Analyticity and continuity in $\mathbb C$ of $f(z)=\frac{ \bar{z}^2}{z} $ if $z \ne0$,$0$ if $z=0$

Question

For $z \in \mathbb C$

$f(z)=\frac{ \bar{z}^2}{z} $ if $z \ne0$,$0$ if $z=0$

I am investigating its analyticity and continuity in $\mathbb C$ or in any open neighbourhood of zero.

I know a complex function if analytic then it satisfies CR equations. So if the function does not satisfy CR equation it will not be analytic. I have checked that $zf(z)$ does not satisfy CR equation. Does that means $f$ is not analytic? Calculation of CR equation for $f$ is difficult. Is there any straight forward check?

If $f$ analytic then $f$ will be continuous too.

Answer 1

You can see that $f(z)$ is continuous at $z=0$ because

$lim_{z\rightarrow 0} f(z)=f(0)=0$ along any path through $z=0$

For differentiability at $z=0$

$f'(0)=lim_{z\rightarrow 0}\frac{f(z)-f(0)}{z}=lim_{z\rightarrow 0}(\frac{\overline z}{z})^2$

$=lim_{z\rightarrow 0}(\frac{x-iy}{x+iy})^2=lim_{x\rightarrow 0}(\frac{1-im}{1+im})^2$ (along $y=mx$) which depends upon $m$

Hence $f'(0)$ doesn't exist and $f(z)$ is not analytic at $z=0$.

Answer 2

Since $$\frac{\partial }{\partial \bar z}f(z)=\frac{2\bar z}{z}\neq 0,$$ it's not analytical. For the continuity, it's rather clear since $|z|=|\bar z|$, and thus $$|f(z)|\leq |z|.$$