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Publisher:

Mathematical Association of America

Publication Date:

2010

Number of Pages:

327

Format:

Hardcover

Edition:

2

Series:

MAA Textbooks

Price:

63.95

ISBN:

9780883857649

Category:

Textbook

The Basic Library List Committee recommends this book for acquisition by undergraduate mathematics libraries.

[Reviewed by , on ]

Warren Johnson

01/13/2011

I have twice taught from Joseph Bak and Donald Newman’s *Complex Analysis*, which was recently reviewed here. When I taught the course again last year, I decided to try something else. After some searching I settled on John Howie’s *Complex Analysis*, because it was in the “Yellow Sale” and it had a positive review here. As I told a publisher’s representative at the time, I am still looking for a complex analysis textbook that I can fall in love with. I think the book under review is very good, better than either Bak and Newman or Howie, but it still isn’t quite my ideal. (It is also at a slightly higher level than Bak and Newman or Howie, with an advanced calculus prerequisite, so I’m not sure I could use it for our course.)

Few people did more for mathematics in the 20^{th} century than the author of this book. Besides outstanding research and exposition, he was executive editor of *Mathematical Reviews* in its early years, and later President of the MAA and editor of *The American* *Mathematical Monthly*. One can get a sense of what he was like from the entertaining collection *Lion Hunting & Other Mathematical Pursuits*. I never met him, but my adviser wrote two papers with him, and my father took complex analysis from him in the early 1950s; he told me it was the best class he had at Northwestern.

*Invitation to Complex Analysis *originally appeared in 1987, five years before the author’s death. This second edition contains no new material, but many minor emendations by the author’s son Harold, himself an excellent mathematician and expositor. (In what follows, “Boas” refers to Ralph, not to Harold.) The book has 5 chapters, comprising 35 sections, 141 subsections, 278 Exercises with solutions, and 208 Supplementary exercises without solutions in 230 pages, followed by 75 pages of solutions and a short bibliography. It has more than enough material for one semester, but it might not stretch to a full year without a little supplementing.

Chapter 1 has the basics. It tries (very much to my own taste) to get to Cauchy’s theorem as quickly as possible without neglecting power series. Chapter 2, Applications of Cauchy’s Theorem, is the longest of the five, and it contains 109 of the 208 Supplementary exercises. Chapter 3 is by far the shortest, Analytic Continuation in 16 pages. Chapter 4 discusses Harmonic Functions and Conformal Mapping, and chapter 5 has some Miscellaneous Topics, including Stirling’s formula (proved only for n!, but the extension to the gamma function is stated). The last section is an introduction to the beautiful topic of univalent functions; it includes the Bieberbach conjecture (now de Branges’s theorem) and the proof of a significant special case.

Another miscellaneous topic, Phragmén-Lindelöf Theorems (section 32), is touted by the back cover as a distinctive feature of the book, as indeed it is. The back cover lists two more notions by name as uncommon and intriguing. One is Landau’s o and O notation, which occurs already in section 2B, and in this case “useful” seems to me a more appropriate adjective. (It is also a curious choice in that Boas makes a virtue in his Preface of having avoided special symbols and abbreviations.) “Uncommon” fairly describes the third notion, Overconvergence (section 17D), though it can also be found in Titchmarsh’s classic *The Theory of Functions*. However, I think most readers will find it less intriguing than most of the other topics in the book, and many instructors will be inclined to skip it.

My own candidates for the book’s most distinguishing features are the excellent chapter 4, a treatment of residue calculus well above the industry average, and the Exercises and solutions. Since the Supplementary exercises without solutions are both less numerous and less interesting than the solved Exercises, the book is perhaps more suitable for self study than as a textbook. Nevertheless, I will give it serious consideration the next time I teach the class.

In the remainder of this review I will discuss, out of all proportion to their importance to the book, two things in it that bothered me. Boas gives as Exercise 18.1 what he calls Abel’s convergence theorem, and in a footnote remarks that unfortunately most calculus books never mention it, even though it is one of the most useful convergence tests for series. The result in question is more often called Dirichlet’s test, e.g. in Hardy’s *A Course of Pure Mathematics , *p. 379, or Bromwich’s

Recall that the alternating series test says that if the sequence a_{n} decreases to zero, then the series ∑(–1)^{n} a_{n} converges. (Speaking of things that many calculus books don’t mention, if a_{n} approaches zero from above but doesn’t consistently decrease, then ∑(–1)^{n} a_{n} might or might not converge. It is easy to construct examples by subtracting the even terms of one p-series from the odd terms of another.) Dirichlet’s test says that (–1)^{n} can be replaced by any sequence b_{n} whose partial sums are bounded. Abel’s test says that if ∑b_{n} actually converges, then a_{n} need not decrease to zero; it just has to be positive and decreasing. The standard proofs, involving the Cauchy criterion and summation by parts, are fine for an analysis course but rather more intricate than I would want to do in calculus, and I don’t think, based on my reading of *Lion Hunting &* *Other Mathematical Pursuits* (in particular the essay “Calculus as an Experimental Science,”* *which was also a Vice-Presidential Address of the American Association for the Advancement of Science), that Boas would advocate *proving *these theorems there. Dirichlet’s test is great for discussing the convergence of series like ∑(sin nx)/n, on which the standard convergence tests from calculus fall flat.

