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I haven't yet started my complex analysis course (soon!), but recently (inspired by you guys) I've been looking into holomorphic functions. And wow, they're cool! There's so much stuff that's true about them... But my question is: why is being holomorphic such a strong condition? Is there some intuitive reason why this seemingly simple condition implies so much about a function?

edit: may I add that I am thinking in particular about entire functions. e.g. it is not all obvious to me why being differentiable on the complex plane would imply Picard's theorem, that the function takes every value except at most one.

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    $\begingroup$ For one: $f$ having a complex derivative in some domain $D\in\mathbb{C}$ implies that $f'$ also has a complex derivative in that domain, and so forth. So having one complex derivative implies arbitrarily many of them. $\endgroup$ – Semiclassical Sep 26 '14 at 16:45
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    $\begingroup$ This is a pretty good question and one can only hope it receives some in-depth answer (if nothing pops up after a while, I might give it a try). $\endgroup$ – Did Sep 26 '14 at 16:47
  • $\begingroup$ Let me say in advance that a great answer may attract a bounty ;) $\endgroup$ – Shakespeare Sep 26 '14 at 16:48
  • $\begingroup$ To add to what @Semiclassical said, the real and imaginary parts of a holomorphic function are harmonic functions (satisfying Laplace's equation). Not just that, given one you can find the conjugate harmonic to within a constant. So I think of lots of things having to line up to make a function holomorphic. Not surprising, then, that they imply so much else. $\endgroup$ – user_of_math Sep 26 '14 at 16:52
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    $\begingroup$ In addition to the comments about the Laplace operator, another result I think is important is that, in some sense, entire functions are as close to polynomials as you're going to get. That is, a lot of results which hold for polynomials hold for entire functions. $\endgroup$ – Cameron Williams Sep 29 '14 at 3:22

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I would contend that what makes complex analytic (holomorphic) functions so special is the structure of the complex numbers themselves. The fact that the complex numbers are essentially $\mathbb{R}^2$ and are also a field is a small miracle. This miracle is at the heart of the special behavior of complex analytic functions.

It is the two dimensional nature of the plane and the field multiplication allow us to see the C-R equations. My favorite way to see the C-R equations is to consider $$f'(z) =\lim_{\Delta z\to 0} \frac{f(z+\Delta z)-f(z)}{\Delta z} $$ Taking in turn $\Delta z =\Delta x$ and $\Delta z =i \Delta y$ the C-R equations just appear. This requires two dimensions for the two approaches and utilizes that $\mathbb{C}$ is a field when dividing by $\Delta z$. (Other chain rule type proofs of the C-R use both these facts, but are somewhat more subtle about it.)

Surely the Cauchy-Integral Formula, which tells us that a holomorphic function is in fact $C^\infty$ deserves mention, in fact it too is a consequence (less directly) of the two dimensional nature of $\mathbb{C}$ and field structure. The standard proof is to use Cauchy's Integral Theorem, which says that if $f$ is a holomorphic on a simply connected domain, then $$\int_\gamma f =0$$ for any sufficiently nice closed $\gamma$ curves in the domain. This itself is seen by applying Green's (2-d real structure again) and the C-R (field structure).

There are many, many more special and cool behaviors of holomorphic functions, but all of them seem to sit on the miracle of the two structures of the complex numbers. Just to advertise a few, Identity Principle (being 2-d connected is easier than being 1-d connected), zero counting theorems like Rouche's (line integrals sometimes have cool answers), the open mapping theorem (zero counting theorems are super cool), Liouville's Theorem (Cauchy integral theorem again), on and on and on.

