6a. Functions, continuity

[math] \newcommand{\mathds}{\mathbb}[/math]

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We discuss in this chapter and in the next two ones the theory of complex functions [math]f:\mathbb C\to\mathbb C[/math], in analogy with the theory of the real functions [math]f:\mathbb R\to\mathbb R[/math]. We will see that many results that we know from the real setting extend to the complex setting, but there will be quite a number of new phenomena too. We will need, in order to get started:

Definition

The distance between two complex numbers is the usual distance in the plane between them, namely:

[[math]] d(x,y)=|x-y| [[/math]]

With this, we can talk about convergence, by saying that [math]x_n\to x[/math] when [math]d(x_n,x)\to 0[/math].

Here the fact that [math]d(x,y)=|x-y|[/math] is indeed the usual distance in the plane is clear for [math]y=0[/math], because we have [math]d(x,0)=|x|[/math], by definition of the modulus [math]|x|[/math]. As for the general case, [math]y\in\mathbb C[/math], this comes from the fact that the distance in the plane is given by:

[[math]] d(x,y)=d(x-y,0)=|x-y| [[/math]]

Observe that in real coordinates, the distance formula is quite complicated, namely:

[[math]] \begin{eqnarray*} d(a+ib,c+id) &=&|(a+ib)-(c+id)|\\ &=&|(a-c)+i(b-d)|\\ &=&\sqrt{(a-c)^2+(b-d)^2} \end{eqnarray*} [[/math]]

However, for most computations, we will not need this formula, and we can get away with the various tricks regarding complex numbers that we know. As a first result now, regarding [math]\mathbb C[/math] and its distance, that we will need in what follows, we have:

Proposition

The complex plane [math]\mathbb C[/math] is complete, in the sense that any Cauchy sequence converges.

Show Proof

Consider indeed a Cauchy sequence [math]\{x_n\}_{n\in\mathbb N}\subset\mathbb C[/math]. If we write [math]x_n=a_n+ib_n[/math] for any [math]n\in\mathbb N[/math], then we have the following estimates:

[[math]] |a_n-a_m|\leq\sqrt{(a_n-a_m)^2+(b_n-b_m)^2}=|x_n-x_m| [[/math]]

[[math]] |b_n-b_m|\leq\sqrt{(a_n-a_m)^2+(b_n-b_m)^2}=|x_n-x_m| [[/math]]

Thus both the sequences [math]\{a_n\}_{n\in\mathbb N}\subset\mathbb R[/math] and [math]\{b_n\}_{n\in\mathbb N}\subset\mathbb R[/math] are Cauchy, and since we know that [math]\mathbb R[/math] itself is complete, we can consider the limits of these sequences:

[[math]] a_n\to a\quad,\quad b_n\to b [[/math]]

With [math]x=a+ib[/math], our claim is that [math]x_n\to x[/math]. Indeed, we have:

[[math]] \begin{eqnarray*} |x_n-x| &=&\sqrt{(a_n-a)^2+(b_n-b)^2}\\ &\leq&|a_n-a|+|b_n-b| \end{eqnarray*} [[/math]]

It follows that we have [math]x_n\to x[/math], as claimed, and this gives the result.

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Talking complex functions now, we have the following definition:

Definition

A complex function [math]f:\mathbb C\to\mathbb C[/math], or more generally [math]f:X\to\mathbb C[/math], with [math]X\subset\mathbb C[/math] being a subset, is called continuous when, for any [math]x_n,x\in X[/math]:

[[math]] x_n\to x\implies f(x_n)\to f(x) [[/math]]

Also, we can talk about pointwise convergence of functions, [math]f_n\to f[/math], and about uniform convergence too, [math]f_n\to_uf[/math], exactly as for the real functions.

