Real part of function vanishes Consider the function
$$
f(a+bi) = e^{-bi} + e^{\left(\frac{a\sqrt3}{2} + \frac{b}{2}\right)i} + e^{\left(-\frac{a\sqrt3}{2} + \frac{b}{2}\right)i}.
$$
Using a Mathematica plot, I can see that the real part of $f$ vanishes on circles in the complex plane, see below. Is there a way to rigorously show this?

They look like circles, but they are not. If you look very closely, they are a little bit flatened at their vertical extremes. I will present a far more convincing mathematical argument below.
Note that
$$
f(z) = e^{-Im(z)i} + e^{-Im(z\bar\omega)i} + e^{-Im(z\omega)i}.
$$
Then translate it to cartesian coordinates, obtaining
$$
f(a+bi) = e^{-bi} + e^{\left(\frac{a\sqrt3}{2} + \frac{b}{2}\right)i} + e^{\left(-\frac{a\sqrt3}{2} + \frac{b}{2}\right)i}.
$$
Then, the real part will be
\begin{align*}
Re(f(a+bi)) & = \cos(-b) + \cos\left(\frac{a\sqrt3}{2} + \frac{b}{2}\right) + \cos\left(-\frac{a\sqrt3}{2} + \frac{b}{2}\right) \\
& = \cos(b) + 2\cos\left(\frac{a\sqrt3}{2}\right)\cos\left(\frac{b}{2}\right).
\end{align*}
So we are looking for $(a,b)\in\mathbb R^2$ such that
$$
\cos(b) + 2\cos\left(\frac{a\sqrt3}{2}\right)\cos\left(\frac{b}{2}\right)=0.
$$
If we solve it for $a=0$, according to Wolfram alpha, we get
$$
b = 4\left(\pi n \pm \arctan\left(\sqrt{2\sqrt{3}-3}\right)\right)
$$
where the solutions closest to $0$ are
$$
b^{\pm} = \pm 4\arctan\left(\sqrt{2\sqrt{3}-3}\right).
$$
If we solve it for $b=0$, according to Wolfram alpha, we get
$$
a = \frac{4(3\pi n \pm \pi)}{3\sqrt{3}}
$$
where the solutions closest to $0$ are
$$
a^{\pm} = \pm \frac{4\pi}{3\sqrt{3}}.
$$
By the plot, it seems that the points $(a^{\pm},0)$, $(0,b^{\pm})$ are in a same circumpherence. Let us see that they are not.
Suppose the points $(a^{\pm},0)$ and $(0,b^{\pm})$ are in a circumpherence $C$ of equation $(x-x_0)^2 + (y-y_0)^2 = r^2$ (a priori, we are not even assuming that $C$ is centered at the origin).
Using that $(a^{\pm},0)$ are in $C$ and $a^-=-a^+\neq0$, we get
$$
\left\{\begin{array}{l} (a^+-x_0)^2 + y_0^2 = r^2 \\
(-a^+-x_0)^2 + y_0^2 = r^2\end{array}\right. \Rightarrow a^+x_0 = 0 \Rightarrow x_0=0,
$$
and substituting it, we get
$$
(a^+)^2 + y_0^2 = r^2.
$$
Analogously, we show that $y_0=0$ and
$$
(b^+)^2 = r^2.
$$
Putting all together, we conclude that $x_0=y_0=0$, so $C$ is centered at the origin, and its radius satisfies
$$
r= a^+ = b^+\ \ \Rightarrow\ \ \arctan\left(\sqrt{2\sqrt{3}-3}\right) = \frac{\pi}{3\sqrt{3}}.
$$
A simple computation (I used Wolfram alpha again) shows that the last equality is false.
