Let $g(z)=1+e^z+e^{\alpha z}$ and $A=\{\Re(z) \mid g(z)=0\}$. Prove that $\overline{A}=[a,b]$. Let $g:\mathbb{C} \to \mathbb{C}$ and $\alpha\in \mathbb{R}\setminus \mathbb{Q}$ such that $g(z)=1+e^z+e^{\alpha z}$ and let $A=\{\Re(z) \mid g(z)=0\}$.(Note that $\Re(z)$ is the real part of a complex number $z$ )

Prove $\overline{A}=[a,b]$ for some $a,b \in \mathbb{R}$.

let  $1<\alpha $ (for case $0<\alpha<1,\alpha<0$  we have similar way) $$
r_{0}=\inf \left\{r\mid 1<e^{r}+e^{\alpha r}\right\}, \quad r_{1}=\sup \left\{r \mid e^{\alpha r}<1+e^{r}\right\}
$$
so $r_{0}<0 <r_{1}$ now for every $z$ such that $g(z)=0$ we have :
$$
1 \leq e^{\operatorname{Re}(z)}+e^{\alpha \operatorname{Re}(z)}, \quad e^{\alpha \operatorname{Re}(z)} \leq 1+e^{\operatorname{Re}(z)}
$$ then $r_{0}<\Re(z) <r_{1}$ so $\bar{A} \subset\left[r_{0}, r_{1}\right]$ i want to show $\left[r_{0}, r_{1}\right] \subset \bar{A}$.
 A: As in the OP, I will assume that $\alpha > 1$ is an irrational number. Also, let $r_0$ and $r_1$ be as in the OP, so that $A \subseteq [r_0, r_1]$. Now we prove:

Claim. $\overline{A} = [r_0, r_1]$.

As the first step, we show that approximate zeros of $g(z)$ with arbitrary real part in $[r_0, r_1]$ can be found:

Lemma 1. Let $r \in [r_0, r_1]$ and $\varepsilon > 0$. Then there exists $z_0$ such that
$$ \operatorname{Re}(z_0) = r \qquad\text{and}\qquad \left|g(z_0)\right| < \varepsilon. $$

Proof of Lemma 1. Since $\alpha$ is irrational, for each $\theta \in \mathbb{R}$, the sequence
$$ g\bigl(r + i\theta + \tfrac{2\pi i k}{\alpha}\bigr) = 1 + e^r e^{2\pi i(k/\alpha) + i\theta} + e^{\alpha r}e^{i\theta}, \qquad k \in \mathbb{Z}$$
is dense in the circle of radius $e^{r}$ centered at $1 + e^{\alpha r}e^{i\theta}$. This tells that $g(r+i\mathbb{R})$ is a dense subset of the union $A$ of all such circles given by
\begin{align*}
A
&= \{ 1 + e^r e^{i\phi} + e^{\alpha r} e^{i\theta} : \phi, \theta \in \mathbb{R} \} \\
&= \{w \in \mathbb{C} : \left|e^{\alpha r} - e^{r}\right| \leq |w - 1| \leq e^{\alpha r} + e^r \}.
\end{align*}
Moreover, from the choice of $r_0$ and $r_1$, we have $\left|e^{\alpha r} - e^{r}\right| \leq 1 \leq e^{\alpha r} + e^r$. This shows that $0 \in A$ and hence is a limit point of the set $g(r+i\mathbb{R})$, proving the desired claim. $\square$
The figure below depicts the image of $g(0.7+i[-300,300])$ when $\alpha=\sqrt{2}$. The origin is represented by the black dot.

Now we are ready to prove the claim. To this end, we will show that true zeros of $g(z)$ can be found near the approximate zeros discovered in Lemma 1 in a uniform manner. The upshot of this argument is the next lemma, which immediately implies the desired claim that $\overline{A} = [r_0, r_1]$.