Observe that, since [math]x_n\to x[/math] in the complex sense means that [math](a_n,b_n)\to(a,b)[/math] in the usual, real plane sense, a function [math]f:\mathbb C\to\mathbb C[/math] is continuous precisely when it is continuous when regarded as real function, [math]f:\mathbb R^2\to\mathbb R^2[/math]. But more on this later in this book. At the level of examples now, we first have the polynomials, [math]P\in\mathbb C[X][/math]. We already met such polynomials in chapter 5, so let us recall from there that we have:

Theorem

Each polynomial [math]P\in\mathbb C[X][/math] can be regarded as a continuous function [math]P:\mathbb C\to\mathbb C[/math]. Moreover, we have the formula

[[math]] P(x)=a(x-r_1)\ldots(x-r_n) [[/math]]

with [math]a\in\mathbb C[/math], and with the numbers [math]r_1,\ldots,r_n\in\mathbb C[/math] being the roots of [math]P[/math].

Show Proof

This is something that we know from chapter 5, the idea being that one root can be always constructed, by reasoning by contradiction, and doing some analysis around the minimum of [math]|P|[/math], and then a recurrence on the degree [math]n\in\mathbb N[/math] does the rest.

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Next in line, we have the rational functions, which are defined as follows:

Theorem

The quotients of complex polynomials [math]f=P/Q[/math] are called rational funtions. When written in reduced form, with [math]P,Q[/math] prime to each other,

[[math]] f=\frac{P}{Q} [[/math]]

is well-defined and continuous outside the zeroes [math]P_f\subset\mathbb C[/math] of [math]Q[/math], called poles of [math]f[/math]:

[[math]] f:\mathbb C-P_f\to\mathbb C [[/math]]

In addition, the rational functions, regarded as algebraic expressions, are stable under summing, making products and taking inverses.

Show Proof

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

(1) First of all, we can surely talk about quotients of polynomials, [math]f=P/Q[/math], regarded as abstract algebraic expressions. Also, the last assertion is clear, because we can indeed perform sums, products, and take inverses, by using the following formulae:

[[math]] \frac{P}{Q}+\frac{R}{S}=\frac{PS+QR}{QS}\quad,\quad \frac{P}{Q}\cdot\frac{R}{S}=\frac{PR}{QS}\quad,\quad \left(\frac{P}{Q}\right)^{-1}=\frac{Q}{P} [[/math]]

(2) The question is now, given a rational function [math]f[/math], can we regard it as a complex function? In general, we cannot say that we have [math]f:\mathbb C\to\mathbb C[/math], for instance because [math]f(x)=x^{-1}[/math] is not defined at [math]x=0[/math]. More generally, assuming [math]f=P/Q[/math] with [math]P,Q\in\mathbb C[/math], we cannot talk about [math]f(x)[/math] when [math]x[/math] is a root of [math]Q[/math], unless of course we are in the special situation where [math]x[/math] is a root of [math]P[/math] too, and we can simplify the fraction.

(3) In view of this discussion, in order to solve our question, we must avoid the situation where the polynomials [math]P,Q[/math] have common roots. But this can be done by writing our rational function [math]f[/math] in reduced form, as follows, with [math]P,Q\in\mathbb C[X][/math] prime to each other:

[[math]] f=\frac{P}{Q} [[/math]]

(4) Now with this convention made, it is clear that [math]f[/math] is well-defined, and continuous too, outside of the zeroes of [math]f[/math]. Now since these zeroes can be obviously recovered from the knowledge of [math]f[/math] itself, as being the points where “[math]f[/math] explodes”, we can call them poles of [math]f[/math], and so we have a function [math]f:\mathbb C-P_f\to\mathbb C[/math], as in the statement.

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As a comment here, the term “pole” does not come from the Poles who invented this, but rather from the fact that, when trying to draw the graph of [math]f[/math], or rather imagine that graph, which takes place in [math]2+2=4[/math] real dimensions, we are faced with some sort of tent, which is suspended by infinite poles, which lie, guess where, at the poles of [math]f[/math].

