TITLE: How is $\mathbb{C}$ different than $\mathbb{R}^2$?
QUESTION [45 upvotes]: I'm taking a course in Complex Analysis, and the teacher mentioned that if we do not restrict our attention to analytic functions, we would just be looking at functions from $\mathbb{R}^2$ to $\mathbb{R}^2$.  
What I don't understand is why this is not true when we do restrict our attention to analytic functions. I understand that complex analytic functions have different properties than real functions on $\mathbb{R}^2$, but I don't understand why this is so.  If I look at a complex number $z$ as a vector in $\mathbb{R}^2$, then isn't differentiability of $w=f(z)$ in $\mathbb{C}$ defined the same way
as differentiability of $(u,v)=F(x,y)$ in $\mathbb{R}^2$?

REPLY [10 votes]: To go down a level from differentiability: the root of the difference between $\mathbb{C}$ and $\mathbb{R}^2$ comes from the multiplicative structure on $\mathbb{C}$.  Look at the definition of differentiation itself: $\lim_{h\rightarrow 0} h^{-1}\cdot \left(f(z+h)-f(z)\right)$ - there's a multiplication here, by the multiplicative inverse of the (complex) number $h$, that simply can't be performed in $\mathbb{R}^2$ without giving it a field structure.  There isn't 'a' derivative of a function from $\mathbb{R}^2 \mapsto \mathbb{R}^2$, just two partials; the multiplicative structure of $\mathbb{C}$ is then what forces the Cauchy-Riemann constraints on those partial derivatives and allows for a definition of the derivative as a single function from $\mathbb{C}\mapsto\mathbb{C}$.