Let's examine a cubic complex function $F(z) = z^3 + e_1z^2 + e_2z + e_3$ with $z$ in the complex numbers. Suppose this function zeros out at two points, $m$ and $n$, which lie inside the boundary of the unit circle in the complex plane. We need to show that along the straight line between $m$ and $n$, there's a point, let's call it $u$, where the real part of $F$'s derivative at point $u$ becomes zero.
To address this, we utilize the derivative $F'(z)$, focusing on its real part across the interval connecting $m$ and $n$. The fundamental theorem of algebra supports the existence of roots for $F(z)$ within the complex plane. Given $F(z)$ zeroes at $m$ and $n$ inside the unit circle, we consider the continuity of $F'(z)$ across a linear path in the complex plane.
By parameterizing the segment from $m$ to $n$ with $z(t) = m + t(n - m)$, where $0 \leq t \leq 1$, we observe $F'(z)$ along this parameterized line. The Intermediate Value Theorem states that if a continuous function transitions across a range of values, it must pass through every intermediate value, including zero.
$f(t)=Re (P(a+t(b-a))$ is real valued function defined on $[0,1]$. It is continuously differentiable and $f(0)=Re P(a)=0, f(1)=Re P(b)=0$. By Rolle's Theroen ther exist $t \in (0,1)$ such that $f'(t)=0$. But $f'(t)=Re \frac d {dt} P(a+t(b-a))=Re P'(a+t(b-a)) (b-a)$ by Chain Rule. Hence, $(b-a) Re P'(w)=0$, where $w=a+t(b-a)$. Cancel $b-a$ to finish. ($w$ belongs to the line segment from $a$ to $b$).
Note: This works for any differentiable function $P$ and any two complex numbers $a$ and $b$ with $P(a)=P(b)=0$.