Getting back now to Theorem 6.5, as stated, that is obviously a mixture of algebra and analysis. So, let us first further clarify the algebra part. We know that the rational functions are stable under summing, making products and taking inverses, and this makes the link with the following notion, from number theory and abstract algebra:

Definition

A field is a set [math]F[/math] with a sum operation [math]+[/math] and a product operation [math]\times[/math], subject to the following conditions:

• [math]a+b=b+a[/math], [math]a+(b+c)=(a+b)+c[/math], there exists [math]0\in F[/math] such that [math]a+0=0[/math], and any [math]a\in F[/math] has an inverse [math]-a\in F[/math], satisfying [math]a+(-a)=0[/math].
• [math]ab=ba[/math], [math]a(bc)=(ab)c[/math], there exists [math]1\in F[/math] such that [math]a1=a[/math], and any [math]a\neq0[/math] has a multiplicative inverse [math]a^{-1}\in F[/math], satisfying [math]aa^{-1}=1[/math].
• The sum and product are compatible via [math]a(b+c)=ab+ac[/math].

As basic examples of fields, we have the rational numbers [math]\mathbb Q[/math], the real numbers [math]\mathbb R[/math], and the complex numbers [math]\mathbb C[/math]. Some further examples of fields of numbers, which are more specialized, and useful in number theory, can be constructed as well. In view of this, it is useful to think of any field [math]F[/math] as being a “field of numbers”, and this because the elements [math]a,b,c,\ldots\in F[/math] behave under the operations [math]+[/math] and [math]\times[/math] exactly as the usual numbers do.

In what regards the various spaces of functions, such as the polynomials [math]\mathbb C[X][/math], or the continuous functions [math]C(\mathbb R)[/math], these certainly have sum and product operations [math]+[/math] and [math]\times[/math], but are in general not fields, because they do not satisfy the following field axiom:

[[math]] f\neq 0\implies\exists f^{-1} [[/math]]

However, and here comes our point, Theorem 6.5 tells us that the rational functions form a field. This is quite interesting, and opposite to the general spirit of analysis and function spaces, which are in general not fields. Let us record this finding, as follows:

Definition

We denote by [math]\mathbb C(X)[/math] the field of rational functions

[[math]] f=\frac{P}{Q}\quad,\quad P,Q\in\mathbb C[X] [[/math]]

with the usual sum and product operations [math]+[/math] and [math]\times[/math] for the rational functions.

To be more precise, this is some sort of reformulation of Theorem 6.5, or rather of the algebraic content of Theorem 6.5, telling us that the rational functions form indeed a field. And to the question, how can a theorem suddenly become a definition, the answer is that this is quite commonplace in mathematics, and especially in algebra.

Back now to analysis, let us point out that, contrary to what the above might suggest, everything does not always extend trivally from the real to the complex case. For instance, we have the following result, that we already talked about a bit in chapter 5:

Proposition

We have the following formula, valid for any [math]|x| \lt 1[/math],

[[math]] \frac{1}{1-x}=1+x+x^2+\ldots [[/math]]

but, for [math]x\in\mathbb C-\mathbb R[/math], the geometric meaning of this formula is quite unclear.

Show Proof

Here the formula in the statement holds indeed, by multiplying and cancelling terms, exactly as in the real case, with the convergence being justified by:

[[math]] \left|\sum_{n=0}^\infty x^n\right|\leq\sum_{n=0}^\infty|x|^n=\frac{1}{1-|x|} [[/math]]

As for the last assertion, this is something rather informal, which hides however many interesting things, that we discussed in some detail in chapter 5.

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Getting now to more complicated functions, such as [math]\sin[/math], [math]\cos[/math], [math]\exp[/math], [math]\log[/math], again many things extend well from real to complex, the basic theory here being as follows:

Theorem

The functions [math]\sin,\cos,\exp,\log[/math] have complex extensions, given by

[[math]] \sin x=\sum_{l=0}^\infty(-1)^l\frac{x^{2l+1}}{(2l+1)!}\quad,\quad \cos x=\sum_{l=0}^\infty(-1)^l\frac{x^{2l}}{(2l)!} [[/math]]

[[math]] e^x=\sum_{k=0}^\infty\frac{x^k}{k!}\quad,\quad \log(1+x)=\sum_{k=1}^\infty(-1)^{k+1}\frac{x^k}{k} [[/math]]

with [math]|x| \lt 1[/math] needed for [math]\log[/math], which are continuous over their domain, and satisfy the formulae [math]e^{x+y}=e^xe^y[/math] and [math]e^{ix}=\cos x+i\sin x[/math].

