True/false regarding C-R equation

Question

Is the following statement is True /false ?

The function $f$ defined by $$f(z) = \begin{cases} \frac {Im(z^2)}{\bar z} \ \text{if } z\neq 0 \\ 0 \ \text {if z=0} \end{cases}$$

satisfies the cauchy reimann equation at origin and also $f(z)$ is differentiable at origin?

My attempt : i think it will be true because for cauchy reimann equation here $Re(z)=0 $ so $f(z)$ will become constant that is satisfied the C-R equation and for differentiability here its is already given that $f(z)=0$ so $f'(z) =0$

Is its true ?

Any hints/solution will be appreciated

thanks u

Answer 1

Note that, if $x,y\in\mathbb R$,$$f(x+yi)=\frac{2xy}{x-yi}=\frac{2x^2y}{x^2+y^2}+\frac{2xy^2}{x^2+y^2}i.$$So, $u(x,y)=\dfrac{2x^2y}{x^2+y^2}$ and $v(x,y)=\dfrac{2xy^2}{x^2+y^2}$. So, both $u$ and $v$ are the null function when $x=0$ or $y=0$ and therefore $$\dfrac{\partial u}{\partial x}(0,0)=\dfrac{\partial u}{\partial y}(0,0)=\dfrac{\partial v}{\partial x}(0,0)=\dfrac{\partial v}{\partial y}(0,0)=0.$$So, yes, $(0,0)$ is a solution of the Cauchy-Riemann equations.

However,$$\lim_{z\to0}\frac{f(z)-f(0)}z=\lim_{z\to0}\frac{\operatorname{Im}(z^2)}{\lvert z\rvert^2}$$and this limit doesn't exist. For instance, see what happens if $z$ is of the form $x+xi$, with $x\in(0,\infty)$. So, $f$ is not differentiable at $0$.

Answer 2

Observe that the limit $\lim_{z\rightarrow 0}\frac{f(z)-f(0)}{z}=\lim_{(x,y)\rightarrow (0,0)}\frac{2xy}{x^2+y^2}$ doesn't exist uniquely (depends on $m$) on the line $y=mx$ through origin, hence $f'(0)$ doesn't exist.