# 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 $\mathbb{R}^2$. – J. M. Sep 21 '10 at 1:26

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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The essential point of complex analysis is that the answer to this question is a resounding no. Differentiability of a function $f$ in the complex sense requires taking a limit in which the limiting variable is a complex number:

$f'(z) = \lim_{w\to z} \frac{f(w) - f(z)}{w-z}$.

For this limit to exist, one needs convergence uniformly in all directions (think through the $\epsilon$-$\delta$ definition of this limit). On the other hand, a function from $\mathbb{R}^2\to \mathbb{R}^2$ (or more generally $\mathbb{R}^n\to \mathbb{R}^m$) is differentiable (in the sense explained in Ryan's answer) at all points in an open neighborhood if and only if all the partial derivatives exist at every point and are continuous on the given neighborhood. Of course that forces all the directional derivatives to exist, but doesn't require the same sort of uniformity from different directions.

This Wikipedia page is a good place to learn some more: http://en.wikipedia.org/wiki/Holomorphic_function

The Cauchy-Riemann equations, as Ryan's answer explains, relate the directional derivatives of the real and complex parts of a holomorphic function and give rise to all sorts of interesting properties.

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The last sentence of your first paragraph isn't entirely correct. I guess it could be a matter of terminology, but usually to call a function differentiable you would at least expect it to be Gâteaux differentiable, and I would think that Fréchet differentiable would be more common. The latter is still much weaker than holomorphic. – Jonas Meyer Sep 21 '10 at 1:18
I wasn't precise enough in what I wrote. Hopefully the edited version will be better... – Dan Ramras Sep 21 '10 at 3:21
That is still not the same as differentiable in the sense of Ryan's answer; he uses Fréchet differentiability, whereas a function can have continuous partials everywhere without being differentiable in that sense. A function can even have continuous directional derivatives everywhere without being Fréchet differentiable. Some examples appear on the Wikipedia page: en.wikipedia.org/wiki/Fr%C3%A9chet_derivative – Jonas Meyer Sep 21 '10 at 4:48
Okay, maybe I'm confused, but on that Wikipedia page is the statement I was trying to make: "If all partial derivatives of f exist and are continuous, then f is Fréchet differentiable (and, in fact, C1)." I don't have any books in front of me to compare with. Is there a difference between what I wrote above and what I just quoted? It's of course essential to assume continuity on the partial derivatives. – Dan Ramras Sep 21 '10 at 5:06
Good point, and actually maybe my nitpicking was off: the examples they give are actually not continuous, but the partials are, if I'm reading it correctly (they give functions that are differentiable and thus continuous along lines, but not along other curves). However, Fréchet differentiability also implies continuity, and it's fair to say that you intended continuity. My point was not apt, and I basically agree with your post. – Jonas Meyer Sep 21 '10 at 5:22
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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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