$ so that $f_1(R) \le f_1(1), \forall R \ge 1$. Indeed $\frac{a}{b} < 1$ is equivalent to

$$\cos \theta + \gamma_3 \sin \theta < \cos 2\theta + \gamma_3 \sin 2\theta \Leftrightarrow |\gamma_3| > \frac{|\sin \frac{3\theta}{2}|}{\cos \frac{3\theta}{2}},$$

which is true since

$$|\gamma_3| = \frac{\delta_0 + \sin \frac{3\theta_0}{2}}{\cos \frac{3\theta_0}{2}} > \frac{|\sin \frac{3\theta}{2}|}{\cos \frac{3\theta}{2}} + \varepsilon_0 \quad \text{where} \quad \varepsilon_0 = \frac{\delta_0}{\cos \frac{3\theta_0}{2}}.$$
Proof. Notice that $\gamma_1^2 = 1$, $\gamma_1 \operatorname{Im}(\lambda^2 - \lambda) \ge 0$. We have

$$[\operatorname{Re}(\lambda^2 - \lambda) + \gamma_1 \operatorname{Im}(\lambda^2 - \lambda)]^2 = [\operatorname{Re}(\lambda^2 - \lambda)]^2 + [\operatorname{Im}(\lambda^2 - \lambda)]^2 + 2\gamma_1 \operatorname{Re}(\lambda^2 - \lambda) \operatorname{Im}(\lambda^2 - \lambda) \\ \ge [\operatorname{Re}(\lambda^2 - \lambda)]^2 + [\operatorname{Im}(\lambda^2 - \lambda)]^2 = |\lambda^2 - \lambda|^2,$$

which yields $\operatorname{Re}(\lambda^2 - \lambda) + \gamma_1 \operatorname{Im}(\lambda^2 - \lambda) \ge |\lambda(\lambda - 1)|$. Finally,

$$|\lambda - 1| = |R \cos \theta - 1 + iR \sin \theta| = \sqrt{R^2 + 1 - 2R \cos \theta} \ge \sqrt{2 - 2 \cos \theta} = 2 \left| \sin \frac{\theta}{2} \right|$$

since the function $R \mapsto R^2 + 1 - 2R \cos \theta$, for $R \ge 1$, is increasing.

(ii) In this case we have $\frac{\theta}{2} \in [\frac{\theta_0}{2}, \frac{\pi}{2} - \frac{\theta_0}{2}] \cup [-\frac{\pi}{2} + \frac{\theta_0}{2}, -\frac{\theta_0}{2}]$ so $|\sin \frac{\theta}{2}| \ge \sin \frac{\theta_0}{2}$. We also notice that $\gamma_2^2 = 1$ and $\gamma_2 \operatorname{Im}(\lambda^2 - \lambda) \le 0$. Similar to (i), we have $|\operatorname{Re}(\lambda^2 - \lambda) + \gamma_2 \operatorname{Im}(\lambda^2 - \lambda)| = -\operatorname{Re}(\lambda^2 - \lambda) - \gamma_2 \operatorname{Im}(\lambda^2 - \lambda) \ge |\lambda(\lambda - 1)| \ge 2|\sin(\theta/2)|$, that implies the conclusion.

(iii) Note that $\cos 2\theta > 0, -\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$, and $\sin 2\theta$ has the same sign as $\theta$ and $\gamma_3$, so we have

$$\begin{aligned} \operatorname{Re}(\lambda^2 - \lambda) + \gamma_3 \operatorname{Im}(\lambda^2 - \lambda) &= R(R \cos 2\theta - \cos \theta + \gamma_3 R \sin 2\theta - \gamma_3 \sin \theta) \\ &\ge \cos 2\theta - \cos \theta + \gamma_3 \sin 2\theta - \gamma_3 \sin \theta \\ &= -2 \sin \frac{3\theta}{2} \sin \frac{\theta}{2} + 2\gamma_3 \cos \frac{3\theta}{2} \sin \frac{\theta}{2} \\ &= 2 \sin \frac{\theta}{2} \left( \gamma_3 \cos \frac{3\theta}{2} - \sin \frac{3\theta}{2} \right). \end{aligned}$$

