6b. Holomorphic functions

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Let us study now the differentiability of the complex functions [math]f:\mathbb C\to\mathbb C[/math]. Things here are quite tricky, but let us start with a straightforward definition, as follows:

Definition

We say that a function [math]f:X\to\mathbb C[/math] is differentiable in the complex sense when the following limit is defined for any [math]x\in X[/math]:

[[math]] f'(x)=\lim_{t\to0}\frac{f(x+t)-f(x)}{t} [[/math]]

In this case, we also say that [math]f[/math] is holomorphic, and we write [math]f\in H(X)[/math].

As basic examples, we have the power functions [math]f(x)=x^n[/math]. Indeed, the derivative of such a power function can be computed exactly as in the real case, and we get:

[[math]] \begin{eqnarray*} (x^n)' &=&\lim_{t\to0}\frac{(x+t)^n-x^n}{t}\\ &=&\lim_{t\to0}\frac{nx^{n-1}t+\binom{n}{2}x^{n-2}t^2+\ldots+t^n}{t}\\ &=&\lim_{t\to0}\frac{nx^{n-1}t}{t}\\ &=&nx^{n-1} \end{eqnarray*} [[/math]]

We will see later more computations of this type, similar to those from the real case. To summarize, our definition of differentiability seems to work nicely, so let us start developing some theory. The general results from the real case extend well, as follows:

Proposition

We have the following results:

[math](f+g)'=f'+g'[/math].
[math](\lambda f)'=\lambda f'[/math].
[math](fg)'=f'g+fg'[/math].
[math](f\circ g)'=f'(g)g'[/math].

Show Proof

These formulae are all clear from definitions, following exactly as in the real case. Thus, we are led to the conclusions in the statement.

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As an obvious consequence of (1,2) above, any poynomial [math]P\in\mathbb C[X][/math] is differentiable, with its derivative being given by the same formula as in the real case, namely:

[[math]] P(x)=\sum_{k=0}^nc_kx^k\implies P'(x)=\sum_{k=1}^nkc_kx^{k-1} [[/math]]

More generally, any rational function [math]f\in\mathbb C(X)[/math] is differentiable on its domain, that is, outsides its poles, because if we write [math]f=P/Q[/math] with [math]P,Q\in\mathbb C[X][/math], we have:

[[math]] f'=\left(\frac{P}{Q}\right)'=\frac{P'Q-PQ'}{Q^2} [[/math]]

Let us record these conclusions in a statement, as follows:

Proposition

The following happen:

Any polynomial [math]P\in\mathbb C[X][/math] is holomorphic, and in fact infinitely differentiable in the complex sense, with all its derivatives being polynomials.
Any rational function [math]f\in\mathbb C(X)[/math] is holomorphic, and in fact infinitely differentiable, with all its derivatives being rational functions.

Show Proof

This follows indeed from the above discussion.

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Let us look now into more complicated complex functions that we know. And here, surprise, things are quite tricky, the result being as follows:

Theorem

The following happen:

[math]\sin,\cos,\exp,\log[/math] are holomorphic, and in fact are infinitely differentiable, with their derivatives being given by the same formulae as in the real case.
However, functions like [math]\bar{x}[/math] or [math]|x|[/math] are not holomorphic, and this because the limit defining [math]f'(x)[/math] depends on the way we choose [math]t\to0[/math].

Show Proof

There are several things going on here, the idea being as follows:

(1) Here the first assertion is standard, because our functions [math]\sin,\cos,\exp,\log[/math] have Taylor series that we know, and the derivative can be therefore computed by using the same rule as in the real case, similar to the one for polynomials, namely:

[[math]] f(x)=\sum_{k=0}^\infty c_kx^k\implies f'(x)=\sum_{k=1}^\infty kc_kx^{k-1} [[/math]]

(2) Regarding now the function [math]f(x)=\bar{x}[/math], the point here is that we have:

[[math]] \frac{f(x+t)-f(x)}{t}=\frac{\bar{x}+\bar{t}-\bar{x}}{t}=\frac{\bar{t}}{t} [[/math]]

But this limit does not converge with [math]t\to0[/math], for instance because with [math]t\in\mathbb R[/math] we obtain 1 as limit, while with [math]t\in i\mathbb R[/math] we obtain [math]-1[/math] as limit. In fact, with [math]t=rw[/math] with [math]|w|=1[/math] fixed and [math]r\in\mathbb R[/math], [math]r\to0[/math], we can obtain as limit any number on the unit circle:

[[math]] \lim_{r\to0}\frac{f(x+rw)-f(x)}{rw}=\lim_{r\to0}\frac{r\bar{w}}{rw}=\bar{w}^2 [[/math]]

(3) The situation for the function [math]f(x)=|x|[/math] is similar. To be more precise, we have:

[[math]] \frac{f(x+rw)-f(x)}{rw}=\frac{|x+rw|-|x|}{r}\cdot\bar{w} [[/math]]

Thus with [math]|w|=1[/math] fixed and [math]r\to0[/math] we obtain a certain multiple of [math]\bar{w}[/math], with the multiplication factor being computed as follows:

[[math]] \begin{eqnarray*} \frac{|x+rw|-|x|}{r} &=&\frac{|x+rw|^2-|x|^2}{(|x+rw|+|x|)r}\\ &\simeq&\frac{xr\bar{w}+\bar{x}rw}{2|x|r}\\ &=&Re\left(\frac{x\bar{w}}{|x|}\right) \end{eqnarray*} [[/math]]

Now by making [math]w[/math] vary on the unit circle, as in (2) above, we can obtain in this way limits pointing in all possible directions, so our limit does not converge, as stated.

