Consider the real value function $f(x) = 2x^2 - 2x - 6 = (x-1)^2 + x^2 - 5$. $f(x)$ is symmetric around $x = 1/2$.
However, if we consider a complex function with real coefficients, $f(z) = (z-1)^2 + z^2$, we cannot say that this is symmetric about the line $Re(z) = 1/2$. In this case, the zeros $(1/2 \pm i/2)$ are on the line $z = 1/2$ and symmetric about the real axis.
Consider another complex function formed out of four complex roots $1/2 \pm i, 1/2\pm 2i$. I am choosing it like this to make $f(1-z) = f(z)$. Here $f(z) = (z-1/2)^4 + 5(z-1/2)^2 + 4$. The zeros are symmetric about the real axis and on the line $z = 1/2$.
I claim that all complex functions with real coefficients and with only complex roots that exhibit $f(1-z) = f(z)$ have zeros on the line $z = 1/2$. Functions like $\sin (2\pi z)$ satisfy $f(1-z) = f(z)$ but have real zeros.
Question: Can someone show a counter example to my claim?
Thanks
Consider the function
$$f(z) = \left( z - \frac{1}{2} \right)^4 + 1.$$
The equation $f(z) = 0$ is equivalent to $\left( z - \frac{1}{2} \right)^4 = -1$, so the solutions are the four vertices of a tilted square centered at $\frac{1}{2}$, namely
$$\begin{align*} & \frac{1}{2} + \frac{\phantom{-}1+i}{\sqrt{2}}, \qquad \frac{1}{2} + \frac{-1+i}{\sqrt{2}}, \\[1ex] & \frac{1}{2} + \frac{-1-i}{\sqrt{2}}, \qquad \frac{1}{2} + \frac{\phantom{-}1-i}{\sqrt{2}}. \end{align*}$$
So $f$ has no real roots and $f(z) = f(1-z)$.
Also note that each polynomial $f(z)$ with real coefficients such that $f(z) = f(1-z)$ will have roots both symmetric with respect to the line $\Im z = 0$ (because of the real coeffiecients $f(z) = 0 \iff f(\overline{z}) = 0$) and symmetric with respect to the point $z = \frac{1}{2}$, which makes for a symmetry with repsect to the line $\Re z = \frac{1}{2}$.