Then we consider two cases: if $0 < \theta < \theta_0$ then $\gamma_3 > 0$, $|\sin \frac{\theta}{2}| = \sin \frac{\theta}{2} > 0$, $0 < \frac{3\theta}{2} < \frac{3\theta_0}{2} < \frac{\pi}{2}$ and $\gamma_3 \cos \frac{3\theta}{2} - \sin \frac{3\theta}{2} > \gamma_3 \cos \frac{3\theta_0}{2} - \sin \frac{3\theta_0}{2} = \delta_0$; if $-\theta_0 < \theta < 0$ then $-\gamma_3 > 0$, $|\sin \frac{\theta}{2}| = -\sin \frac{\theta}{2} > 0$, $-\frac{\pi}{2} < -\frac{3\theta_0}{2} < \frac{3\theta}{2} < 0$ and $-\gamma_3 \cos \frac{3\theta}{2} + \sin \frac{3\theta}{2} > -\gamma_3 \cos \frac{3\theta_0}{2} - \sin \frac{3\theta_0}{2} = \delta_0$.

Next, we will show that $\frac{|\operatorname{Re}(\lambda-1)+\gamma_3 \operatorname{Im}(\lambda-1)|}{\operatorname{Re}(\lambda^2-\lambda)+\gamma_3 \operatorname{Im}(\lambda^2-\lambda)}$ is bounded. First,

$$\frac{|\operatorname{Re}(\lambda - 1) + \gamma_3 \operatorname{Im}(\lambda - 1)|}{\operatorname{Re}(\lambda^2 - \lambda) + \gamma_3 \operatorname{Im}(\lambda^2 - \lambda)} = \frac{|(\cos \theta + \gamma_3 \sin \theta)R - 1|}{R[(\cos 2\theta + \gamma_3 \sin 2\theta)R - (\cos \theta + \gamma_3 \sin \theta)]} \\ \le \frac{|(\cos \theta + \gamma_3 \sin \theta)R - 1|}{(\cos 2\theta + \gamma_3 \sin 2\theta)R - (\cos \theta + \gamma_3 \sin \theta)}.$$

Since $\gamma_3$ does not depend on $R$, let us study $f_1(R) := \left(\frac{aR-1}{bR-a}\right)^2$ where $a := \cos \theta + \gamma_3 \sin \theta$ and $b := \cos 2\theta + \gamma_3 \sin 2\theta$. We observe that:

* $a > 0$ and $b > 0$. Indeed, $\cos \theta > 0, \cos 2\theta > 0$, and $\theta$ and $\gamma_3$ have the same sign.

* $bR - a > 0$ since $\operatorname{Re}(\lambda^2 - \lambda) + \gamma_3 \operatorname{Im}(\lambda^2 - \lambda) > 0$, thus $R > \frac{a}{b}$.

* $a^2 > b$ (equivalently $\frac{a}{b} > \frac{1}{a}$), since $a^2 = \cos^2 \theta + \gamma_3^2 \sin^2 \theta + \gamma_3 \sin 2\theta > \cos^2 \theta - \sin^2 \theta + \gamma_3 \sin 2\theta = b$.

Now, $f'_1(R) = 2 \cdot \frac{aR-1}{bR-a} \cdot \frac{b-a^2}{(bR-a)^2} < 0$ for $R > \frac{a}{b} > \frac{1}{a}$ and we would like to have $\frac{a}{b} < 1$ so that $f_1(R) \le f_1(1), \forall R \ge 1$. Indeed $\frac{a}{b} < 1$ is equivalent to

$$\cos \theta + \gamma_3 \sin \theta < \cos 2\theta + \gamma_3 \sin 2\theta \Leftrightarrow |\gamma_3| > \frac{|\sin \frac{3\theta}{2}|}{\cos \frac{3\theta}{2}},$$

which is true since

$$|\gamma_3| = \frac{\delta_0 + \sin \frac{3\theta_0}{2}}{\cos \frac{3\theta_0}{2}} > \frac{|\sin \frac{3\theta}{2}|}{\cos \frac{3\theta}{2}} + \varepsilon_0 \quad \text{where} \quad \varepsilon_0 = \frac{\delta_0}{\cos \frac{3\theta_0}{2}}.$$
Then we study

$$f_1(1) = \left( \frac{\cos \theta - 1 + \gamma_3 \sin \theta}{\cos 2\theta - \cos \theta + \gamma_3(\sin 2\theta - \sin \theta)} \right)^2 = \left( \frac{-\sin \frac{\theta}{2} + \gamma_3 \cos \frac{\theta}{2}}{-\gamma_3 \sin \frac{3\theta}{2} + \gamma_3^2 \cos \frac{3\theta}{2}} \right)^2 \gamma_3^2.$$

