Demystifying complex numbers

Question

At the end of this month I start teaching complex analysis to 2nd year undergraduates, mostly from engineering but some from science and maths. The main applications for them in future studies are contour integrals and Laplace transform, but this should be a "real" complex analysis course which I could later refer to in honours courses. I am now confident (after this discussion, especially Gauss’s complaints given in Keith’s comment) that the name "complex" is discouraging to average students.

Why do we need to study numbers which do not belong to the real world?

We all know that the thesis is wrong and I have in mind some examples where the use of complex variable functions simplify solving considerably (I give two below). The drawback is that all them assume some knowledge from students already.

So I would be happy to learn elementary examples which may convince students that complex numbers and functions of a complex variable are useful. As this question runs in the community wiki mode, I would be glad to see one example per answer.

Thank you in advance!

Here are the two promised examples. I was reminded of the second one by several answers and comments about trigonometric functions (and also by the notification that "the bounty on your question Trigonometry related to Rogers–Ramanujan identities expires within three days"; it seems to be harder than I expected).

Example 1. What is the Fourier expansion of the (unbounded) periodic function $$ f(x)=\ln\Bigl\lvert\sin\frac x2\Bigr\rvert\ ? $$

Solution. The function $f(x)$ is periodic with period $2\pi$ and has poles at the points $2\pi k$, $k\in\mathbb Z$.

Consider the function on the interval $x\in[\varepsilon,2\pi-\varepsilon]$. The series $$ \sum_{n=1}^\infty\frac{z^n}n, \qquad z=e^{ix}, $$ converges for all values $x$ from the interval. Since $$ \Bigl\lvert\sin\frac x2\Bigr\rvert=\sqrt{\frac{1-\cos x}2} $$ and $\operatorname{Re}\ln w=\ln\lvert w\rvert$, where we choose $w=\frac12(1-z)$, we deduce that $$ \operatorname{Re}\Bigl(\ln\frac{1-z}2\Bigr)=\ln\sqrt{\frac{1-\cos x}2} =\ln\Bigl\lvert\sin\frac x2\Bigr\rvert. $$ Thus, $$ \ln\Bigl\lvert\sin\frac x2\Bigr\rvert =-\ln2-\operatorname{Re}\sum_{n=1}^\infty\frac{z^n}n =-\ln2-\sum_{n=1}^\infty\frac{\cos nx}n. $$ As $\varepsilon>0$ can be taken arbitrarily small, the result remains valid for all $x\ne2\pi k$.

Example 2. Let $p$ be an odd prime number. $\newcommand\Legendre{\genfrac(){}{}}$For an integer $a$ relatively prime to $p$, the Legendre symbol $\Legendre ap$ is $+1$ or $-1$ depending on whether the congruence $x^2\equiv a\pmod{p}$ is solvable or not. Using the elementary result (a consequence of Fermat's little theorem) that $$ \Legendre ap \equiv a^{(p-1)/2}\pmod p, \tag{*}\label{star} $$ show that $$ \Legendre 2p=(-1)^{(p^2-1)/8}. $$

Solution. In the ring $\mathbb Z+\mathbb Zi=\Bbb Z[i]$, the binomial formula implies $$ (1+i)^p\equiv1+i^p\pmod p. $$ On the other hand, $$ (1+i)^p =\bigl(\sqrt2e^{\pi i/4}\bigr)^p =2^{p/2}\biggl(\cos\frac{\pi p}4+i\sin\frac{\pi p}4\biggr) $$ and $$ 1+i^p =1+(e^{\pi i/2})^p =1+\cos\frac{\pi p}2+i\sin\frac{\pi p}2 =1+i\sin\frac{\pi p}2. $$ Comparing the real parts implies that $$ 2^{p/2}\cos\frac{\pi p}4\equiv1\pmod p, $$ hence from $\sqrt2\cos(\pi p/4)\in\{\pm1\}$ we conclude that $$ 2^{(p-1)/2}\equiv\sqrt2\cos\frac{\pi p}4\pmod p. $$ Then using the elementary result \eqref{star}: $$ \Legendre2p \equiv2^{(p-1)/2} \equiv\sqrt2\cos\frac{\pi p}4 =\begin{cases} 1 & \text{if } p\equiv\pm1\pmod8, \cr -1 & \text{if } p\equiv\pm3\pmod8, \end{cases} $$ which is exactly the required formula.