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  • $\begingroup$ This synthesizes other answers and comments very well - thank you! $\endgroup$ – Shakespeare Oct 2 '14 at 11:38
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    $\begingroup$ I think the simplest way to make the connection with Green's theorem is to express the CR equations as the condition $\frac{\partial}{\partial \overline{z}}f(z,\overline{z})=0$. Then $d(f \,dz)=f'(z)\,dz\wedge dz=0$, so $\int_{\partial D}f(z)\,dz=\int_D 0 = 0.$ $\endgroup$ – Semiclassical Oct 2 '14 at 14:12
  • $\begingroup$ @Semiclassical: that's a very nice proof, but most people are probably not familiar with differential geometry enough to know Stokes' theorem. Though I am inclined to believe that the geometry of complex numbers is the reason OP asked about. $\endgroup$ – tomasz Oct 2 '14 at 15:00
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    $\begingroup$ My hope was to draw attention to the fact that you can't really get any of the nice theorems available in holomorphic function theory without being able to divide. This is the primary difference between holomorphic and real 2-d differentiable functions. $\endgroup$ – Joe Manlove Oct 2 '14 at 17:14
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I think the nicest way to 'picture' this intuitively without having started your course is the 'amplitwist' concept coined by Tristan Needham (I'd recommend his book, Visual Complex Analysis, if this idea interests you).

Essentially, to quote from his exposition:

Analytic mappings are precisely those whose local effect is an amplitwist: all the infinitesimal complex numbers emanating from a single point are amplified and twisted the same amount.

This notion can be tightened up and made rigorous (see Needham), but it doesn't need rigour to provide illumination. Unfortunately I can't give you a direct intuition transplant, but I'd encourage you to play around a little with functions in $ \mathbf{C} $ and $ \mathbf{R}^2 $ to see if you can understand both what the above condition means, and why it might be significantly more restrictive than mere $ \mathbf{R}^2 $ analyticity.

So to directly answer some of your questions:

  • Is there some intuitive reason why this seemingly simple condition implies so much about a function?

Yes -- and it's because it's not a simple condition! Essentially because of the required interplay between real and complex parts that is shown either algebraically in the C-R equations, or geometrically in the 'amplitwist' idea (there are amplitwist diagrams about online, and hundreds in Needham's book -- some might help).

  • it is not all obvious to me why being differentiable on the complex plane would imply Picard's theorem

You're certainly not alone there. I think everyone is a little surprised by Picard's ('Great'!) theorem. I'm perpetually slightly surprised by it. But then, many of the major results in complex analysis are highly non-obvious. In some sense analytic functions are a major object of study precisely because their definition leads to a great number of 'nice' and interesting properties.

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    $\begingroup$ Ok, another good idea! I also found this to be a good resource, it really highlighted, if quite lengthily, what you were trying to say. usf.usfca.edu/vca/PDF/amplitwist.pdf. I have an issue with the amplitwist idea though - it may make me REALLY understand what a complex derivative is, but it doesn't make it at first sight more obvious why being holomorphic does indeed imply so much about a function. e.g. why does being $f$ being "amplitwistable" mean $f'$ is also "amplitwistable"? $\endgroup$ – Shakespeare Sep 29 '14 at 17:20
  • $\begingroup$ Very nice reference! (+1) $\endgroup$ – Markus Scheuer Sep 29 '14 at 17:54
  • $\begingroup$ @Shakespeare Yes, that PDF is an extract from Needham's Visual Complex Analysis. It's not really obvious from the amplitwist characterisation that once-differentiable implies analytic, I'm afraid. Essentially, locally a differentiable function maps discs to (rotated, expanded, and/or translated) discs with the derivative as the amplitwist. I think second-differentiability does make sense if you think about discs and limits -- in some sense it's clear that unpleasant/pathological behaviour is forbidden by the discs -> discs condition -- but it's not immediate. More machinery required! $\endgroup$ – JCW Oct 3 '14 at 17:30
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I'd say the the main thing is that you have the Cauchy integral formula.

Intuitively, integrating something makes it have one more derivative. Being able to write a function as something involving it's own integral at all gives it being differentiable. Being able to write it in a way that reacts well to differentiation gives you infinitely many derivatives. It's sort of similar to how a differentiable solution to $f'= f$, or even say $f' = f^2 + (\cos x )\,f$ has to be infinitely differentiable.