We have:

• $(-\sin \frac{\theta}{2} + \gamma_3 \cos \frac{\theta}{2})^2 \le 1 + \gamma_3^2$ by Cauchy-Schwartz inequality;

• $\gamma_3^2 = |\gamma_3|^2 > \frac{\gamma_3 \sin \frac{3\theta}{2}}{\cos \frac{3\theta}{2}} + \varepsilon_0 |\gamma_3|$ that leads to

$$-\gamma_3 \sin \frac{3\theta}{2} + \gamma_3^2 \cos \frac{3\theta}{2} > \varepsilon_0 \cos \frac{3\theta}{2} |\gamma_3| = \delta_0 |\gamma_3|;$$

hence $f_1(1) \le \frac{1+\gamma_3^2}{\delta_0^2}$ and this yields $\frac{|\mathrm{Re}(\lambda-1)+\gamma_3 \mathrm{Im}(\lambda-1)|}{\mathrm{Re}(\lambda^2-\lambda)+\gamma_3 \mathrm{Im}(\lambda^2-\lambda)} \le \frac{\sqrt{1+\gamma_3^2}}{\delta_0}$.

Finally, we show that if $0 < \theta_0 < \frac{\pi}{4}$ then $\frac{|\gamma_3 \mathrm{Re}(\lambda-1)-\mathrm{Im}(\lambda-1)|}{\mathrm{Re}(\lambda^2-\lambda)+\gamma_3 \mathrm{Im}(\lambda^2-\lambda)}$ is bounded. Indeed, we have

$$\begin{aligned} \frac{|\gamma_3 \mathrm{Re}(\lambda - 1) - \mathrm{Im}(\lambda - 1)|}{\mathrm{Re}(\lambda^2 - \lambda) + \gamma_3 \mathrm{Im}(\lambda^2 - \lambda)} &= \frac{|(\gamma_3 \cos \theta - \sin \theta)R - \gamma_3|}{R[(\cos 2\theta + \gamma_3 \sin 2\theta)R - (\cos \theta + \gamma_3 \sin \theta)]} \\ &\le \frac{|(\gamma_3 \cos \theta - \sin \theta)R - \gamma_3|}{(\cos 2\theta + \gamma_3 \sin 2\theta)R - (\cos \theta + \gamma_3 \sin \theta)}. \end{aligned}$$

Since $\gamma_3$ does not depend on $R$, let us study $f_2(R) := \left(\frac{cR-\gamma_3}{bR-a}\right)^2$ where $c := \gamma_3 \cos \theta - \sin \theta$ and $a, b$ as above. We observe that:

• $\gamma_3 b - ca$ and $\theta$ have the same sign. Indeed, $\gamma_3 b - ca = (\gamma_3^2 + 1) \sin \theta \cos \theta$. Consequently, we always have $(\gamma_3 b - ca)\gamma_3 > 0$.

• We always have $\frac{\gamma_3}{c} > 1$. Indeed, if $\theta > 0$ then $c > 0$ since $\gamma_3 = \frac{\delta_0+\sin \frac{3\theta_0}{2}}{\cos \frac{3\theta_0}{2}} > \frac{\sin \theta}{\cos \theta}$, also $\frac{\gamma_3}{c} = \frac{\gamma_3}{\gamma_3 \cos \theta - \sin \theta} > 1$; if $\theta < 0$ then $c < 0$ since $-\gamma_3 = \frac{\delta_0+\sin \frac{3\theta_0}{2}}{\cos \frac{3\theta_0}{2}} > -\frac{\sin \theta}{\cos \theta}$, also $\frac{\gamma_3}{c} = \frac{-\gamma_3}{-\gamma_3 \cos \theta + \sin \theta} > 1$.

