# Smooth and not analytic

Can someone show me, without reference to Taylor series, why a complex function can be smooth but not analytic? I do not understand it intuitively or visually either. I would like an explanation which simply refers to the definition of analytic functions as functions for which the complex derivative exists everywhere in the domain.

• Actually analytic functions are usually defined to be the functions which are equal to their Taylor series as far as I know. Complex differentiable functions on the other hand are said to be holomorphic. That being said, it is a standard result of complex analysis that these two classes of functions are precisely the same.
– MSDG
Oct 19, 2018 at 18:38
• I know, but I haven't seen the proof yet, so I wanted an explanation not referring to that fact. Oct 19, 2018 at 18:48
• @HBHSU: As Sobi has said, there is no exmplanation because in complex variable, smooth and analytic mean the same. You cannot understand why some functions are smooth but not analytic because such functions don't exist. Oct 19, 2018 at 22:57
• @Dog_69: "Smooth" in this context does not mean "complex differentiable to all orders." It simply means that the real and imaginary parts of a function are differentiable to all orders. You're obfuscating the actual question the OP has by refusing to acknowledge that while "smooth" means one thing for real variable functions, it is simply not used to describe the complex differentiability of functions of a complex variable, for which the terms "analytic" or "holomorphic" or "complex differentiable" are used. Oct 21, 2018 at 0:21

The best way to understand this, I would argue, is to consider a complex function as nothing more than a function $$f(x,y) = \big(u(x,y),v(x,y)\big)$$, and viewing complex analysis as an extension of the ideas from real variables. The definition of a function as being complex differentiable is equivalent to it satisfying the Cauchy-Riemann equations: $${\partial u\over \partial x} = {\partial v\over \partial y},\qquad{\partial u\over \partial y} = -{\partial v\over \partial x}.$$ So now, if you want an example of a smooth function that is not analytic, merely find a function $$f(x,y) = \big(u(x,y),v(x,y)\big)$$ where both $$u$$ and $$v$$ are smooth (infinitely differentiable in the sense of real variables), but that do not satisfy the Cauchy-Riemann equations! Simple as that.
For example, the function $$f(x,y) = (x,-y)$$ is smooth in the sense that $$u(x,y) = x$$ and $$v(x,y) = -y$$ are both smooth, but it is just a short check to see that $$f$$ doesn't satisfy the Cauchy-Riemann equations. In complex notation, $$f(z) = \overline z$$ is the conjugation map that sends a complex number to its conjugate.