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The above result is quite surprising, because we are so used, from the real case, to the notion of differentiability to correspond to some form of “smoothness” of the function, and to be more precisely, “smoothness at first order”. Or, if you prefer, to correspond to the “non-bumpiness” of the function. So, we are led to the following dilemma: \begin{dilemma} It's either that [math]\bar{x}[/math] and [math]|x|[/math] are smooth, as the intuition suggests, and we are wrong with our definition of differentiability. Or that [math]\bar{x}[/math] and [math]|x|[/math] are bumpy, while this being not very intuitive, and we are right with our definition of differentiability. \end{dilemma} And we won't get discouraged by this. After all, this is just some empty talking, and if there is something to rely upon, mathematics and computations, these are the computations from the proof of Theorem 6.17. So, moving ahead now, based on that computations, let us formulate the following definition, coming as a complement to Definition 6.14:

Definition

A function [math]f:X\to\mathbb C[/math] is called differentiable:

In the real sense, if the following two limits converge, for any [math]x\in X[/math]:
[[math]] f_1'(x)=\lim_{t\in\mathbb R\to0}\frac{f(x+t)-f(x)}{t}\quad,\quad f_i'(x)=\lim_{t\in i\mathbb R\to0}\frac{f(x+t)-f(x)}{t} [[/math]]
In a radial sense, if the following limit converges, for any [math]x\in X[/math], and [math]w\in\mathbb T[/math]:
[[math]] f_w'(x)=\lim_{t\in w\mathbb R\to0}\frac{f(x+t)-f(x)}{t} [[/math]]
In the complex sense, if the following limit converges, for any [math]x\in X[/math]:
[[math]] f'(x)=\lim_{t\to0}\frac{f(x+t)-f(x)}{t} [[/math]]

If [math]f[/math] is differentiable in the complex sense, we also say that [math]f[/math] is holomorphic.

We can see now more clearly what is going on. We have [math](3)\implies(2)\implies(1)[/math] in general, and most of the functions that we know, namely the polynomials, the rational functions, and [math]\sin,\cos,\exp,\log[/math], satisfy (3). As for the functions [math]\bar{x},|x|[/math], these do not satisfy (3), and do not satisfy (2) either, but they satisfy however (1). It is possible to say more about all this, and we will certainly come back to this topic, later in this book.

Back to business now, all the examples of holomorphic functions that we have are infinitely differentiable, and this raises the question of finding a function such that [math]f'[/math] exists, while [math]f''[/math] does not exist. Quite surprisingly, we will see that such functions do not exist. In order to get into this latter phenomenon, let us start with:

Theorem

Each power series [math]f(x)=\sum_nc_nx^n[/math] has a radius of convergence

[[math]] R\in[0,\infty] [[/math]]

which is such that [math]f[/math] converges for [math]|x| \lt R[/math], and diverges for [math]|x| \gt R[/math]. We have:

[[math]] R=\frac{1}{C}\quad,\quad C=\limsup_{n\to\infty}\sqrt[n]{|c_n|} [[/math]]

Also, in the case [math]|x|=R[/math] the function [math]f[/math] can either converge, or diverge.

Show Proof

This follows from the Cauchy criterion for series, from chapter 1, which says that a series [math]\sum_nx_n[/math] converges if [math]c \lt 1[/math], and diverges if [math]c \gt 1[/math], where:

[[math]] c=\limsup_{n\to\infty}\sqrt[n]{|x_n|} [[/math]]

Indeed, with [math]x_n=|c_nx^n|[/math] we obtain that the convergence radius [math]R\in[0,\infty][/math] exists, and is given by the formula in the statement. Finally, for the examples and counterexamples at the end, when [math]|x|=R[/math], the simplest here is to use [math]f(x)=\sum_nx^n[/math], where [math]R=1[/math].

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Back now to our questions regarding derivatives, we have:

Theorem

Assuming that a function [math]f:X\to\mathbb C[/math] is analytic, in the sense that it is a series, around each point [math]x\in X[/math],

[[math]] f(x+t)=\sum_{n=0}^\infty c_nt^n [[/math]]

it follows that [math]f[/math] is infinitely differentiable, in the complex sense. In particular, [math]f'[/math] exists, and so [math]f[/math] is holomorphic in our sense.

Show Proof

Assuming that [math]f[/math] is analytic, as in the statement, we have:

[[math]] f'(x+t)=\sum_{n=1}^\infty nc_nt^{n-1} [[/math]]

Moreover, the radius of convergence is the same, as shown by the following computation, using the Cauchy formula for the convergence radius, and [math]\sqrt[n]{n}\to1[/math]:

[[math]] \frac{1}{R'} =\limsup_{n\to\infty}\sqrt[n]{|nc_n|} =\limsup_{n\to\infty}\sqrt[n]{|c_n|} =\frac{1}{R} [[/math]]

Thus [math]f'[/math] exists and is analytic, on the same domain, and this gives the result.

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General references

Banica, Teo (2024). "Calculus and applications". arXiv:2401.00911 [math.CO].