Now, $f'_2(R) = 2 \cdot \frac{c}{\gamma_3 R - a} \cdot \frac{(\gamma_3 b - ca)\gamma_3}{(bR-a)^2}$, so, thanks to the above results, $f_2(R)$ decreases for $1 \le R < \frac{\gamma_3}{c}$ and increases for $R > \frac{\gamma_3}{c}$. Moreover, like for $f_1(1)$, we can estimate

$$f_2(1) = \left( \frac{-\cos \frac{\theta}{2} - \gamma_3 \sin \frac{\theta}{2}}{-\gamma_3 \sin \frac{3\theta}{2} + \gamma_3^2 \cos \frac{3\theta}{2}} \right)^2 \gamma_3^2 \le \frac{1 + \gamma_3^2}{\delta_0^2}$$

and $\lim_{R \to +\infty} f_2(R) = \left( \frac{\gamma_3 \cos \theta - \sin \theta}{\cos 2\theta + \gamma_3 \sin 2\theta} \right)^2 \le \frac{1+\gamma_3^2}{\cos 2\theta_0}$. Therefore

$$\frac{|\gamma_3 \mathrm{Re}(\lambda - 1) - \mathrm{Im}(\lambda - 1)|}{\mathrm{Re}(\lambda^2 - \lambda) + \gamma_3 \mathrm{Im}(\lambda^2 - \lambda)} \le \max \left( \frac{\sqrt{1 + \gamma_3^2}}{\delta_0}, \frac{\sqrt{1 + \gamma_3^2}}{\cos 2\theta_0} \right).$$

(iv) For $\theta \in (\pi - \theta_0, \pi) \cup (-\pi, -\pi + \theta_0)$, we have $\cos 2\theta > 0$ since $2\theta \in (\frac{3\pi}{2}, 2\pi) \cup (-2\pi, -\frac{3\pi}{2})$, while $\cos \theta < 0$. Hence $\mathrm{Re}(\lambda^2 - \lambda) = R(R \cos 2\theta - \cos \theta) > 0$. $\square$
Then we study

$$f_1(1) = \left( \frac{\cos \theta - 1 + \gamma_3 \sin \theta}{\cos 2\theta - \cos \theta + \gamma_3(\sin 2\theta - \sin \theta)} \right)^2 = \left( \frac{-\sin \frac{\theta}{2} + \gamma_3 \cos \frac{\theta}{2}}{-\gamma_3 \sin \frac{3\theta}{2} + \gamma_3^2 \cos \frac{3\theta}{2}} \right)^2 \gamma_3^2.$$

We have:

* $(-\sin \frac{\theta}{2} + \gamma_3 \cos \frac{\theta}{2})^2 \le 1 + \gamma_3^2$ by Cauchy-Schwartz inequality;

* $\gamma_3^2 = |\gamma_3|^2 > \frac{\gamma_3 \sin \frac{3\theta}{2}}{\cos \frac{3\theta}{2}} + \varepsilon_0 |\gamma_3|$ that leads to

$$-\gamma_3 \sin \frac{3\theta}{2} + \gamma_3^2 \cos \frac{3\theta}{2} > \varepsilon_0 \cos \frac{3\theta}{2} |\gamma_3| = \delta_0 |\gamma_3|;$$

hence $f_1(1) \le \frac{1+\gamma_3^2}{\delta_0^2}$ and this yields $\frac{|\mathrm{Re}(\lambda-1)+\gamma_3 \mathrm{Im}(\lambda-1)|}{\mathrm{Re}(\lambda^2-\lambda)+\gamma_3 \mathrm{Im}(\lambda^2-\lambda)} \le \frac{\sqrt{1+\gamma_3^2}}{\delta_0}$.

