Is $z+\overline{z}$ an Analytic function? How can I immediately demonstrate that $z+\overline z$ is not an analytic function?
 A: $z+\overline{z}$ is a real non constant function of z. The only real valued functions on $\mathbb{C}$ that are analytic are constant (open mapping propty).
A: Cauchy-Riemann equations don't hold. So no. It's not analytic.
$$z=u+iv=2x+0i$$
$$\frac{\partial u}{\partial x}=2\neq \frac{\partial v}{\partial y}=0$$
A: Your function is equal to $2x$ in the real axis. By analytic continuation, it should be $2z$ in the whole complex line. But this is clearly not true.
A: The zeros of an analytic function are isolated, but the zeros of $z+\overline{z}$ is the whole imaginary axis.
A: Note that a complex function is analytic if and only if it is holomorphic.
The function $f(z)=z+\bar{z}=2Re(z)$ , but the function $g(z)=Re(z)$ is not holomorphic.
A: One equivalent way of defining analytic is that 
$$\frac{\partial f}{\partial \bar{z}} =0 $$
or
$$\frac{\partial f}{\partial z} = \frac{\partial f}{\partial x}. $$
So since 
$$\frac{\partial (z+\bar{z})}{\partial \bar{z}} =1, $$ it's not analytic.
FYI these are some good properties to remember for exams.
