A non constant complex valued function with zero derivative I need an example of a non constant complex valued function whose derivative is zero at every point in in the complex plane.
 A: A complex valued function on a connected set with derivative $0$ is a constant. (Refer Stein - Shakarachi Complex analysis Chapter 1 for proof.)
If the function is defined on disconnected regions then its derivative is zero need not imply function is constant.
For example take $f$ to be $1$ on $|z|<1$ and $2$ on $|z-3|<1$. Then $f$ is non-constant but $f'(z)=0$.
A: If by derivative you mean:
$$
\frac{df(z)}{dz} = 0,
$$
then simply take $f(z)=\bar z$.
A: A function  with zero derivative (anywhere) is always constant.
Let $f:\mathbb{C}\to\mathbb{C}$ a function with zero derivative at any $z_0\in\mathbb{C}$.
well write $f(z)=f(x+iy)=u(x,y)+iv(x,y)$ and $z_0=x_0+iy_0$.
So the derivative exists and $f'(z_0)=lim_{n\to\infty}\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z_0}$, and from its existence we conclude $f'(z_0)=lim_{\Delta z\to0}\frac{f(z_0+\Delta x)-f(z_0)}{\Delta x}=\frac{\partial u}{\partial x}(x_0,y_0)+i\frac{\partial v}{\partial x}(x_0,y_0)=0$ and...
$f'(z_0)=lim_{\Delta z\to0}\frac{f(z_0+i\Delta y)-f(z_0)}{i\Delta y}=-i\frac{\partial u}{\partial y}(x_0,y_0)+\frac{\partial v}{\partial y}(x_0,y_0)=0$
Thus for any $x,y\in\mathbb{R}$,  $0=\frac{\partial u}{\partial x}=\frac{\partial u}{\partial y}$ so $u:\mathbb{R}^2\to\mathbb{R}$ is constant.
Moreover $0=\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y}$ so $v:\mathbb{R}^2\to\mathbb{R}$ is constant.
Hence $u+iv=f$ is constant. 
I think it holds ...