It's also useful because the formula reacts so well to differentiation that bounds on $f$ imply bounds on every derivative that are good enough to bound the error term in Taylor expansions, giving rise to holomorphicity being equivalent to being analytic.

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    $\begingroup$ Also, gowers.wordpress.com/2007/09/19/… gives natural motivation/context for thinking about vanishing path integrals in the first place. (See expii.com/e/768 for one of the points of Gowers' analogy made explicit.) Alternatively, the Cauchy integral formula (and hence much else of complex analysis) can be traced back to the fundamental fact that $\int_0^{2\pi} e^{i\theta} d\theta = 0$ (i.e. the basis of Fourier analysis, or its discrete analog, the roots of unity filter). $\endgroup$ – Victor Wang Sep 27 '14 at 7:15
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One thing that strikes me as important is that being holomorphic (i.e. analytic in the complex plane) is a much stronger property than on the real line alone.

A complex differentiable function must satisfy the Cauchy-Riemann equations. What the C-R equations essentially capture is that for the derivative to be defined in the complex sense, it must exist and be the same from all directions. This restricts us to a very special set of functions, e.g. the conjugate function $f(z)=\bar{z}$ is not complex-differentiable (see here)

Whereas, if we restrict ourselves to the real line you essentially have only one direction (the real line) along which the derivative is required to exist, and the "interplay" between real and imaginary parts is lost (see Cauchy-Riemann equations)

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    $\begingroup$ obviously it's stronger, but when you distill it down to the Cauchy-Riemann equations it still doesn't really look like all that much. This easily followed line of reasoning is entirely false... What I'm trying to say that yes, this is the base reason, but it's not really intuitive, and perhaps a more intuitive expansion of it can be made. (Or maybe that's just me, and this really is the best answer!) $\endgroup$ – Shakespeare Sep 26 '14 at 17:15
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    $\begingroup$ The infinitude of directions the derivative must agree on is a pretty good reason to expect nice behavior. It's easy to find multivariable rational functions which fail even continuity because of this. But that still doesn't quite separate it from the real variables case. It's the specific algebraic properties of the complex numbers that cause the magic to happen. The directions are connected not just by limit considerations but also the properties of complex multiplication. That's what the Cauchy-Riemann equations tell us. $\endgroup$ – zibadawa timmy Sep 26 '14 at 17:22
  • $\begingroup$ @zibadawatimmy that is all a good point; I just feel like I'm still missing something though... perhaps an intuitive expansion on the consequences of the Cauchy-Riemann equations (i.e. being holomorphic) would help clear it up? $\endgroup$ – Shakespeare Sep 26 '14 at 17:31
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    $\begingroup$ @Semiclassical An interesting thing to think about is that the CR equations are an example of an elliptic differential operator. Many of the nice properties of holomorphic functions follow simply from the fact that they are solutions of an elliptic operator. $\endgroup$ – Miha Habič Sep 26 '14 at 20:32
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    $\begingroup$ I don't really like the "infinite number of directions" argument. With the same argument, one would be lead to conclude that every differentiable function $f : \Bbb{R}^2 \to \Bbb{R}$ should enjoy similar properties to holomorphic functions, because here one can also approach each point from an infinite number of directions. Of course, the "derivative" is allowed to be a matrix here instead of a complex number (a $\Bbb{C}$-linear map). $\endgroup$ – PhoemueX Sep 29 '14 at 21:10
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What really helped me to understand how special holomorphic functions are is that, if $U \subset \Bbb C$ is some region (open and connected, e. g. $U = \Bbb C$), then every holomorphic function $f\colon U \to \Bbb C$ is uniquely determined by its values on any subset of $U$ that contains a limit point.

