Is $f(z) = z \bar{z}$ differentiable at $0$?

Question

I am trying to work out when $f(z) = z \bar{z}$ is complex differentiable. Looking at the Cauchy-Riemann equations, they are not satisfied away from $0$, but I think you just get $0 = 0$ and $0=0$ at $0$. Since $f$ depends on the conjugate of $z$, I am inclined to think it shouldn't be differentiable, but I can't seem to formalise this. Please help!

Answer 1

$f$ is real-differentiable at $(0,0)$ because $f(x,y)=x^{2}+y^{2}$ is a polynomial. Now $\dfrac{\partial f}{\partial\overline{z}}(z)=z$ and hence $\dfrac{\partial f}{\partial\overline{z}}(0,0)=0$, so Cauchy-Riemann is satisfied, so $f$ is complex-differentiable at $(0,0)$.

Perhaps one goes directly with definition that $\left|\dfrac{z\overline{z}-0}{z-0}\right|=|\overline{z}|=|z|\rightarrow 0$ as $z\rightarrow 0$.

Answer 2

Let us work with the Cauchy-Riemann equations (CRE) in $\mathbb{R}^{2}$.

We write $f:\,\mathbb{R}^{2}\to\mathbb{R}^{2}$ with $f\left(x,y\right)\,:=\,\left(f^{x}\left(x,y\right),\, f^{y}\left(x,y\right)\right)$.

Now we have $$f^{x}\left(x,y\right)=x^{2}+y^{2}$$ $$f^{y}\left(x,y\right)\equiv0$$

The CRE are now:

$$\frac{\mathtt{d}}{\mathtt{d}x}f^{x}\left(x,y\right)=2x\;\overset{!}{=}\;\frac{\mathtt{d}}{\mathtt{d}y}f^{y}\left(x,y\right)=0$$ $$\frac{\mathtt{d}}{\mathtt{d}y}f^{x}\left(x,y\right)=2y\;\overset{!}{=}\;-\frac{\mathtt{d}}{\mathtt{d}x}f^{y}\left(x,y\right)=0$$

From that you can see that indeed the complex derivative exists at $\left(x,y\right)=\left(0,0\right)$ and only there. As there exists no neighbourhood of $\left(0,0\right)$ where the CRE hold, $f$ however is nowhere holomorphic / analytic.

Answer 3

You have

$$f(x,y)=x^2+y^2.$$

So using the Cauchy Riemann equations, one has that $u(x,y)=x^2+y^2, v(x,y)=0$. Thus by $u_x=v_y, u_y=-v_x$, one has

$$2x=0=2y \Rightarrow x=0=y.$$

Thus $f$ is only differentiable at the origin, $(0,0)$.