This is going to be completely obvious, but I can't seem to get a satisfying answer on my own. Any help would be much appreciated.
I thought I understood the definitions of complex vs real analytic functions, but an example I found has proved me wrong. What does it mean for a function to be a real analytic non-holomorphic function from $\mathbb{C} \rightarrow \mathbb{C}$. It seems to imply having a series expansion but I don't understand how that doesn't make it holomorphic.
Best Answer
A function $f:\mathbb{R}^n \to \mathbb{R}^m$ is real-analytic if each component $f_i$ ($i = 1,\ldots,m$) can be written as a power series in the variables $x_j$ ($j = 1,\ldots,n$) with non-zero radius of convergence. The function $f:\mathbb{R}^2 \to \mathbb{R}^2$ given by $$ f(x,y) = (x,-y)$$ is hence real-analytic on account that \begin{align} f_1(x,y) = x = &\sum_{r,s = 0}^\infty a_{rs} x^r y^s \quad \text{with} \quad a_{10} = 1, \,a_{rs} = 0 \quad\text{for}\quad rs \neq 10\\ f_2(x,y) = -y = &\sum_{r,s = 0}^\infty b_{rs} x^r y^s \quad \text{with} \quad b_{01} = -1, \,b_{rs} = 0 \quad\text{for}\quad rs \neq 01 \end{align} are convergent sums for all $x,y$.
A function $g : \mathbb{C} \to \mathbb{C}$ is complex-analytic if it can be written as a power series in $z$ with non-zero radius of convergence. The function $g:\mathbb{C}\to \mathbb{C}$ given by $$g(z) = \bar{z}$$ is not complex-analytic – for instance, it doesn't satisfy the Cauchy-Riemann equations.
Finally, note that under the identification $\mathbb{R}^2 \simeq \mathbb{C}$, we have $f \simeq g$.