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Question 1: What do holomorphic functions 'look like'? Not really sure what is meant by this, but I heard this was asked this in an interview for graduate school admissions.

What I have in mind is that

  • A. holomorphic functions are conformal, infinitely complex differentiable or something (not sure of the precise wording, but i guess it's like 'complex smooth'), equivalent to complex analytic (therefore we can just drop the qualifier 'complex') and have harmonic real and imaginary parts (i didn't learn in elementary complex analysis, but in miranda's book 'harmonic' is defined precisely as this. in elementary complex analysis, i learned holomorphic implies harmonic parts). Not really sure how to graph any of these, but the conformal thing is on wiki.

  • Here's the picture:

  • enter image description here

  • B. when you're asked what continuous functions $f: \mathbb R \to \mathbb R$ 'look like', intuitively/heuristically, you can draw a part of it on a paper without lifting your writing utensil from the paper. so you probably just do something that looks like sin/cos or a polynomial. (i remember in calc 1, the 1st continuous functions we're introduced to are these smooth functions sin/cos and polynomials.) I think this is kind of what the question means.

  • C. from the very definition itself, i just think to draw some circles like $g: \mathbb C \to \mathbb C$ holomorphic at $z=w$ means complex differentiable in some disc/disk around $z=w$. but i guess i'm just making drawings on its domain rather drawing the function itself.

Question 2: what books answer this?

Question 3: In relation to (B) above I actually find this question kind of weird. I mean does it make sense to ask what differentiable real functions $f: \mathbb R \to \mathbb R$ 'look like' ? twice differentiable functions $f: \mathbb R \to \mathbb R$ 'look like'? thrice? smooth functions $f: \mathbb R \to \mathbb R$? real-analytic functions?

Question 4: What do poles look like? what do poles 'look like'? Please provide references.


Update: Based on the comments below, this version of the Maximum modulus principle seems to be relevant:

Let $D \subseteq \mathbb C$ be bounded, non-empty and open. Suppose $f: \overline D \to \mathbb C$ is continuous on $\overline D$ and holomorphic on $D$. Then $|f|$ attains a maximum at some point in boundary of $D$ (i guess topological boundary, and i guess equal to $\overline D \setminus D$).

The comments Severin Schraven include:

It makes quite a lot of sense to ask this question in my opinion. The point is that holomorphic functions have a lot of structure and force them to have certain features. (...) The maximum principle is a very strong property. General continuous functions can be extremely wild!

A comment of Moishe Kohan:

Strictly speaking, the question is meaningless unless the meaning of "looks like" is specified. For this reason, no self-respecting textbook will discuss this, nor an interviewer at a grad school interview. They may ask: "What local properties of holomorphic functions do you know?" This would be a reasonable "long list" question.

God bless you, Moishe Kohan.

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    $\begingroup$ It makes quite a lot of sense to ask this question in my opinion. The point is that holomorphic functions have a lot of structure and force them to have certain features. The key here is the maximum principle if you are familiar with that. Try to draw a function that satisfies this, then you will understand :) $\endgroup$ Mar 15, 2021 at 11:10
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    $\begingroup$ Exactly. The maximum principle is a very strong property. General continuous functions can be extremely wild! $\endgroup$ Mar 15, 2021 at 11:24
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    $\begingroup$ Yes, that is what I mean :) $\endgroup$ Mar 15, 2021 at 11:36
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    $\begingroup$ holomorphic functions look locally like powers - if $f$ holomorphic in some domain $U$ then for each $w \in U$ there is a small disc around $w$ contained in $U$ and unique $n \ge 1$ for which the function "looks like" $f(z)=a+b(z-w)^n$ with $b \ne 0$ and $a,b$ depending on $w$; this seemingly innoucous statement is actually very powerful and gives among other things, the maximum modulus principle, the discreteness of zeroes, results about conformality etc; $\endgroup$
    – Conrad
    Mar 15, 2021 at 13:29
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    $\begingroup$ Strictly speaking, the question is meaningless unless the meaning of "looks like" is specified. For this reason, no self-respecting textbook will discuss this, nor an interviewer at a grad school interview. They may ask: "What local properties of holomorphic functions do you know?" This would be a reasonable "long list" question. $\endgroup$ Apr 19, 2021 at 16:57

1 Answer 1

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Setup

A holomorphic function $f$ may be constant (and every constant function is holomorphic), in which case its effect on the input plane would be to send it all to a single point, and its graph in four real dimensions would look like a translation of the input plane. Constant functions are exceptional in many ways, so for the rest of this discussion, suppose that that $f$ is not constant.

