Derivability of $|z|^2$

Question

I’m a sophomore learning complex analysis. I read this in my text book:

When a function is derivable at a point, it’s not necessary to be analytic at this point.

For example: $|z|^2$ is derivable at $z_0=0$, whereas not analytic at at $z_0=0$.

My question:

Isn’t it $|z|^2=z^2$? So...I may think it’s just derivable as well as analytic!

What have I missed? I’m totally confused...

Any help would be sincerely appreciated! Thanks!

Answer 1

Using the C-R equations, $$z = x + yi\longmapsto |z|^2 = x^2 + y^2 = u + iv$$ isn't complex differentiable at any $z\ne 0$ because $z\ne 0\implies x\ne 0$ or $y\ne 0$ and: $$x\ne 0 \implies u_x = 2x \ne 0 = v_y,$$ $$y\ne 0\implies u_y = 2y \ne 0 = -v_x.$$

Answer 2

Since $\dfrac{\partial}{\partial\overline{z}}|z|^{2}=\dfrac{\partial}{\partial\overline{z}}z\cdot\overline{z}=z\dfrac{\partial\overline{z}}{\partial\overline{z}}=z\ne 0$ for any $z\ne 0$ closed to $0$, so $|z|^{2}$ is not analytic at $0$.

Answer 3

For $f$ to be analytic at $0$ it has to be differentiable in a neighborhood of $0$. The function $|z|^{2}$ is not differntiable at $\epsilon$ for any $\epsilon >0$: consider $\lim_{h \to 0} \frac {|\epsilon +h|^{2} -\epsilon ^{2}} h$. If you take the limit through positive ral values of $h$ you get $2\epsilon$ wheras if you take limit through purely imaginary values of $h$ you get $0$.