For $z \in \mathbb{C}$, the only point where the derivative for $f(z)=|z|^{2}$ exist is the origin.

Question

Let $z=x+iy$, as $f(z)=z\bar{z}$ we have $f(z)=x^2+y^2$. Then we can define:

$$f(x,y)=u(x,y)+i v(x,y).$$

Where $u(x,y)=x^2$ and $v(x,y)=y^2$. To see $f$ has derivative in the origin I can see the Cauchy-Riemann equations are satisfied as for $(0,0)\in \mathbb{R}^2 \cong \mathbb{C}$:

$$\frac{\partial{u}}{\partial{x}}(0,0)=2(0)=\frac{\partial{v}}{\partial{y}}(0,0)$$

And $$\frac{\partial{u}}{\partial{y}}(0,0)=0=-\frac{\partial{v}}{\partial{x}}(0,0)$$. So the derivative is the origin exist. How can I formally prove that if $(x_{0},y_{0}) \neq (0,0)$ then the derivative for $f(z)=x^2+y^2$ doesnt exist? Thanks!

Answer 1

It looks as if you didn't get the definitions of $u(x,y)$ and $v(x,y)$ right. We should have $u(x,y)=x^2+y^2$ and $v(x,y)=0$.

Now try the Cauchy Riemann equations. We get $2x=0\implies x=0$ and $2y=0\implies y=0$ as the only possibilities.