Show Proof

This is a mixture of trivial and non-trivial results, as follows:

(1) We already know about [math]e^x[/math] from chapter 5, the idea being that the convergence of the series, and then the continuity of [math]e^x[/math], come from the following estimate:

[[math]] |e^x|\leq\sum_{k=0}^\infty\frac{|x|^k}{k!}=e^{|x|} \lt \infty [[/math]]

(2) Regarding [math]\sin x[/math], the same method works, with the following estimate:

[[math]] |\sin x|\leq\sum_{l=0}^\infty\frac{|x|^{2l+1}}{(2l+1)!}\leq\sum_{k=0}^\infty\frac{|x|^k}{k!}=e^{|x|} [[/math]]

(3) The same goes for [math]\cos x[/math], the estimate here being as follows:

[[math]] |\cos x|\leq\sum_{l=0}^\infty\frac{|x|^{2l}}{(2l)!}\leq\sum_{k=0}^\infty\frac{|x|^k}{k!}=e^{|x|} [[/math]]

(4) Regarding now the formulae satisfied by [math]\sin,\cos,\exp[/math], we already know from chapter 5 that the exponential has the following property, exactly as in the real case:

[[math]] e^{x+y}=e^xe^y [[/math]]

We also have the following formula, connecting [math]\sin,\cos,\exp[/math], again as before:

[[math]] \begin{eqnarray*} e^{ix} &=&\sum_{k=0}^\infty\frac{(ix)^k}{k!}\\ &=&\sum_{k=2l}\frac{(ix)^k}{k!}+\sum_{k=2l+1}\frac{(ix)^k}{k!}\\ &=&\sum_{l=0}^\infty(-1)^l\frac{x^{2l}}{(2l)!}+i\sum_{l=0}^\infty(-1)^l\frac{x^{2l+1}}{(2l+1)!}\\ &=&\cos x+i\sin x \end{eqnarray*} [[/math]]

(5) In order to discuss now the complex logarithm function [math]\log[/math], let us first study some more the complex exponential function [math]\exp[/math]. By using [math]e^{x+y}=e^xe^y[/math] we obtain [math]e^x\neq0[/math] for any [math]x\in\mathbb C[/math], so the complex exponential function is as follows:

[[math]] \exp:\mathbb C\to\mathbb C-\{0\} [[/math]]

Now since we have [math]e^{x+iy}=e^xe^{iy}[/math] for [math]x,y\in\mathbb R[/math], with [math]e^x[/math] being surjective onto [math](0,\infty)[/math], and with [math]e^{iy}[/math] being surjective onto the unit circle [math]\mathbb T[/math], we deduce that [math]\exp:\mathbb C\to\mathbb C-\{0\}[/math] is surjective. Also, again by using [math]e^{x+iy}=e^xe^{iy}[/math], we deduce that we have:

[[math]] e^x=e^y\iff x-y\in 2\pi i\mathbb Z [[/math]]

(6) With these ingredients in hand, we can now talk about [math]\log[/math]. Indeed, let us fix a horizontal strip in the complex plane, having width [math]2\pi[/math]:

[[math]] S=\left\{x+iy\Big|x\in\mathbb R,y\in[a,a+2\pi)\right\} [[/math]]

We know from the above that the restriction map [math]\exp:S\to\mathbb C-\{0\}[/math] is bijective, so we can define [math]\log[/math] as to be the inverse of this map:

[[math]] \log=\exp^{-1}:\mathbb C-\{0\}\to S [[/math]]

(7) In practice now, the best is to choose for instance [math]a=0[/math], or [math]a=-\pi[/math], as to have the whole real line included in our strip, [math]\mathbb R\subset S[/math]. In this case on [math]\mathbb R_+[/math] we recover the usual logarithm, while on [math]\mathbb R_-[/math] we obtain complex values, as for instance [math]\log(-1)=\pi i[/math] in the case [math]a=0[/math], or [math]\log(-1)=-\pi i[/math] in the case [math]a=-\pi[/math], coming from [math]e^{\pi i}=-1[/math].

