I'm defining a function $f(z) = z^3 + 1$, and I will let two points $$z_1 = \frac{-1+i\sqrt3}{2}\quad\text{and}\quad z_2 =\frac{-1-i\sqrt3}{2}.$$
I am trying to show that there is no point $w$ on the line segment from $z_1$ to $z_2$ such that $$f(z_2) - f(z_1) = f'(w)(z_2-z_1).$$
I'm not too sure how I can take a point w and make a function out of it, and even find its derivative. Any help? I'm not too sure where i would even start.
We know from calculus that a point on the line segment would have the form $$-\frac{1}{2} - \frac{i\sqrt{3}}{2}(1-2t),$$ where $0 \le t \le 1$.
$f(z_1)=f(z_2)=2$, but $f'(z)=3z^2=0$ only when $z=0$, and $z=0$ does not lie in the segment $$ \left[\frac{-1-i\sqrt{3}}{2},\frac{-1+i\sqrt{3}}{2}\right]. $$ Note that $$ \frac{-1-i\sqrt{3}}{2}=\cos(2\pi/3)+i\sin(2\pi/3), $$ and hence $$ \left(\frac{-1-i\sqrt{3}}{2}\right)^{\!3}=\big(\cos(2\pi/3)+i\sin(2\pi/3)\big)^3= \cos(3\cdot 2\pi/3)+i\sin(3\cdot 2\pi/3)=1. $$