How are complex functions different from functions of the form $\mathbb{R}^2\rightarrow \mathbb{R}^2$?

Question

I have been trying to intuitively understand complex analysis for over an year now and I still seem to get stuck at the basics. How are complex functions different from a function from $\mathbb{R}^2$ to $\mathbb{R}^2$? Does it have to do with the difference in product of vectors? Will I get the same results if I just stick to studying multivariable functions? And why are analytic functions so important? Why do we not talk about analytic functions from $\mathbb{R}^2$ to $\mathbb{R}^2$? I apologise for asking so many questions but any help or tip would be highly appreciated. I know the theorems and all but I need to know what they mean (perhaps) geometrically. Please try to give an intuitive and basic explanation. Thank you.

Answer 1

There are many things to address here, but I'll take a few, and maybe others will bring other important points to your attention.

1. The only structural difference between $\mathbb{C}$ and $\mathbb{R}^2$ is the operation of complex multiplication. $\mathbb{R}^2$ is not endowed with any operation that resembles this, and thus we can construct functions from $\mathbb{C}$ into $\mathbb{C}$ that cannot be reviewed as function from $\mathbb{R}^2$ to $\mathbb{R}^2$.

2. As Tito Eliatron has already commented, the notion of complex differentiability is very different from the notion of differentiability in $\mathbb{R}^2$. Complex differentiability is a much stronger condition than real differentiability. As a matter of fact, you can compare the notion of analyticity to the notion of a divergence and curle free vector field in $\mathbb{R}^2$. That is how restrictive the condition of complex differentiability is.

Answer 2

The main difference arrives when you try to define differentiability.

In the complex case, you can do in the same way as in the real case: $$f'(z)=\lim_{w\to0}\frac{f(z)-f(w)}{z-w}$$ And you can divide by $z-w$ as $\Bbb C$ is a field.

But in $\Bbb R^2\to\Bbb R^2$ case, you cannot proceed in the same way, because you cannot divide by a vector. So, the solution is to divide by the modulus

Answer 3

The difference between $\mathbb{R^2}$ and $\mathbb{C}$ is that in $\mathbb{C}$ we define multiplication of complex numbers. This operation turns it into a field, and allows us to define a derivative of a complex function like this:

$f'(z)=\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}$

We can't define it like this in $\mathbb{R^2}$, because there we don't have a multiplication, and thus the division by $h$ makes no sense. The complex derivative is really different from a derivative in $\mathbb{R^2}$. For a complex function $f=u+iv$ to be differentiable at a point, the functions $u,v$ must have partial derivatives at that point and also they have to satisfy the Cauchy-Riemann equations: $u_x=v_y, u_y=-v_x$. As you can imagine, the number of functions which satisfy this is very limited. So complex differentiability is a much more complicated term than a usual derivative in the sense of $\mathbb{R^2}$. And if you studied complex analysis you know that it leads to a very big and beautiful theory-analytic functions have very surprising properties.