Is $f(z)= z^2\bar{z}$ not differentiable anywhere?

Question

Question: prove that $\frac{d}{dz}(z^2\bar{z})$ does not exists anywhere.

My attempt: my intention is that it is differentiable at origin.

Since, $$f'(0)=\lim_{z\to 0}\frac{f(z)-f(0)}{z-0}$$

$$=\lim_{z\to 0}\frac{z^2\bar{z}}{z}$$

$$=0$$

Hence, I think that, it must be differentiable at origin. question is from standard reputed textbook but is wrong Or I am incorrect somewhere?

However, at other points it is not differentiable but I am unable to prove it. :-(

Please help me....

Answer 1

The Cauchy Riemann equation $u_{x} = v_{y}$ implies $$3x^{2}+y^{2} = x^{2}+3y^{2} \implies x^{2}+y^{2}=0 \implies z=0$$ so that $f$ is not differentiable if $z \ne 0$. Conversely, you correctly prove that $f$ is differentiable at $z=0$.

Answer 2

Yes, $f$ is differntiable only at $z = 0$, In any other point:

$z\longmapsto z^2\bar z$ differentiable $\implies z\longmapsto\frac{z^2\bar z}{z^2} = \bar z$ differentiable

and the conjugation isn't differentiable in any point.