Zeros of complex function Consider the function
$f(z)=e^{z}+\varepsilon_1e^{\varepsilon_1 z}+\varepsilon_2e^{\varepsilon_2 z}$ of a complex variable $z=x+i y$, where
$\varepsilon_1=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$, $\varepsilon_2=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$.
Numerical calculations show that all zeros of the function $f(z)$ are located on the lines $y=0$, $y=\pm \sqrt{3} x$.
Are there any ideas how to prove it theoretically?

Since $f(\varepsilon_1 z) = \varepsilon_2f(z)$, it is enough to study $f$ on a "third" of the complex plane, for example the infinite cone centered on the negative real axis with an angle of $2\pi/3$.
There, $|\exp(z)|$ will quickly get negligible compared to the other two exponentials :
$f(z) = \exp(z) + \exp(2i\pi/3+z\frac{-1+\sqrt {-3}}2) +  \exp(-2i\pi/3+z\frac{-1-\sqrt {-3}}2) \\
= \exp(z) + 2\exp(-z/2)\cos(2\pi/3+z\sqrt3/2)$
If, for $k \in \Bbb N$ (including $0$) you pick $z_k = -(2/\sqrt 3)(k+2/3)\pi $, so that $z_k\sqrt 3 / 2 + 2\pi/3 = -k\pi$, you get 
$f(z_k) = \exp(z_k) + 2(-1)^k\exp(-z_k)$.
Since $z_k$ is negative, the first term is smaller than the second (quite an understatement), so this has the sign of $(-1)^k$. 
Then by the Intermediate Value Theorem, you get a sequence of zeroes on the negative real axis, one on each interval $(z_{k+1} ; z_k)$ (it should be exponentially close to $(z_{k+1}+z_k)/2$).
Additionally, by expanding $f$ as a power series around zero, you get that there is a double zero at $0$.
To show that there aren't any other zero, we use the argument principle on $f$, or rather, on $g(z) = 2\exp(-z/2)\cos(2\pi/3+z\sqrt 3/2)$.
If $z$ is on the vertical line $z_k+iy$, a symmetry argument shows that the cosine $\cos(-k\pi+iy\sqrt3/2)$ stays real (but gets larger and larger as $y$ grows), and so $|g(z_k+iy)| \ge |g(z_k)| = 2\exp(-z_k/2)$, and $arg(g(z_k+iy)) = -y/2 + \arg(g(z_k))$.
Threfore, when $z$ moves from $z_k-i\sqrt 3 z_k$ down to $z_k+ i\sqrt 3 z_k$, the argument of $g(z)$ increases by $-\sqrt 3 z_k = 2(k+2/3)\pi$.
Because $|f-g| <<< |g|$, the argument of $f(z)$ does the same thing up to an exponentially small error. In fact, since $f$ is a positive real on the positive real axis, with the equation from earlier we know exactly what the argument is at the endpoints modulo $2\pi$, and therefore the argument of $f(z)$ also increases by exactly $2(k+2/3)\pi$.
Now we use the symmetry $f(\varepsilon_1z) = \varepsilon_2f(z)$ to get the variation of the argument on the two rotated line segments are also exactly $2(k+2/3)\pi$, and so the variation on the completed triangle is three times that, $(3k+2)(2\pi)$.
From this we get that the number of zeroes of $f$ inside each triangle is $3k+2$, and we are done.
