In Spiegel Complex variables book and in Wolfram Alpha it says the function $z^2 \bar{z}$ derivative does not exist anywhere in the complex plane.
But at $z_0=0$
$\frac{z^2 \bar{z}}{z}= x^2+y^2$ then the limit when $z \rightarrow 0$ would be $0$. So the derivative exists.
What I'm doing wrong?
Best Answer
There is nothing wrong with what you've done. $z\lvert z\rvert^2$ is complex-differentiable at $0$: in fact, as you've pointed out, $$\lim_{z\to 0}\left\lvert\frac{z\lvert z\rvert^2-0\cdot\lvert 0\rvert^2-0\cdot z }{z}\right\rvert=\lim_{z\to 0}\lvert z\rvert^2=0$$