Answer 1

"Why do we need to study numbers which do not belong to the real world?"

Having been through the relevant mathematical mill, I subsequently engaged with Geometric Algebra (a Clifford Algebra interpreted strictly geometrically).

Once I understood that the square of a unit bivector is -1 and then how rotors worked, all my (conceptual) difficulties evaporated.

I have never had a reason to use (pure) complex numbers since and I suspect that most engineering/physics/computing types would avoid them if they were able.

Likely you have the above group mixed together with pure mathematicians that feel quite at home with the non-physical aspects of complex numbers and wouldn't dream of asking such an impertinent question:-)

Answer 2

An interesting example of usage of complex numbers can be found in http://arxiv.org/abs/math/0001097 (Michael Eastwood, Roger Penrose, Drawing with Complex Numbers).

Answer 3

I noticed this old question because it got bumped recently, and am surprised that the original, historical motivation for complex numbers—namely, a formula for solving a cubic equation—does not seem to be mentioned in any of the answers. There is now a lovely Veritasium video that tells the story in a way that I think would appeal to any second-year undergraduate student.

Complex numbers are not well-motivated by the quadratic equation, because their appearance just indicates the absence of (real) solutions, so you can dismiss them. But when you try to find a formula for the cubic equation in terms of radicals, complex numbers unavoidably force themselves upon you even when the solutions are real.

Answer 4

It seems that many of the answers fall into two categories: those that are applications of Euler's formula $e^{i\theta}=\cos\theta + i\sin\theta$, and those that are related to the fundamental theorem of algebra.

To supply one other instance of the latter, I'd go with Bézout's theorem from algebraic geometry (allegedly one of Gauss's earlier proofs of the FTA was by way of Bézout's, which smells like circular reasoning).

The students may remember conics, or at least circles. How many points can a circle intersect with a line? 0, 1, or 2. What about an ellipse with a line? Same. Two ellipses? Up to 4. How come two circles never intersect in 3 or 4 points. Well it may require a little bit of homogeneous coordinates, but we may cheat by saying that the points $(\infty, i\infty)$ and $(\infty, -i\infty)$ are on every circle. It may help to compare with the situation of two hyperbolas whose asymptotes have slopes $\pm 1$.

Answer 5

Here's a visual thing I handed out to students in a much more elementary class than the one the question mentions:
http://web.archive.org/web/20130701140646/http://www.math.umn.edu/~hardy/1031/handouts/March.3.pdf

Answer 6

Since the audience for such a class must be to a large degree E.E.s and M.E.s, I think they would appreciate a Reader's Digest version of the paper "Steinmetz and the Concept of Phasor: A Forgotten Story" by A. E. A. Araújo & D. A. V. Tonidandel.

For the human interest aspect of the story, there is much material from which to choose some vignettes. Biographies (videos (see comment), papers, books) of the two main characters in the paper—Steinmetz and Heaviside—contain many very entertaining technically-related tales of the two eccentrics, and some background on their contributions to the development of signal transmission via cables and the ether in the case of Heaviside and of the electric motor in the case of Steinmetz would accentuate the importance of their ideas, along with an overall portrait of Electric City, at which Steinmetz was the acclaimed wizard, and the laying of the trans-Atlantic cables, in which Heaviside played an important role, despite obstructions created by some authorities.

This epoch of engineering—the genesis of the age of electricity—lies between that of the steam engine and of the computer and IT. I think the engineers would appreciate both the technical and historical perspectives. (You could also slip in the importance of securing patents for financial security—every engineer has an eye on that—using H & S as contrasting examples.)

Answer 7

Start with a line segment $AB$, and pick a point $P$ on this. On $AP$ and $PB$, create two equilateral triangles. You get two new points, $A_2$ and $B_2$. Put $P_2$ on the line between these, such that the ratios $AP:PB$ and $A_2P_2:P_2B_2$ are the same. Repeat. This has a limit point, $P^*$. Show that this $P^*$ lies on a circle with $AB$ as diameter.

This is a neat problem, where complex numbers do help quite a lot; although one can of course use linear algebra also.

Answer 8

Complex numbers make working with polynomials much easier, including purely real-valued questions about polynomials.