(8) Finally, assuming [math]|x| \lt 1[/math], we can consider the following series, which converges:

[[math]] f(x)=\sum_{k=1}^\infty(-1)^{k+1}\frac{x^k}{k} [[/math]]

We have then [math]e^{f(x)}=1+x[/math], and so [math]f(x)=\log(1+x)[/math], when [math]1+x\in S[/math].

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As an interesting consequence of the above result, which is of great practical interest, we have the following useful method, for remembering the basic math formulae: \begin{method}\ Knowing [math]e^x=\sum_kx^k/k![/math] and [math]e^{ix}=\cos x+i\sin x[/math] gives you

[[math]] \sin(x+y)=\sin x\cos y+\cos x\sin y [[/math]]

[[math]] \cos(x+y)=\cos x\cos y-\sin x\sin y [[/math]]

right away, in case you forgot these formulae, as well as

[[math]] \sin x=\sum_{l=0}^\infty(-1)^l\frac{x^{2l+1}}{(2l+1)!}\quad,\quad \cos x=\sum_{l=0}^\infty(-1)^l\frac{x^{2l}}{(2l)!} [[/math]]

again, right away, in case you forgot these formulae. \end{method} To be more precise, assume that we forgot everything trigonometry, which is something that can happen to everyone, in the real life, but still know the formulae [math]e^x=\sum_kx^k/k![/math] and [math]e^{ix}=\cos x+i\sin x[/math]. Then, we can recover the formulae for sums, as follows:

[[math]] \begin{eqnarray*} e^{i(x+y)}=e^{ix}e^{iy} &\implies&\cos(x+y)+i\sin(x+y)=(\cos x+i\sin x)(\cos y+i\sin y)\\ &\implies&\begin{cases} \cos(x+y)=\cos x\cos y-\sin x\sin y\\ \sin(x+y)=\sin x\cos y+\cos x\sin y \end{cases} \end{eqnarray*} [[/math]]

And isn't this smart. Also, and even more impressively, we can recover the Taylor formulae for [math]\sin,\cos[/math], which are certainly difficult to memorize, as follows:

[[math]] \begin{eqnarray*} e^{ix}=\sum_k\frac{(ix)^k}{k!} &\implies&\cos x+i\sin x=\sum_k\frac{(ix)^k}{k!}\\ &\implies&\begin{cases} \cos x=\sum_{l=0}^\infty(-1)^l\frac{x^{2l}}{(2l)!}\\ \sin x=\sum_{l=0}^\infty(-1)^l\frac{x^{2l+1}}{(2l+1)!} \end{cases} \end{eqnarray*} [[/math]]

Finally, in what regards [math]\log[/math], there is a trick here too, which is partial, namely:

[[math]] \begin{eqnarray*} \log(\exp x)=x &\implies&\log\left(1+x+\frac{x^2}{2}+\ldots\right)=x\\ &\implies&\log(1+y)=y-\frac{y^2}{2}+\ldots \end{eqnarray*} [[/math]]

To be more precise, [math]\log(1+y)\simeq y[/math] is clear, and with a bit more work, that we will leave here as an instructive exercise, you can recover [math]\log(1+y)=y-y^2/2[/math] too. Of course, the higher terms can be recovered too, with enough work involved, at each step.