Lemma 2. There exists a constant $\varepsilon_0 > 0$, depending only on $\alpha$, such that the following hold: For any $r \in [r_0, r_1]$ and $\varepsilon \in (0, \varepsilon_0)$, there exists $z_1$ such that
$$ \left|\operatorname{Re}(z_1) - r\right| < \varepsilon \qquad\text{and}\qquad g(z_1) = 0. $$

Proof of Lemma 2. We introduce several constants that depend only on $\alpha$:

*

*$\text{(1)}$ Let $c_1 = \sup\{|g''(z)| : r_0 - 1 \leq \operatorname{Re}(z) \leq r_1 + 1\}$. It is clear that $c_1 \leq e^{r_1+1} + \alpha^2 e^{\alpha(r_1+1)} < \infty$.


*$\text{(2)}$ Let $c_2 = \alpha - (\alpha - 1)e^{r_1}$. Noting that $\alpha = \frac{\log(1+e^{r_1})}{r_1}$ and $r_1 > 0$, it is not hard to show that $c_2 > 0$. This constant is relevant to our estimation through the inequality: If $\operatorname{Re}(z) \in [r_0, r_1]$, then
\begin{align*}
|g'(z)|
= |e^z + \alpha e^{\alpha z}|
&= |-\alpha -(\alpha-1)e^z + \alpha g(z)|\\
&\geq \alpha - (\alpha-1) \operatorname{Re}(e^z) - \alpha |g(z)| \\
&\geq c_2 - \alpha |g(z)|.
\end{align*}


*$\text{(3)}$ We choose $\varepsilon_0 = \min\{1, c_2/c_1 \}$.


*$\text{(4)}$ Choose $c_3 > 0$ so as to satisfy $(\alpha \varepsilon_0+1)c_3 < \frac{1}{2}c_2$.
Now we return to the proof. Let $r \in [r_0, r_1]$ and $\varepsilon \in (0, \varepsilon_0)$. Also, use Lemma 1 to choose $z_0$ such that $\operatorname{Re}(z_0) = r$ and $|g(z_0)| \leq c_3 \varepsilon$. Also, define
$$h(z) = g(z_0) + g'(z_0)(z - z_0) . $$
Then for $z$ satisfying $|z - z_0| = \varepsilon$, Lagrange's form of the remainder gives
\begin{align*}
\left| g(z) - h(z) \right|
&= \left| (z - z_0)^2 \int_{0}^{1} (1 - t)g''(z_0 + t(z - z_0)) \, \mathrm{d}t \right| \\
&\leq \tfrac{1}{2}c_1\varepsilon^2. \tag{by (1) and (3)}
\end{align*}
Moreover, still assuming that $|z - z_0| = \varepsilon$, we have
\begin{align*}
|h(z)|
&\geq |g'(z_0)|\varepsilon - |g(z_0)| \\
&\geq c_2 \varepsilon - (\alpha \varepsilon + 1) |g(z_0)| \tag{by (2)} \\
&\geq c_2 \varepsilon - (\alpha \varepsilon_0 + 1) c_3 \varepsilon \tag{by the choice of $z_0$} \\
&> c_2 \varepsilon - \tfrac{1}{2}c_2 \varepsilon \tag{by (4)} \\
&\geq \tfrac{1}{2}c_1\varepsilon^2 \tag{by (3)} \\
&\geq \left| g(z) - h(z) \right|.
\end{align*}
So by Rouché's theorem, $g(z)$ and $h(z)$ has the same number of zeros inside the circle $|z - z_0| = \varepsilon$. Moreover, since the zero of $h(z)$ satisfies
$$ h(z) = 0
\qquad\implies\qquad
|z - z_0|
= \left|-\frac{g(z_0)}{g'(z_0)}\right|
\leq \frac{c_3\varepsilon}{c_2 - \alpha c_3\varepsilon}
< \varepsilon \tag{by (4)}, $$
hence $h(z)$ has a zero inside the circle $|z - z_0| = \varepsilon$. Therefore the same must be true for $g(z)$ and the conclusion follows. $\square$
