# why is this complex function not analytic anywhere?

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?

• It has complex derivative at $(0,0)$, but to be analytic it should have those at all points in some neighborhood of the point. Mar 19, 2019 at 0:15

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.

• Thank You very much. I had studied somewhere if a function is single valued then, satisfaction of C-R equations is supposed to be the sufficient condition for analiciticy.
– Shay
Mar 19, 2019 at 0:38
• @Shay In the question you forgot the condition about it being at every point, and here you forgot the condition about differentiability. Note, differentiability, not just existence of the partial derivatives that show up in the C-R equations. Pointing it out because this one can also be missing making analyticity fail. Mar 19, 2019 at 0:59
• @Kavi Please don't confuse analytic and holomorphic in those kind of questions Mar 19, 2019 at 10:46

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.

• Thank you very much
– Shay
Mar 19, 2019 at 14:48

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 $$\mathbb C \setminus \{0\}$$

• Thank You very much
– Shay
Mar 19, 2019 at 14:48
• @Shay You are welcome. Mar 19, 2019 at 22:00