# How is $\mathbb{C}$ different than $\mathbb{R}^2$?

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 $\mathbb{R}^2$?

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The standard example is complex conjugation. This is not analytic, although a smooth function $\mathbb{R}^2 \to \mathbb{R}^2$. –  Martin Brandenburg Sep 21 '10 at 1:11
Somewhat related: mathoverflow.net/questions/3819/… –  Jonas Meyer Sep 21 '10 at 1:25
There's nothing corresponding to "maximum modulus" on $\mathbb{R}^2$. –  Guess who it is. Sep 21 '10 at 1:26
When you say "that complex analytic functions [...] than real functions" do you mean real analytic functions? –  Freeze_S Jan 26 '14 at 12:44

For a function $f : \mathbb R^2 \to \mathbb R^2$ "differentiable" at a point $x \in \mathbb R^2$ means you have a linear approximation $f'_x$ which satisfies

$$\lim_{y \to x} \frac{f(x)-f(y)-f'_x(x-y)}{|x-y|} = 0$$

Saying that $f$ is complex analytic is the constraint that $f'_x$ is a complex linear function for all $x$. "Complex linear" means that not only is it true that $f'_x(av+bw)=af'_x(v)+bf'_x(w)$ for $a, b \in \mathbb R$, but it also holds for $a,b \in \mathbb C$.

One way to say $L$ is "Complex linear" is that $L$ is a regular (real linear) function plus $L(iv)=iL(v)$ for all vectors $v$. Stated in terms of the derivative, the directional derivative of $f$ in the direction $(0,1)$ is $i$ times the directional derivative in the direction $(1,0)$. In component notation these are the "Cauchy-Riemann" equations.

$$\frac{\partial f}{\partial y} = i \frac{\partial f}{\partial x}$$

where if you write $f(z) = f(x+iy) = u(x,y) + iv(x,y)$ translates to

$$\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}, \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$$

The formula:

$$\frac{\partial f}{\partial y} = i \frac{\partial f}{\partial x}$$

gives the key picture. A complex linear map means that the map looks like the composite of a rotation together with a re-scaling map $v \longmapsto av$ where $a \in \mathbb R$. These maps are called "conformal". The nice things about conformal linear maps is they preserve all angles -- they do not always preserve length. So the "nice" thing about complex differentiable maps is that if you have any collection of curves in the plane, and you apply your complex differentiable function to them, it preserves the angles of intersection of your curves. That's a very special property.

edit: An instructive example would be to think through two different functions from $\mathbb R^2$ to $\mathbb R^2$. The first function:

$$(x,y) \longmapsto x(\cos(y),\sin(y))$$

and the second function

$$(x,y) \longmapsto e^x(\cos(y),\sin(y))$$

The first function preserves the angles between the coordinate grid lines -- curves like $x=a$ and $y=b$ in the coordinate plane. The first function is not complex differentiable but the second is! So this means that the second function preserves all angles (not just the coordinate lines $x=a, y=b$). Can you spot curves in the domain which intersect in some angle $\theta$, but when after you compose them with the 1st function, their angle of intersection is not $\theta$?

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Wow, thanks. This is going to take some more thought, but I'm starting to get it. –  jake Sep 21 '10 at 2:03

Analytic functions map tiny disks to tiny disks. (Of course that's not rigorous, but you could make it rigorous by putting in the right limit language.) Analytic functions can shift, stretch, and rotate disks, but they can't flip disks over.

Smooth functions of two real variables can map disks to ellipses. That is, they can stretch a disk more in one direction than in another. Complex analytic functions can't do that.

Complex conjugation is not analytic because it flips disks over.

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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 $\mathbb{C}\mapsto\mathbb{C}$.

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But that applies to functions $f:\mathbb{R}\to\mathbb{R}$ too. But that seems not enough for differentiability to imply analyticity. –  Freeze_S Jan 24 '14 at 1:32