Moving ahead, Theorem 6.9 leads us into the question on whether the other formulae that we know about [math]\sin,\cos[/math], such as the values of these functions on sums [math]x+y[/math], or on doubles [math]2x[/math], extend to the complex setting. Things are quite tricky here, and in relation with this, we have the following result, which is something of general interest:

Proposition

The following functions, called hyperbolic sine and cosine,

[[math]] \sinh x=\frac{e^x-e^{-x}}{2} \quad,\quad \cosh x=\frac{e^x+e^{-x}}{2} [[/math]]

are subject to the following formulae:

• [math]e^x=\cosh x+\sinh x[/math].
• [math]\sinh(ix)=i\sin x[/math], [math]\cosh(ix)=\cos x[/math], for [math]x\in\mathbb R[/math].
• [math]\sinh(x+y)=\sinh x\cosh y+\cosh x\sinh y[/math].
• [math]\cosh(x+y)=\cosh x\cosh y+\sinh x\sinh y[/math].
• [math]\sinh x=\sum_l\frac{x^{2l+1}}{(2l+1)!}[/math], [math]\cosh x=\sum_l\frac{x^{2l}}{(2l)!}[/math].

Show Proof

The formula (1) follows from definitions. As for (2), this follows from:

[[math]] \sinh(ix)=\frac{e^{ix}-e^{-ix}}{2}=\frac{\cos x+i\sin x}{2}-\frac{\cos x-i\sin x}{2}=i\sin x [[/math]]

[[math]] \cosh(ix)=\frac{e^{ix}+e^{-ix}}{2}=\frac{\cos x+i\sin x}{2}+\frac{\cos x-i\sin x}{2}=\cos x [[/math]]

Regarding now (3,4), observe first that the formula [math]e^{x+y}=e^x+e^y[/math] reads:

[[math]] \cosh(x+y)+\sinh(x+y)=(\cosh x+\sinh x)(\cosh y+\sinh y) [[/math]]

Thus, we have some good explanation for (3,4), and in practice, these formulae can be checked by direct computation, as follows:

[[math]] \frac{e^{x+y}-e^{-x-y}}{2}= \frac{e^x-e^{-x}}{2}\cdot\frac{e^y+e^{-y}}{2}+ \frac{e^x+e^{-x}}{2}\cdot\frac{e^y-e^{-y}}{2} [[/math]]

[[math]] \frac{e^{x+y}+e^{-x-y}}{2}= \frac{e^x+e^{-x}}{2}\cdot\frac{e^y+e^{-y}}{2}+ \frac{e^x-e^{-x}}{2}\cdot\frac{e^y-e^{-y}}{2} [[/math]]

Finally, (5) is clear from the definition of [math]\sinh[/math], [math]\cosh[/math], and from [math]e^x=\sum_k\frac{x^k}{k!}[/math].

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Finally, we can talk as well about powers, in the following way:

\begin{fact} Under suitable assumptions, we can talk about [math]x^y[/math] with [math]x,y\in\mathbb C[/math], and in particular about the complex functions [math]a^x[/math] and [math]x^a[/math], with [math]a\in\mathbb C[/math]. \end{fact} To be more precise, in what regards [math]x^y[/math], we already know from chapters 1-2 that things are quite tricky, even in the real case. In the complex case the same problems appear, along with some more, but these questions can be solved by using the above theory of [math]\exp,\log[/math]. To be more precise, in order to solve the first question, we can set:

[[math]] x^y=e^{y\log x} [[/math]]

We will be back to these functions later, when we will have more tools for studying them. In fact, all of a sudden, we are now into quite complicated mathematics, and we cannot really deal with the problems left open above, with bare hands. More later.

At the level of the general theory now, the main tool for dealing with the continuous functions [math]f:\mathbb R\to\mathbb R[/math] was the intermediate value theorem. In the complex setting, that of the functions [math]f:\mathbb C\to\mathbb C[/math], we do not have such a theorem, at least in its basic formulation, because there is no order relation for the complex numbers, or things like complex intervals. However, the intermediate value theorem in its advanced formulation, that with connected sets, extends of course, and we have the following result:

Theorem

Assuming that [math]f:X\to\mathbb C[/math] with [math]X\subset\mathbb C[/math] is continuous, if the domain [math]X[/math] is connected, then so is its image [math]f(X)[/math].

Show Proof

This follows exactly as in the real case, with just a bit of discussion being needed, in relation with open and closed sets, and then connected sets, inside [math]\mathbb C[/math].

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General references

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