$ f:\mathbb C\rightarrow \mathbb C$ defined by $f(z)=0$, if $Re(z)=0$ or $Im(z)=0$ and $f(z)=z$, otherwise. I was thinking about the following problem :
Define $ f:\mathbb C\rightarrow \mathbb C$ by 
$$f(z)=\begin{cases}0 & \text{if } Re(z)=0\text{ or }Im(z)=0\\z & \text{otherwise}.\end{cases}$$
Then the set of points where $f$ is analytic is:

(a) $\{z:Re(z)\neq 0$ and $Im(z)\neq 0\}$,
  (b) $\{z:Re(z)\neq 0$ or $Im(z)\neq 0\}$,
  (c) $\{z:Re(z)\neq 0\}$,
  (d) $\{z:Im(z)\neq 0\}$.

I think $f(z)=z$ if $f$ is defined on the set $\{z:Re(z)\neq 0$ and $Im(z)\neq 0\}$ and in that case $f$ is analytic. But I am not sure about the other options. So, choice (a) is right. Am I going in the right direction? Please help. Thanks in advance for your time.
 A: It is not even continuous elsewhere except at $z=0$.
At $z=0$, taking derivatives along the axis and along the diagonal produces different results.
A: (a) is correct. 
For $(a,0)\in\mathbb C$ with $a\neq0, u_x=0\neq 1=v_y$ at $(a,0)$ and for
$(0,a)\in\mathbb C$ with $a\neq0, u_y=1\neq 0=v_y$ at $(a,0)$. Further $f$ is not differentiable at $(0,0)$.
Since $C-R$ equation doesn't get satisfied anywhere in the real and imaginary axis (minus origin) it's not differentiable there. As a result $(f$ is not differentiable at $0),$ $f$ is differentiable nowhere on the abscissa or the ordinate. 
$f$ being the identity map elsewhere, is analytic thereon.
Now consider the sets indicated in all $4$ options.
A: I take $f(z)=u+iv$ and $z=x+iy$ where $x= \Re(z), y=\Im(z).$   In case ,$\Re(z), \Im(z) \ne 0$, then we can take $f(z)=z$ so that $$u+iv=x+iy$$ and so $$u=x,\,\,v=y$$. Whence $u_x=1=v_y,\,\, u_y=0=-v_x$. So, C-R equations are satisfied and hence option (a) is correct.    
But if we take $f(z)=iy ,y \ne 0\,\,$ or of the form $f(z)=x,\,x \ne 0\, $ then C-R equations are not satisfied and hence other options can be eliminated.
