My new book, Complex Analysis: An Introduction, is nearly finished. To help my students with revision I created a list of common mistakes and this forms a chapter in the book. As a lecturer with many years of experience of teaching the subject I have seen these mistakes appear again and again in examinations. I’m sure that, due to pressure, we’ve all written nonsense in an exam which under normal conditions we wouldn’t have. Nonetheless, many of these errors occur every year and I suspect something deeper is going on. What follows is not intended to be a criticism of my students, who, luckily for me, are generally hard-working and intelligent. Nor is it an attempt to mock or ridicule them. Instead the aim is to identify common mistakes so that they are not made in the future. And if this post seems negative in tone, the a later one is more positive as it delves into techniques that improve understanding. Imaginary numbers cannot be compared The first mistake is the probably the most common: the comparison of imaginary numbers. For example, students write for a complex number. This cannot be right. If were what does it mean for to be less than ? What is usually intended is the modulus of , i.e., . The point is, unlike real numbers, we cannot order the complex numbers. For example, which is bigger or ? This is difficult to decide! Since complex numbers can be identified with the plane ordering them is equivalent to ordering the points of the plane and clearly this can’t be done — at least not in any useful or meaningful way. One last point needs to be made. Although is incorrect, note that expressions like can be true if is...

## What is the Best Proof of Cauchy’s Integral Theorem?...

posted by Kevin Houston

My book on Complex Analysis is now available! You can find it at xtothepowerofn.com. The material below is there along with other sample chapters on Common Mistakes and on Improving Understanding. Today’s post may look as though I’m going all Terry Tao on you with a long post with lots of mathematical symbols. It’s really about the learning and teaching of Cauchy’s integral theorem from undergraduate complex analysis, so isn’t for everyone. If it’s not your cup of tea/coffee, then pop over here for some entertainment. Cauchy’s Integral Theorem Cauchy’s Integral Theorem is one of the greatest theorems in mathematics. There are many ways of stating it. Here’s just one: Cauchy’s Integral Theorem: Let be a domain, and be a differentiable complex function. Let be a closed contour such that and its interior points are in . Then, . Here, contour means a piecewise smooth map . In my years lecturing Complex Analysis I have been searching for a good version and proof of the theorem. My definition of good is that the statement and proof should be short, clear and as applicable as possible so that I can maintain rigour when proving Cauchy’s Integral Formula and the major applications of complex analysis such as evaluating definite integrals. Many of the proofs in the literature are rather complicated and so time is lost in lectures proving lemmas that that are never needed again. Here’s a version which I think has a good balance between simplicity and applicability. I’ve highlighted the difference with the version above. Cauchy’s Integral Theorem (Simple version): Let be a domain, and be a differentiable complex function. Let be a simple closed contour made of a finite number of lines and arcs such that and its interior points are in . Then, . Here an important point is that the curve is simple, i.e., is injective except at the start and end points. This means that we have a Jordan curve and so the curve has well-defined interior and exterior and both are connected sets. With this version I believe one can prove all the major theorems in an introductory course. I would be interested to hear from anyone who knows a simpler proof or has some thoughts on this one. Proof of Simple Version of Cauchy’s Integral Theorem Let denote the interior of , i.e., points with non-zero winding number and for any contour let denote its image. First we need a lemma. Lemma Let be a simple closed contour made of a finite number of lines and arcs in the domain with . Let be a square in bounding and be analytic. Then for any there exists a subdivision of into a grid of squares so that for each square in the grid with there exists a such that for all Proof of Lemma The set up looks like the following. For a contradiction we will assume the statement is false. Let and divide into 4 equal-sized squares. At least one of these squares will not satisfy the required condition in the lemma. Let be such a square. Repeat the process to produce an infinite sequence of squares with .By the Nested Squares Lemma (which is just a generalization of the Nested Interval Theorem) there exists . As is differentiable there exists such that for . But as the size of the squares becomes arbitrarily small there must exist such that is contained in the disk . This is a contradiction. Main part of proof Given there exists a grid of squares covering . Let be the set of squares such that and let be the set of distinguished points in the lemma. Define by Then as is differentiable, is continuous (and hence integrable). Without loss of generality we can assume that is positively oriented. Let be the union of positively oriented contours giving the boundary of . Since is made of a finite...

## Cool proof of the Fundamental Theorem of Algebra...

posted by Kevin Houston

Despite being on holiday I can’t resist looking for cool proofs. This one is not so much cool as interesting in a Why-didn’t-I-think-of-that way. The fundamental theorem of algebra — that any polynomial has a complex root — is well known to be a theorem of analysis rather than algebra and many proofs are known. The proof I use in my course relies on Liouville’s Theorem and the Maximum Modulus Theorem. However, there is a more direct route using only Cauchy’s Integral Formula and the Estimation Lemma. I found the basic idea in Complex Proofs of Real Theorems by Peter Lax and Lawrence Zalcman. Here’s my modified version: Every polynomial , , , has a root in . The proof is as follows. Suppose not and derive a contradiction. Since for all the function defined by is differentiable on all of . Now, for , So there exists an such that implies that . Thus for such . As is differentiable on all of we have for all , by Cauchy’s Integral Formula, (and where denotes a circle of radius centred at the origin), The right hand side can be made as small as we like by taking large enough. Thus . This is impossible so we have a contradiction and therefore has a...