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?

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    $\begingroup$ 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 $\mathbb R^3$. $\endgroup$ Aug 31, 2012 at 18:10
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    $\begingroup$ Complex numbers are a field, while $\mathbb{R}^2$ is not. $\endgroup$
    – copper.hat
    Aug 31, 2012 at 18:18
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    $\begingroup$ 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. $\endgroup$ Aug 13, 2013 at 21:36
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    $\begingroup$ Regarding $\mathbb{C}$ vs. $\mathbb{R}^2$: math.dartmouth.edu/~m43s12/notes/class1.pdf $\endgroup$
    – cic
    Sep 7, 2014 at 7:46

6 Answers 6


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.

  • $\begingroup$ 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. Can you explain little bit? Where will I find the detailed explanation? Please help me. $\endgroup$
    – user464147
    Nov 18, 2017 at 13:52
  • $\begingroup$ Why is this a stronger requirement? If a real linear approximation exists, then a complex one exists too, because R is a subset of C. But if a complex linear approximation exists, it doesn't mean that a real one exists. Why is it wrong? $\endgroup$ Dec 31, 2019 at 15:14

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

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

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.


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).


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 ${\mathbb R}^3$.


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 $\mathbb{R}^2$.


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 $\mathbb{C}$.

So, for instance, $z^2=x^2-y^2+2ixy$ is a polynomial in $z$, but $\bar z=x-iy$ is not. However, $f(x,y)=(x^2-y^2,xy)$ and $g(x,y)=(x,-y)$ are both legitimate polynomial functions from $\mathbb{R}^2$ to $\mathbb{R}^2$.


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