Finally, we show that if $0 < \theta_0 < \frac{\pi}{4}$ then $\frac{|\gamma_3 \mathrm{Re}(\lambda-1)-\mathrm{Im}(\lambda-1)|}{\mathrm{Re}(\lambda^2-\lambda)+\gamma_3 \mathrm{Im}(\lambda^2-\lambda)}$ is bounded. Indeed, we have

$$\begin{aligned} \frac{|\gamma_3 \mathrm{Re}(\lambda - 1) - \mathrm{Im}(\lambda - 1)|}{\mathrm{Re}(\lambda^2 - \lambda) + \gamma_3 \mathrm{Im}(\lambda^2 - \lambda)} &= \frac{|(\gamma_3 \cos \theta - \sin \theta)R - \gamma_3|}{R[(\cos 2\theta + \gamma_3 \sin 2\theta)R - (\cos \theta + \gamma_3 \sin \theta)]} \\ &\le \frac{|(\gamma_3 \cos \theta - \sin \theta)R - \gamma_3|}{(\cos 2\theta + \gamma_3 \sin 2\theta)R - (\cos \theta + \gamma_3 \sin \theta)}. \end{aligned}$$

Since $\gamma_3$ does not depend on $R$, let us study $f_2(R) := \left(\frac{cR-\gamma_3}{bR-a}\right)^2$ where $c := \gamma_3 \cos \theta - \sin \theta$ and $a, b$ as above. We observe that:

* $\gamma_3 b - ca$ and $\theta$ have the same sign. Indeed, $\gamma_3 b - ca = (\gamma_3^2 + 1) \sin \theta \cos \theta$. Consequently, we always have $(\gamma_3 b - ca)\gamma_3 > 0$.

* We always have $\frac{\gamma_3}{c} > 1$. Indeed, if $\theta > 0$ then $c > 0$ since $\gamma_3 = \frac{\delta_0+\sin \frac{3\theta_0}{2}}{\cos \frac{3\theta_0}{2}} > \frac{\sin \theta}{\cos \theta}$, also $\frac{\gamma_3}{c} = \frac{\gamma_3}{\gamma_3 \cos \theta - \sin \theta} > 1$; if $\theta < 0$ then $c < 0$ since $-\gamma_3 = \frac{\delta_0+\sin \frac{3\theta_0}{2}}{\cos \frac{3\theta_0}{2}} > -\frac{\sin \theta}{\cos \theta}$, also $\frac{\gamma_3}{c} = \frac{-\gamma_3}{-\gamma_3 \cos \theta + \sin \theta} > 1$.

Now, $f'_2(R) = 2 \cdot \frac{c}{\gamma_3 R - a} \cdot \frac{(\gamma_3 b - ca)\gamma_3}{(bR-a)^2}$, so, thanks to the above results, $f_2(R)$ decreases for $1 \le R < \frac{\gamma_3}{c}$ and increases for $R > \frac{\gamma_3}{c}$. Moreover, like for $f_1(1)$, we can estimate

$$f_2(1) = \left( \frac{-\cos \frac{\theta}{2} - \gamma_3 \sin \frac{\theta}{2}}{-\gamma_3 \sin \frac{3\theta}{2} + \gamma_3^2 \cos \frac{3\theta}{2}} \right)^2 \gamma_3^2 \le \frac{1 + \gamma_3^2}{\delta_0^2}$$

and $\lim_{R \to +\infty} f_2(R) = \left(\frac{\gamma_3 \cos \theta - \sin \theta}{\cos 2\theta + \gamma_3 \sin 2\theta}\right)^2 \le \frac{1+\gamma_3^2}{\cos 2\theta_0}$. Therefore

$$\frac{|\gamma_3 \mathrm{Re}(\lambda - 1) - \mathrm{Im}(\lambda - 1)|}{\mathrm{Re}(\lambda^2 - \lambda) + \gamma_3 \mathrm{Im}(\lambda^2 - \lambda)} \le \max \left( \frac{\sqrt{1 + \gamma_3^2}}{\delta_0}, \frac{\sqrt{1 + \gamma_3^2}}{\cos 2\theta_0} \right).$$

(iv) For $\theta \in (\pi - \theta_0, \pi) \cup (-\pi, -\pi + \theta_0)$, we have $\cos 2\theta > 0$ since $2\theta \in (\frac{3\pi}{2}, 2\pi) \cup (-2\pi, -\frac{3\pi}{2})$, while $\cos \theta < 0$. Hence $\mathrm{Re}(\lambda^2 - \lambda) = R(R \cos 2\theta - \cos \theta) > 0$. $\square$