In particular, any holomorphic function $f \colon \Bbb C \to \Bbb C$ is given by its values on $\{ 0 \} \cup \{ \frac1n \ | \ n \in \Bbb N \}$.

First of all, this makes it believable that the integral of a holomorphic function along some closed curve encapsulates so much information about the behavior of $f$ inside the curve (since $f$ is uniquely determined by its values on the curve!).

Secondly, it says that knowing $f$ in some very small region (it can be any disk or it can even be countable) automatically tells you what $f$ looks like at every other point. Hence a holomorphic function actually only has some "countable set of freedom"!

Edit: As requested, here is a proof: I assume that we know that every holomorphic function $f\colon U \to \Bbb C$ is analytic (i. e., it can be expanded in a power series around every point of its domain). I will further assume that we know that two power series around the same point $z$ are equal if they have the same values on any sequence $(z_n)$ converging to $z$ (where $z_n \ne z$ for all $n$). If you want, I can say more about these two facts.

Let $f, g\colon U \to \Bbb C$ be two holomorphic functions, let $A \subset U$ be the set of points where $f$ and $g$ are equal and let $L$ be its limit points. We will show that $L$ is both closed and open. This implies (since $U$ is connected) that either $L = \varnothing$ or $L = U$, thereby proving the above proposition.

$L$ is closed: It is easy to see that $A$ is closed (since $A = (f - g)^{-1}(0)$ and $f - g$ is continuous). This already proves that $L$ is closed.

$L$ is open: Let $z \in L$. Then $z$ is in $U$ (since $A$ is closed, $z \in A \subset U$), so we can find a ball $B$ around $z$ in which we can expand $f$ and $g$ in power series. Furthermore, since $z$ is a limit point of $A$, there is a sequence $(z_n)$ converging to $z$ with $z_n \in A \backslash \{ z \}$ for all $n$. By definition of $A$ we have $f(z_n) = g(z_n)$, so the power series are equal, that is, $f = g$ on $B$. But this implies $B \subset L$, so $L$ is open. $\square$

As a side note, we actually only need to know that $f$ and $g$ are equal on some set whose limit point is contained in $U$. However, the continuity of $f$ and $g$ shows that this is equivalent to the above formulation.

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  • $\begingroup$ This is an interesting answer, a very good one in fact - I was not familiar with this. To make it more complete, could you please provide a proof and/or intuitive motivation? $\endgroup$ – Shakespeare Sep 29 '14 at 15:20
  • $\begingroup$ Ok, that was a fascinating proof! Pitched just at the pace I like, too. If one can intuitively accept that a holomorphic function is infinitely differentiable then this is an intuitive result (now you've explained it) IMO, so it's a valid part of the answer to my question. Thanks :) $\endgroup$ – Shakespeare Sep 29 '14 at 17:47
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Yet another viewpoint:

The Cauchy-type integral $$\varphi\longmapsto I(\varphi)$$ $$I(\varphi)(z)=\int_\gamma\frac{\varphi(w)}{w-z}$$ transforms continuous functions in analytic functions (locally power series). Holomorphic (one-time $\Bbb C$-differentiable) functions are fixed points of the operator.

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    $\begingroup$ Out of curiousity, do higher cycles of this map have a known characterization? (E.g. Fixed points are to holomorphic functions as two-cycles are to...?) $\endgroup$ – Semiclassical Oct 2 '14 at 13:36
  • $\begingroup$ @Semiclassical, Very good question! I will think on it. $\endgroup$ – Martín-Blas Pérez Pinilla Oct 2 '14 at 13:37
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    $\begingroup$ @Semiclassical, As $I(\varphi)$ is analytic, $I(I(\varphi))=I(\varphi)$ and the fixed points of $I$ and $I^2$ are the same. $\endgroup$ – Martín-Blas Pérez Pinilla Oct 2 '14 at 13:54
  • $\begingroup$ So all cycles have primitive period of 1. That's a tad underwhelming but probably not surprising. Finding a variation on this map with higher cycles are nontrivial would be interesting, though I've no idea how one would go about that. $\endgroup$ – Semiclassical Oct 2 '14 at 13:57
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Why holomorphic functions are so much nicer the differentiable functions in $\mathbb R$ or $\mathbb R^n$?

