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After I have taken a course in complex analysis I am still wondering exactly which property of $\mathbb{C}$ makes complex analysis so different. Since we actually proved the Fundamental Theorem of Algebra using methods from complex analysis it can not be because $\mathbb{C}$ is algebraically closed.

Here is my try in figuring it out. I think there a two different factors at play.

  1. The complex multiplication makes $(\mathbb{C}, +, \cdot)$ into an associative Banachalgebra. Since we are trying to compare it to real analysis or analysis on $\mathbb{R}^n$ we best think of $(\mathbb{C}, +, \cdot)$ as $\mathcal{C} = (\mathbb{R}^2, +, [\cdot, \cdot])$ whereas $[\cdot, \cdot]: \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}^2$ is defined by $[(a, b), (c, d)] = (ac - bd, ad + bc)$. So $[\cdot, \cdot]$ is bilinear, assoziative and commutative (in addition every element has an inverse). Having this additional structure let us make sense of power series. But even more so, it gives us a canonical way of identifying the Algebra $CDer = \{A \in L(\mathbb{R}^2, \mathbb{R}^2) \mid A \text{ is a scaling followed by a rotation}\}$ with pointwise addition and composition as multiplication with $C$. This is done by the map sending $f \in CDer$ to the vector $a \in \mathcal{C}$ such that $\forall x \in \mathcal{C}: f(x) = [a, x]$. Having this identification at hand allows us to think of higher derivatives in a simple way.
  2. The Identification in the previous point only worked, because we restricted the usual notion of being differentiable in $\mathbb{R}^2$. Instead we required a function to be differentiable in $\mathbb{R}^2$ and in addition the linear approximation had to be a scaling followed by a rotation. In the case of $\mathcal{C}$ the vector space $CDer$ has exactly the same dimensio as $\mathcal{C}$.

This leads to the following questions:

A. Could one not just do the same fruitful analysis on $\mathbb{R}^n$ by just finding the right map $[\cdot, \cdot]: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n$

B. Which properties must $[\cdot, \cdot]$ have? Note to have the identification as in 1. we must have associativity and bilinearity. I have read that there is also some analysis done in the Quarternions which would suggest that commutativity is not that important.

C. How do we have to restrict the differential to get such identification?

Anyways I am thankful for any thoughts.

baleine6
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    You may want to see the following answers of mine: Generalizing the Analyticity of holomorphic functions, Liouville's theorem, Holomorphic functions with values in Banach spaces, implicit and inverse function theorem. There are lots of conceptual similarities in the definitions and basic theorems in both the real and complex case. – peek-a-boo Apr 14 '25 at 01:06
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    One of the biggest differences is of course that if $V,W$ are complex vector spaces, and $T:V\to W$ is a function, then $\Bbb{C}$-linearity is a stronger condition than $\Bbb{R}$-linearity (barring trivial edge cases like $V$ or $W$ being ${0}$). So, the requirement that the derivative $Df_a$ be $\Bbb{C}$-linear enforces the Cauchy-Riemann equations. These are elliptic PDEs, so we get the wonderful regularity theorems. Furthermore, it's a very simple PDE, with solutions obtained by convolving against the Cauchy kernel $\frac{1}{\zeta}$, so these give integral representation formulas. – peek-a-boo Apr 14 '25 at 01:09
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    and the integral representation formula gives us very precise information about the functions involved (mean-value property, rigidity like the various versions of Liouville's theorems etc). But notice that a lot of these facts have more to do with the Cauchy-Riemann equations being elliptic PDEs than having the field be $\Bbb{C}$ (because many of the theorems I quoted hold for harmonic functions as well). Hopefully the links above and these remarks give you some things to think about. – peek-a-boo Apr 14 '25 at 01:11
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    oh and another example of the fact that complex derivatives behave nicely because it's actually integrals which behave nicely – peek-a-boo Apr 14 '25 at 02:27

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It's really tempting to think of complex analysis (in one variable) as just "real analysis in 2 variables, but with some extra notation." But the real secret sauce is in the definition of complex differentiablity. For the complexes, that means the existence of a complex number $u$ such that $$ \lim_{h \to 0} \frac{1}{h} \left[(f (z+h) - f(z)) - u h\right] = 0 $$ while in the case of the real plane, it'd be the existence of a linear map $L$ from $\mathbb R^2$ to $\mathbb R$ such that $$ \lim_{h \to 0} \frac{1}{|h|} \left[(f (z+h) - f(z)) - L(h)\right] = 0 $$

The use of complex multiplication (or division) in that definition drastically reduces the set of differentiable functions, so much so that the ones that remain have many remarkable properties, like "if it's differentiable (everywhere) once, it's infinitely differentiable", and "if it's bounded on the whole complex plane, then it's a constant," etc.

You could do the same thing in $\mathbb R^2$ -- simply define the difference-quotient for derivatives by expanding out the complex version in coordinates, but it'd look weird and unmotivated. And, of course, you'd rule out a ton of interesting (or annoying, depending on your point of view) features of real analysis.

John Hughes
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  • The mod signs shouldn't be there even in the real case. What makes complex differentiability demanding is that, while $df$ is linear in $dx$ and $dy$ or equivalently in $dz$ and $dz^\ast$, we need the $dz^\ast$ coefficient to vanish. – J.G. Apr 13 '25 at 14:27
  • Thank you for your answer. But I still do not understand how this answers any of my questions. Also why can I not think about complex differentiability as differentiability in $\mathbb{R}^2$ with the addition that the function has to satisfy the CR-equations. I am trying to generalize to higher dimensions. – baleine6 Apr 13 '25 at 15:09
  • Somewhere, explicitly or implicitly, multiplication by $i$ has to appear. In higher dimensions, if $V$ is a real vector space, one normally formulates this as a linear map $J$ such that $$J^2=-I.$$ An immediate consequence is that the dimension of $V$ is even. You can proceed from there without ever mentioning complex numbers. But this is obviously too cumbersome. – Deane Apr 13 '25 at 15:30
  • J.G.: in the real case (by which I meant "$R^2$ case") you need to include the length. I should have included the linear map as well; I'll rewrite. – John Hughes Apr 13 '25 at 17:59
  • baleine6: Your leading question was "Which property of ℂ makes complex analysis so different?" I tried to answer that by noting that the existence of division (and the possibility of using this in defining derivatives) is important. If you think I failed to answer, though, please go ahead and downvote. Probably Deane's comment is more what you were looking for. – John Hughes Apr 13 '25 at 18:06