Is the complex function $f(z) = Re(z)$ differentiable? I am preparing to take my first course in complex variables. I am reading some lecture notes online.  They claim that the function $f(z) = Re(z)$ is continuous but NOT differentiable.  I know the definitions of a limit, of continuity, and of a derivative.  I understand why this function is continuous.  I am trying to show that it is NOT differentiable.  
 A: For a function to be differentiable in $\Bbb{C}$, it must satisfy the Cauchy-Riemann equations, that is, if $$f(x,y)=u(x,y)+iv(x,y)$$
it must satisfy $$u_x=v_y\\u_y=-v_x$$
But for $f(z)=\Re(z)=x$ we get $$u_x=1\neq v_y=0$$
So it is not differentiable. 
A: The definition of derivative can be written as
$$ f'(z) = \lim_{h \to 0} \dfrac{f(z+h) - f(z)}{h} $$
which looks just like the real-variable definition, 
but here this is taken in the complex sense, i.e. $h$ is allowed to be a complex number. $h \to 0$ means the distance in the complex plane from $h$ to $0$ goes to $0$ (or equivalently, both real and imaginary parts of $h$ go to $0$).  In order for the limit to exist, you must always get the same value as $h \to 0$ in any manner.
If you take $h$ to be real, $f(z+h) = f(z) + h$ and the quotient is $1$.
If you take $h$ to be imaginary, $f(z+h) = f(z)$ and the quotient is $0$.
The limit as $h \to 0$ doesn't exist: it can't be both $1$ and $0$.  Thus we say $f'(z)$ doesn't exist, and the function is not differentiable.
A: Another way to see it, it is that the real part of a complex number can be written with its conjugate: $Re(x) = \frac{1}{2} (x + x^*)$. Since the conjugate function is the classical example of a non-complex-differentiable function (see for exampe this), it follows that the real part is not complex-differentiable.
