Smooth and not analytic

Question

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.

Answer 1

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.