why is this complex function not analytic anywhere?

Question

I know for $f(z)=z\bar z$, where $\bar z$ means the conjugate of $z$. Cauchy-Riemann equations are satisfied at $(0,0)$. Also, partial derivatives of U and V exist and are continuous everywhere. so why this function is not analytic anywhere?

Answer 1

A function is said to be analytic at a point if it is differentiable in some open disk containing that point. This function is not differentiable at any point other than $0$ so it is not analytic at $0$. Validity of C-R equations at a point do not guarantee analyticity at that point. Try C-R equations at other points.

Answer 2

The reason is that "differentiable" and "analytic", or, perhaps better here, "holomorphic", are not the same thing.

This function is (complex) differentiable at $z = 0$ indeed, and the derivative there is, as expected, 0. Holomorphism adds to simple differentiability at a point that it also must be differentiable not just at that point, but also throughout some extended region centered on that point (i.e. a continuous, two-dimensional cut-out of plane containing the point). Since no other points besides zero permit differentiability, it is not holomorphic at 0, and so not holomorphic anywhere.

Answer 3

There is already an answer by Mr. Rama Murthi, I will just fill in the details:

Definition: A function $f: R\subseteq\mathbb C \rightarrow \mathbb C$ is analytic if and only if $\forall z\in R$, $f(z)$ is $C^2$ in $R$

Meaning $f(z)$ is differentiable at every point in $R$ and $f'(z)$ is continuous in $R$.

Consequently analyticity at a point is defined as:

A function $g(z)$ is analytic at a point $z_0$ if and only if $g(z)$ is $C^2$ in $B_{\epsilon}(z_0)$. Which means, the function is analytic in a small ball of radius $\epsilon$ and center $z_0$.

Bearing those in mind, one can see the function defined as $z\bar z$ is not analytic at $(0,0)$ because it is not analytic in a small ball with a radius $\epsilon$ around $(0,0)$. In fact, from the definition of analyticity it is not analytic anywhere, since it is not differentiable in $\mathbb C \setminus \{0\}$