Why complex differentiability implies so much more than differentiability in $\mathbb R$ or $\mathbb R^n$?

The answer is the Cauchy-Riemann equations.

A holomorphic function $f(z)$ can be also viewed as a function from an open subset of $\mathbb R^2$ to $\mathbb R^2$, i.e., if $f(x+iy)=u(x+iy)+iv(x,y)$, where $u=$Re$\,f$ and $v=$Im$\,f$, then $f$ can be viewed as the mapping $$ (x,y)\longmapsto \big(u(x,y),v(x,y)\big), $$ which is differentiable in the usual Euclidean space sense. But that's not all: its differential is realized by a $2\times 2$ matrix of a very specific form $$ \left(\begin{matrix} u_x & u_y \\ v_x& v_y\end{matrix}\right)= \left(\begin{matrix} u_x & u_y \\ -u_y& u_x\end{matrix}\right). $$ This very restrictive form of the differential is the reason behind all the nice properties of holomorphic function, it forces $u$ and $v$ to be harmonic functions, i.e., $$ \frac{\partial u^2}{\partial x^2}+\frac{\partial u^2}{\partial y^2}= \frac{\partial v^2}{\partial x^2}+\frac{\partial v^2}{\partial y^2}=0, $$ and such functions are infinitely differentiable, and hence, in turn the derivatives of $(u,v)$, i.e., $(u_x,v_x)=(-u_y,v_x)$, are also harmonic and satisfy as well Cauchy-Riemann equations, which make $f'$ holomorphic as well.

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    $\begingroup$ This is more or less the same as pbs's answer. $\endgroup$ – tomasz Oct 2 '14 at 14:57
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Edit [2014-10-25]: Due to respect for K. Knopp (and also for my pleasure) I bought the english translation Theory of Functions and replaced in the answer below my translation of the German classic Funktionentheorie I (1949) with an authorized version.


A few more aspects (and a word of warning at the end)

  • Statements from old masters

Many of the answers above provide valuable information around the question what's so special about these holomorphic functions.

I was curious to find statements from the old masters which could help to provide us with some more insight. I had a look for some classics about complex analysis like those of A.Hurwitz, W. Rudin, J.B. Conway, S. Lang and some more.

For me the most valuable statement was given by K.Knopp in his classic Theory of functions. The interesting thing is that in this book there are some places where he refers to our question. It culminates in the section 7.21 The Identity Theorem for Analytic Functions:

Konrad Knopp (extract from Theory of functions)

Now, it is exceedingly remarkable that by means of the single requirement of differentiability, that is, the requirement of regularity, a class of functions having the following properties is selected from the totality of the most general functions of a complex variable. On the one hand, this class is still very general and includes almost all functions arising in applications. On the other hand, a function belonging to this class possesses such a strong inner bond, that from its behavior in a region, however small, of the $z$-plane one can deduce its behavior in the entire remaining part of the plane. ...

A first theorem in this direction is Cauchy's formula which enables us to deduce the values of the function in the interior of a simple closed path $\mathcal{C}$ from the values along the boundary. A second result of this kind is the statement made in connection with the expansion theorem as to the magnitude of the true circle of convergence of a power series. Indeed, here we have already taken into consideration points of the plane which do not even belong to the original region of definition of the function.

... we are now in a position to derive a result ... which is because of its great importance for the development of the theory of functions, the most fundamental result besides Cauchy's integral theorem.

The identity theorem for analytic functions: If two functions are regular in a region $\mathcal{G}$, and if they coincide in a neighborhood, however small, of a point $z_0$ of $\mathcal{G}$, or only along a path segment, however small, of a point $z_0$ of $\mathcal{G}$, or also only for an infinite number of distinct points with the limit point $z_0$, then the two functions are equal everywhere in $\mathcal{G}$.

