Stuck proving that $\mid z\mid^2$ is not analytic in $z$

Question

I am proving that $f(z) = \mid z\mid^2$ is not an analytic function. So i didn't want to use the Cauchy-Riemann condition or anything but i know that this particular function is diffentiable at only $z=0$ and nowhere else. So I check the differentiability at $z=0$ without any difficulty , just by using the definition of diffentiable complex valued function as in pic :

Now I let a another arbitrary point $z_0 \neq 0 $ and check the differentiability at $z_0$ just by using the existence of this limit $$\lim_{z \to z_o} \frac{f(z)-f(z_0)}{ z - z_0}$$

$\implies$ $\lim_{z \to z0o} \frac{\mid{z}\mid^2-\mid{z_0}\mid^2}{ z - z_0}$

$\implies$ $\lim_{z \to z_0} \frac{(X^2-X_0^2) + ( Y^2 -Y_0^2 )}{ (X-X_0) + (Y-Y_0)\iota}$

But now I am stuck that how can I provde the non-existence of Limit . If I will rationalise then also not getting any satisfactory results. Or choosing two different path is looking impossible because $z_0$ is an unknown point.

Answer 1

Let $x$ belongs to $\mathbb R\setminus \{0\}$.

You have

$$\frac{\vert z_0 +x \vert^2 - \vert z_0\vert^2}{(z_0+x)-z_0}=\frac{x(z_0+\overline{z_0}) +x^2}{x}$$

while

$$\frac{\vert z_0 +ix \vert^2 - \vert z_0\vert^2}{(z_0+ix)-z_0}=\frac{ix(-z_0+\overline{z_0}) +x^2}{ix}$$

You get the desired result as for $z_0 \neq 0$,

$$z_0+ \overline{z_0} \neq -z_0 +\overline{z_0}$$

Answer 2

Here's a different perspective on this problem that finds broad use, and is a computationally compact way of using the Cauchy-Riemann condition. A $C^1$ function $f: \mathbb{C} \to \mathbb{C}$ is holomorphic at $z_0$ if and only if $\overline{\partial} f(z_0) = 0$, where $\overline{\partial} = \frac{1}{2}(\partial/\partial x + i \partial/\partial y)$ is the Wirtinger derivative. Wirtinger derivatives satisfy product and chain rules, and $\overline{\partial} z = 0$, $\overline{\partial} \bar{z} = 1$. Here, $\overline{\partial} |z|^2 = \overline{\partial} (z \bar{z}) = z$, which is nonzero except at $0$.