Local

One way to “look” at a function is to consider “local” properties that can be seen when you look at the effect on a small neighborhood of a point in the domain.

Complex

The derivative $f'$ exists and is also holomorphic, and is not constantly zero since $f$ is not constant, so $f'$ has countably many isolated zeros. And the order of each zero is finite.

For every point $a$ that is not a zero of $f'$, there is a neighborhood of $a$ on which $f$ is conformal (preserving angles including their orientation/sense). And sufficiently close to $a$, $f$ is well-approximated by the action of complex multiplication by $f'(a)$, which has the effect of scaling by $\left|f'(a)\right|$ and rotating by $\mathrm{Arg}\left(f'(a)\right)$ (as a consequence, a small enough circle centered at $a$ is sent to a good approximation of a circle centered at $f(a)$).

For every point $b$ that is a zero of $f'$, since the order of the zero is finite, $f$ acts near $b$ like $f(z)\approx f(b)+c(z-b)^{n+1}$ for some positive integer $n$, so that $f$ has the effect of multiplying angles at $b$ by a factor of $n+1$ (by DeMoivre's Theorem, say). A small enough circle centered at $b$ is sent to a curve that approximates going around $f(b)$ in a circle $n+1$ times.

Components

If we instead want to understand what $f$ looks like by examining its components individually, we note that the components of $f$ are harmonic. This means that $\Delta u=\nabla\cdot\nabla u=0$, so that, by the flux form of Green's theorem, the flux integral of $\nabla u$ around a small circle is approximately zero. But this, combined with the continuity of $\nabla u$ coming from the infinite differentiability of $f$ (say) has the consequence that the average displacement of $u$ around a small circle is approximately zero, by the following calculation (see this Khan academy video for some visual intuition): $${\displaystyle \lim_{r\to0}}\dfrac{1}{2\pi r}\oint u\left(x+r\cos t,y+r\sin t\right)-u(x,y)\,\mathrm{d}t$$ $$=\dfrac{1}{2\pi}{\displaystyle \lim_{r\to0}}\oint\dfrac{u\left(x+r\cos t,y+r\sin t\right)-u(x,y)}{r}\,\mathrm{d}t$$

$$=\dfrac{1}{2\pi}{\displaystyle \lim_{r\to0}}\oint\nabla u\left(x+r\cos t,y+r\sin t\right)\cdot\left\langle \cos t,\sin t\right\rangle \,\mathrm{d}t$$

$$=\dfrac{1}{2\pi}{\displaystyle \lim_{r\to0}}\oint\nabla u\left(x+r\cos t,y+r\sin t\right)\cdot\mathrm{d}\hat{\mathbf{n}}$$

$$=0\text{ since }\Delta u=0$$

This is not a trivial claim because if $f$ is not constant, then either component must not be constant either, by the Cauchy-Riemann Equations.

Globally

If we look more broadly than a tiny neighborhood of a point, we would find that the maximum modulus principle (and the related maximum principle for the harmonic components) limits the way $f$ and $u$ can look even further.

And if $f$ is entire, then Picard's little theorem tells us that all values except perhaps one are attained somewhere.

References

Most of this is in many standard texts on Complex Analysis, like Ahlfors. I can edit in theorem/page references if desired.

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  • $\begingroup$ Totally forgot to log in. Thank God for auto award. thank you very very much Mark S. btw do you agree with Moishe Kohan's comment (that i edited into the OP) ? $\endgroup$ Oct 4, 2021 at 13:13
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    $\begingroup$ @John Smith Kyon, I agree that the phrase "looks like" is quite vague (see the different points of view in my answer), but wouldn't say that I agree that textbooks wouldn't discuss this. In some sense, Needham's Visual Complex Analysis is all about what complex functions "look like" from various perspectives. It would be a very bad exam question, but for an interview question, it sounds intentionally open-ended and fine to me. $\endgroup$
    – Mark S.
    Oct 4, 2021 at 13:29
  • $\begingroup$ thanks Mark S. ! $\endgroup$ Oct 4, 2021 at 13:30
  • $\begingroup$ Mark S., what do you think of the link of Andrew D. Hwang? i mean is it related to anything you said? $\endgroup$ Oct 27, 2021 at 20:44
  • $\begingroup$ Mark S., what do poles look like? $\endgroup$ Oct 27, 2021 at 21:36

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