Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Am I correct in noticing that Complex Analysis seems to be a synonym for analysis of functions $\mathbb R^2 \to \mathbb R^2$?

If this is the case, surely all the results from complex analysis carry over to the study of these $\mathbb R^2 \to \mathbb R^2$ functions. Does anything from complex analysis carry over into the study of functions in even higher dimensions?

Furthermore, is there an area of Mathematics similar to complex analysis that investigates functions, say, $\mathbb R^3 \to \mathbb R^3$ to the same level of detail?

share|cite|improve this question
Complex analysis studies a very particular subset of the functions from $\mathbb R^2\to\mathbb R^2$. There isn't an algebra on $\mathbb R^3$ that has the nice properties that are need to for similar math in $\mathbb R^3$. – Thomas Andrews Aug 31 '12 at 18:10
Complex numbers are a field, while $\mathbb{R}^2$ is not. – copper.hat Aug 31 '12 at 18:18
The answers to this question seem to indicate that complex analysis is the study of complex differentiable functions. Their is some truth to this, but it is not completely correct. Many complex analysts study Quasi-conformal maps of the complex plane, which may be thought of as "small" deformations of complex differentiable functions. – Baby Dragon Aug 13 '13 at 21:36
Regarding $\mathbb{C}$ vs. $\mathbb{R}^2$: – cic Sep 7 '14 at 7:46
up vote 27 down vote accepted

Complex analysis is different from the analysis of functions $R^2 \to R^2$ in that it requires the functions not just to be differentiable, but complex-differentiable, which is a stricter requirement. One way of looking at the requirement is requiring that the functions locally look not only like a linear transformation $R^2 \to R^2$, but like a linear transformation corresponding to multiplication by a complex number, which we can identify with linear transformations of the plane by the obvious multiplication

$$(a,b) \cdot (c,d) = (ac - bc, ad + bc)$$

which is obvious when we write $(a,b) = a + bi$ and $(c,d) = c + di$. We can see this in matrix form if we identify $(a,b) = a+bi$ with the matrix

$$ \begin{bmatrix} a & -b \\ b & a \\ \end{bmatrix}.$$

Such functions have many properties that differentiable functions $R^2 \to R^2$ need not have, which gives complex analysis a much different flavor. For example, if we view complex functions on (simply connected subsets of) the complex plane as vector fields on $R^2$, then complex-differentiable functions will be conservative vector fields. However, if we know something about differentiable functions $R^2 \to R^2$, we then by definition know something about complex-differentiable functions $C \to C$, although we may need to do some translation.

share|cite|improve this answer

No, it isn't. This can most obviously be seen from the fact that the map $z \mapsto \overline z$ is not differentiable as a map from $\mathbb C$ to $\mathbb C$, but the corresponding map $(x,y) \mapsto (x,-y)$ is differentiable as a map from $\mathbb R^2$ to $\mathbb R^2$.

This is because to be complex differentiable you need to have a best complex linear approximation which is a much stronger requirement than to have a best real linear approximation.

share|cite|improve this answer

The functions studied in complex analysis are much more specific than functions from $\Bbb{R}^2$ to itself.

If you like, you could say that complex analysis is the study of functions from $\Bbb{R}^2$ to itself which satisfy a particular system of PDE (namely the Cauchy-Riemann equations). Most people probably wouldn't phrase it that way, because it obscures the motivation; "we should extend the concept of differentiability to complex numbers" is a much more natural goal than "we should look at functions which satisfy some arbitrary PDE system."

Since there isn't a good 3-dimensional analogue of the Cauchy-Riemann equations, there also isn't a good 3-dimensional analogue of complex analysis. There are certain areas which extend, however; for example, the study of harmonic functions on $\Bbb{R}^2$ essentially reduces to complex analysis, but you can also study harmonic functions on $\Bbb{R}^n$ (and they have many of the same properties as their 2-dimensional counterparts).

share|cite|improve this answer

Of course it is true that every function from ${\mathbb C}$ to $\mathbb C$ can be considered as a function from ${\mathbb R}^2$ to ${\mathbb R}^2$, but in Complex Analysis we make use of the additional structure arising from the multiplication of complex numbers. Differentiability in the sense of complex numbers is a much stronger assumption than differentiability for functions from ${\mathbb R}^2$ to ${\mathbb R}^2$. There isn't any comparable structure on ${\mathbb R}^3$.

share|cite|improve this answer

The simple answer is that the complex analysis is over the complex plane $\mathbb{C}$, which is NOT equivalent to the Euclidean plane $\mathbb{R}^2$. The complex plane $\mathbb{C}$ processes additional structure defined by complex multiplication, which is not defined over $\mathbb{R}^2$.

share|cite|improve this answer

No. In some sense, complex analysis studies what happens if you generalize polynomials on $\mathbb{C}$ by allowing them to have infinite degree, but still keeping the requirement that the result must be a function having values in $\mathbb{C}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.