Observe the dramatic construction of this theorem which Knopp uses to even more point out the significance of it! :-)

Note: The formulation of the translated version is ... most fundamental result after Cauchy's integral theorem .... In my opinion this is not correct. The formulation in the german original is neben which means besides (implying an equal ranking, and not a ranking after resp. behind Cauchy's integral theorem).

  • So, for me the essence of the nice behavior of holomorphic functions is that they are already characterized within a region $\mathcal{G}$ by infinitely many different points converging at an accumulation point. The situation is similar to a polynomial of degree $n$ which is already completely determined by $n+1$ different points, irrespectively how near or far they are.

  • And the magic is that all that can be derived from solely requiring differentiability of a function within a region $\mathcal{G}$.

Observe that also the topology namely the region $\mathcal{G}$ is an essential ingredient in this statement.


And now for something completely different:

Similar questions in MSE

There are questions in MSE similar to this one (of course :-)) and I would like to emphasize two of them:

  • The answers in Why are differentiable complex functions infinitely differentiable? are interesting, since some of them also try to provide an intuitive feeling.

  • A fundamental difference between complex and real differentiability is the type of linearity. Approximation of complex functions by a $\mathcal{C}$-linear map in contrast to a approximations of real functions by $\mathcal{R}$-linear maps is the reason that the Cauchy-Riemann differential equations hold in the complex case and providing so the bases for the miracle of the nice behaviour of holomorphic functions. See e.g. How is $\mathcal{C}$ different than $\mathcal{R}^2$?


Word of warning

This is not a part of the answer but a remark and also a word of warning to a reference from the answer of @pbs to the Wolfram page Complex Differentiable.

First I have to admit that many, many pages of Wolfram are a highly valuable source of information and I really appreciate this service. But regrettably sometimes a page is not sufficiently elaborated. And the page Complex Differentiable is one of these rare events.

Three critical aspects:

For me it's annoying that this page creates the impression of presenting a definition of complex differentiable and also referring thereby to the author G. Shilov and his book Elementary Real and Complex Analysis so that the reader could think this definition is cited from this book. This is not the case!

  • Definition of complex differentiable: Shilov defines this term in section 10.11.a: Differentiation in the complex domain as follows

Let $C_z$ be the plane of the complex variable $z=x+iy$,$\ldots$, and let $C_w$ be the plane of the complex variable $w=u+iv$,$\ldots$. Then a function $w=f(z)$ defined on a set $E\in C_z$ and taking values in $C_w$ is called a (complex) function of a complex variable.

Let $z_0\in E$ be a nonisolated point of $E$ so that every neighborhood of $z_0$ contains a point of $E$ other than $z_0$ itself. Then a complex number $A$ is said to be the derivative of the function $w=f(z)$ at the point $z=z_0$ relative to the set $E$, denoted by $f^\prime_E(z_0)$, if, given any $\varepsilon>0$, there exists a $\delta>0$, such that $0<|z-z_0|<\delta,z\in E$ implies $$\left|A-\frac{f(z)-f(z_0))}{z-z_0}\right|<\varepsilon$$ In this case, we say that $w=f(z)$ is differentiable at $z=z_0$, with derivative $A=f^\prime_E(z_0)$, where $A$ is clearly the limit of the difference quotient $\frac{f(z)-f(z_0)}{z-z_0}$ in the direction $z\rightarrow_{E} z_0$ determined by the set of intersections of $E$ with all deleted neighborhoods $0<|z-z_0|<\delta$.

The satisfaction of the Cauchy-Riemann equations which Wolfram cites follows in Shilov's book later from the definition, thereby precisely distinguishing between the different domains of the function under consideration and the different consequences resulting thereby!

