This thread is just to collect some examples...
Given an open domain $\Omega\subseteq\mathbb{C}$.
Consider a holomorphic function $f:\Omega\to\mathbb{C}$.
What would be a counterexample to: $$f(b)-f(a)=f'(c)(b-a)\quad(c\in\Omega)$$ and when does even the estimate fail: $$|f(b)-f(a)|\leq\|f'\|_\infty|b-a|$$
Feel free to post anything related you have in mind!
The exponential function $f(x)=e^x$. $$ f(2\pi i)-f(0) = 0 $$ but $f'(c)\cdot 2\pi i=e^c\cdot 2\pi i$ can never be zero. For the estimate I don't know yet.
The estimate is always true, as long as $\Omega$ is convex.
Because in this case,
\begin{eqnarray*} f(b) - f(a) &=& \int_0^1 (\frac{d}{dt}f(a + t(b-a))) \, dt\\ & = &\int_0^1 f'(a + t(b-a)) \cdot (b-a) \, dt. \end{eqnarray*}
Now simply take the absolute value and use that: $|\int f \, dt| \leq \int |f|\,dt$
Hence, this has not so much to do with $f$ assuming complex values than with the shape of $\Omega \subset \Bbb{C}$.
Disclaimer
I will give only a sketch!!
Reference
This example is taken from: MVT: Logarithm
Counterexample
Given the cutted unit circle: $\Omega:=\mathbb{S}\setminus\{-1\}$
Consider the main branch of the logarithm: $\ln(re^{i\varphi}):=\ln r+i\varphi$
Assuming the estimate holds: $$\Delta\downarrow0:\quad2\pi\leftarrow|2\pi-2\Delta|=\left|\ln e^{i(\pi-\Delta)}-\ln e^{i(-\pi+\Delta)}\right|\\\leq\sup_{z\in\Omega}\left|\frac{1}{z}\right|\cdot\left|e^{i(\pi+\Delta)}-e^{i(\pi-\Delta)}\right|=\left|e^{i2\Delta}-1\right|\to 0$$ But that is a contradiction!