For example, prove every real polynomial factors into linear and quadratics. With Fundamental Theorem of Algebra this is trivial (factors into complex linear polynomials; each complex root appears with its conjugate). A purely real proof would be ugly.

In my opinion, negative numbers do not "exist" either. However, negative numbers make subtraction much easier. Complex numbers fill the analogous role for algebraic manipulations.

Answer 9

Is it too abstract to motivate complex numbers in terms of the equations we can solve depending on whether we choose to work in ${\mathbb N, \mathbb Z, \mathbb Q, \mathbb R, \mathbb C}$? The famous "John and Betty" (Link) takes such an approach.

Answer 10

This video is a good teaser for an introductory course for students majoring in E.E., M.E., C.S., or physics:

What is a fourier series? From heat flow to drawing circles.

Answer 11

This is a bit more about justifying the use of complex analytic functions, rather than complex numbers themselves, and so is perhaps jumping ahead too far for an initial introduction. But I would simply point out a few of the properties of differentiation versus integration of real functions.

On the one hand, derivatives are easy to compute, because they only depend on the local behavior of a function. The product rule and chain rule give us everything we need to compute derivatives of basically anything, provided we can write it in terms of familiar functions.

On the other hand, it is very difficult to make estimates or approximations of $f'$ based purely on estimates of $f$. By itself, an estimate of the form $f(x) \le g(x)$ allows us to say absolutely nothing about $f'$.

With integrals we have the opposite problem. A bound of the form $f(x) \le g(x)$ implies that $\int\limits_{0}^x f(t) dt \le \int\limits_{0}^x g(t) dt$. This enables us to approximate integrals of even complicated functions using simple bounds. However, the global nature of integration makes computing exact integrals or antiderivatives much harder, and there is no general procedure to find closed form antiderivatives of unfamiliar functions other than by inverting the product or chain rule, which only works in special cases.

In complex analysis, as opposed to real analysis, differentiation of $f$ is expressed as an integral of $f$. This means that bounds on $f$ do indeed allow us to bound the derivatives of $f$, giving us refined information about $f$'s local behavior. Conversely, we can evaluate integrals, which inherently depend on the global behavior of $f$ on the domain of integration, in terms of $f$'s local behavior. We can compute even complicated integrals just by evaluating some explicit function (or residue) at a point. Complex analysis lacks the disadvantages of the pure locality of real differentiation, and the pure globality of real integration, so both derivatives and integrals are much richer.

For emphasis you can give them an example of an integral which would be very difficult to compute with real methods, and tell them that soon they will be able to calculate it exactly.

Answer 12

I'm surprised nobody mentioned the way complex numbers actually "came to existence", namely for finding roots of third order polynomials, even when said roots are real.

Answer 13

Complex-step methods can be used to numerically estimate the derivative of real analytic functions with more numerical stability than real-step methods. Consider an analytic function $f(x+iy) = u + iv$.

By Cauchy-Riemann equations,

$$ \begin{align} \frac{\partial u}{\partial x} & = \frac{\partial v}{\partial y} \\ & = \lim_{h \rightarrow 0} \frac{v(x+i(y+h))-v(x+iy)}{h} \end{align} $$

Since we are interested in the real parts of the function, $y = 0$, $u(x) = f(x)$ and $v(x) = 0$. Thus,

$$ \begin{align} \frac{\partial f}{\partial x} &= \lim_{h \rightarrow 0} \frac{Im[f(x+ih)]}{h} \\ & \approx \frac{Im[f(x+ih)]}{h} \text{ (for small h) } \end{align} $$

The advantage to the real counterparts e.g.

$$ \begin{align} \frac{df}{dx} &= \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ & \approx \frac{f(x+h)-f(x)}{h} \text{ (for small h) } \end{align} $$

is that the subtraction in the numerator is eliminated. Thus, the complex-step method avoid catastrophic numerical cancellation (which usually happens when $h$ decreases too much and $f(x+h) - f(x)$ goes below the machine precision).

More details about the complex-step method can be seen in Martins, Joaquim R. R. A.; Sturdza, Peter; Alonso, Juan J., The complex-step derivative approximation, ACM Trans. Math. Softw. 29, No. 3, 245-262 (2003). ZBL1072.65027.

Answer 14

I think the description of complex numbers via the Argand Diagram is great. And we can see addition of complex numbers as translation and multiplication as a combination of rotation and dilation.