  • complex differentiable vs. holomorph:

Despite the statment in Wolframs page, the term complex differentiable is per se NOT THE SAME as holomorphic (or analytic or regular, which may be used interchangeably) and G. Shilov states this clearly in his book. He defines in the section 10.11.f:

A function $f(z)$ is said to be analytic (synonymously, holomorphic or regular) on an open set $G \in \mathcal{C}_z$ if $f(z)$ is differentiable on $G$.

Please note, that we have to clearly distinguish between phrases $f$ is complex differentiable in $z_0$ and $f$ is complex differentiable in $\mathcal{G}$!

The third aspect adresses Wolfram's statement: $f:z\rightarrow\overline{z}$ is not complex differentiable. It would be helpful if Wolfram had pointed out the conditions which are necessary that this statement is true!

We can read in Shilov section 10.12.b:

  • The function $f(z)=\overline{z}=x-iy$ is differentiable at every point $z=z_0$ relative to any ray $E$ drawn from $z_0$, since $$\frac{f(z)-f(z_0)}{z-z_0}=\frac{\overline{z-z_0}}{z-z_0}=-2\mathop{arg}(z-z_0)$$ and hence \begin{align*} f^\prime_E(z_0)=-2\mathop{arg}(z-z_0)\qquad\qquad(z\in E,z\ne z_0)\tag{2} \end{align*} However,(2) shows that $f(z)=\overline{z}$ fails to be differentiable relative to any set containing two distinct rays drawn from $z_0$. Therefore $f(z)=\overline{z}$ fails to be analytic at every point $z=z_0\in C_z$.

Conclusion: We should always check at least two independent sources when learning new terms.

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The C-R equations, Cauchy integral formula and friends are insanely hard for a function to satisfy, I guess we all agree on this! But honestly, my intuition does not work very well with the aforementioned properties.

What really helped me getting an idea about how rigid holomorphic functions are is a corollary of Cauchy's integral formula: the mean value property.

You can look at the MVP as a restatement of Cauchy's formula, but now you can see explicitly that the value of your holomorphic function $f$ at $a$ depends on the value of $f$ on any circle around $a$!

To give a rough idea of the rigidity that this implies, the modulus $|f|$ of a holomorphic function $f$ cannot exhibit a true local maximum that is properly within the domain of $f$. Quoting wikipedia "In other words, either $f$ is a constant function, or, for any point $z_0$ inside the domain of $f$ there exist other points arbitrarily close to $z_0$ at which $|f|$ takes larger values."

Food for thought: (I trying to stick to properties that one can actually visualize instead of purely analytical relations between objects)

  • Def: a conformal map is a function which preserves angles locally.
  • Thm: If $U$ is an open subset of the complex plane, then a function $f: U \rightarrow \mathbb{C}$ is conformal if and only if it is holomorphic and its derivative is everywhere non-zero on $U$.
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Maybe it is also worthwile to think about these concepts in real analysis setup: in one dimensional setting there is one of the most powerful coincidencies between the antiderivative and the usual Riemann (in most cases-Lebesgue) integral. There is no straightforward generalisation of the concept of antiderivative of the function $u:\mathbb{R}^n \to \mathbb{R}$ but the concept of antiderivativie fits better in the framework of vector fields. However the condition for a vector field to be a total derivative of a given function is much more restricitive than only mere conditions involving (various levels of) regularity. Take two arbitrary chosen functions $P(x,y), Q(x,y)$. You have to be lucky to get $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$! And now coming back to complex analysis: take an arbitrary complex valued function $f(z)$ and ask whether this has an antyderivative. It seems that it is not a problem: but try to find the formula for $u(x,y),v(x,y)$-real and imaginary parts of $f(z)$ in terms of real and imaginary parts of $z$. You will find that giving two such functions $u,v$ it will rarely happen that they satisfy Cauchy-Riemann equations. This is pretty much the same as explained above but maybe using a bit different language.

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