# $f:\mathbb C\rightarrow \mathbb C$ 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.

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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.

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(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.

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solution ta bujhlamna as $(a,0)$ te to function ta basically $0$ so all partial derivatives $0$ taina? – Un Chien Andalou May 4 '13 at 1:50
Sorry I was wrong. Corrected. BTW not all the pd's are zero ... as you see: $v_y(a,0)$$=\lim_{h\to 0}\dfrac{v(a,h)-v(a,0)}{h}$$=\lim_{h\to 0}\dfrac{h-0}{h}$$=1. – Sugata Adhya May 4 '13 at 2:36 I take f(z)=u+iv and z=x+iy where x= \Re(z), y=\Im(z). In case ,\Re(z), \Im(z) \ne 0, then we can take f(z)=z so that$$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.

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