We have that $f(z)$ analytic/holomorphic in $\mathbb{C}$. Let $z_1$ and $z_2$ be two points in $\mathbb{C}$. I need to show that there exists $c_1$ and $c_2$ on the line segment joining $z_1$ and $z_2$ such that $Re(f'(c_1))=Re[(f(z_2)-f(z_1))/z_2-z_1]$ and $Im(f'(c_2))=Im[(f(z_2)-f(z_1))/z_2-z_1]$.
I know that the MVT only holds for real-valued functions, so I have barely an idea on how to start the proof. A little nudge to the right direction would be appreciated.
On
I would like to add a bit to Yung-Hsiang Huang's answer.
Usually the real part $\Re f(z)$ of a holomorphic function $f$ is not differentiable(otherwise it's constant). But now we're considering the function $\Re f(\gamma(t))$ where $\gamma$ is a parameterization of the segment $[z_1, z_2]$. And it turns out, that this function is differentiable w.r.t real variable $t$:
$$ \lim_\limits{\Delta t \to 0} \frac{\Re f(\gamma(t + \Delta t)) - \Re(f(\gamma(t))}{\Delta t} $$
You should use the continuity of $\Re : \mathbb{C} \to \mathbb{R}$ to show that such a limit exists:)
You can apply MVT to $F(t) := Re (f(tz_1+(1-t)z_2))$ and $G(t) := Im (f(tz_1+(1-t)z_2))$.