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

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.

• Have you tried evaluating this limit along the horizontal line $y=y_0$ and then the vertical line $x=x_0$? Jan 23 at 21:24
• I just know that $z_o$ is just a non zero point . So how can i approach it by x axis or y axis ? Jan 23 at 21:27
• Write $z$ as $z_0+x+iy$ Jan 24 at 13:44

## 2 Answers

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 $$z_0 \neq 0$$,

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

• Thank you so much dear Jan 23 at 21:45

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 $$0$$.