I have not figured out why Boas names Dirichlet’s test after Abel instead. Perhaps it is just that the lemma on which they both rely is typically named after Abel. Abel’s test is the third theorem in section 2 of his famous paper of 1826 on the binomial series, while Dirichlet’s test is proved in section 101 of his great *Vorlesungen über Zahlentheorie *(available since 1999 in a fine English translation by John Stillwell), in the particular case where a

In a footnote to one of his papers on series in *Mathematische Annalen* (vol. 25, p. 423), Pringsheim says that surely Abel must have known Dirichlet’s test. While I am inclined to agree, I think the name is worth preserving. It may also be worth mentioning the discussion, in Hadamard’s beautiful little book *The Mathematician’s Mind *(originally

I am more troubled by Boas’s treatment of Bernoulli numbers. He devotes the two pages of section 15C to them, which is only enough space for one nontrivial result beyond the definition, and the one he chooses is questionable. Bernoulli numbers first arose in an algorithm of Jakob Bernoulli to find formulas for the sums of the first n k^{th} powers. It was left to Euler to prove that Bernoulli’s method really worked. The most famous application, also due to Euler, is to the evaluation of the Riemann zeta function at the positive even integers. This is not only one of the great theorems of mathematics but also one of the great historical ironies, in that Bernoulli had publicly lamented his failure to evaluate ∑1/n^{2}.

The n^{th} Bernoulli number B_{n} is most conveniently defined as the coefficient of x^{n}/n! in the Taylor series expansion of x/(e^{x} –1). Note that the reciprocal of this generating function has a simple series expansion, so that multiplying the two series together we get a nice recursive formula for the Bernoulli numbers. If we add x/2 to the generating function we get (x/2) coth(x/2), which is an even function. It follows that B_{1}=–1/2 and that the other odd order Bernoulli numbers are all zero. The nonzero Bernoulli numbers alternate in sign. The first several are not too bad, but in general B_{n} is a complicated fraction; for example B_{18}=43867/798. The denominators can be predicted by the von Staudt-Clausen theorem. For example, the denominator of B_{18} is 798 because the factors of 18 are 1,2,3,6,9,18; increasing each of these by one we have 2,3,4,7,10,19, of which 2,3,7,19 are prime, and the product of these four numbers is 798.

By letting x=2iz we now have an expansion of z cot z in terms of Bernoulli numbers. One can then use the trig identities csc z = cot(z/2) – cot z and tan z = cot z – 2 cot(2z) to obtain expansions of csc z and tan z, and it is the latter that Boas chooses to present (by a slightly different route than the one outlined here). Perhaps this is because, strictly speaking, tan z is the only trig function that has a nice Taylor series expansion in terms of Bernoulli numbers, as the expansions of cot z and csc z include the term 1/z.

There is no similar trick for expanding sec z because it is an even function, whereas tan z, cot z, and csc z are all odd, but if one were determined to expand sec z in terms of Bernoulli numbers, one could multiply the series for csc z by the one for tan z. As Euler already realized, it is better to start from scratch: let E_{2n} denote the coefficient of z^{2n}/(2n)! in the Taylor series expansion of sec z. Still better, following Scherk, is to let E_{n} be the coefficient of z^{n}/n! in the Taylor series expansion of sec z + tan z, so that E_{2n-1} is the coefficient of z^{2n-1}/(2n-1)! in the Taylor series expansion of tan z, which makes E_{2n-1} a multiple of B_{2n}.

The problem with the result Boas presents is that, looking at it, one would never guess that the coefficients of z^{2n-1}/(2n-1)! in the Taylor series expansion of tan z are actually positive integers (although one knows this *a priori *since they are derivatives of tan z at z=0). Indeed, E_{n} is the number of up-down permutations (alternately rising and falling, starting with a rise) of {1,2,…,n}. For example, the up-down permutations of {1,2,3,4} are 1324, 1423, 2314, 2413, and 3412, so E_{4}=5. Since the derivative of sec z + tan z is sec z (sec z + tan z), there is a nice recursive formula for E_{n}, which affords a simple induction proof of this combinatorial interpretation. The expansion of tan z in terms of Bernoulli numbers is not completely without interest, since one can deduce number theoretic properties of B_{n} by comparing it with the good expansion, but by itself it is unilluminating.

If one is going to mention Bernoulli numbers at all in a serious course on complex analysis, then I think one has to do the application to ∑1/n^{2k}. This requires Euler’s partial fractions expansion of the cotangent, which is itself a nice application of residues and Liouville’s theorem. Then we just have to expand the partial fractions into geometric series, interchange orders of summation, and compare with the power series for z cot z. An instructor might choose not to cover this, but a mathematician of Boas’s class ought to have included it.

Warren Johnson (warren.johnson@conncoll.edu) is visiting assistant professor of mathematics at Connecticut College.

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