# Demystifying complex numbers

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 of course 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 after Gauss complaints given in Keith's comment) that the name "complex" is quite discouraging to average students.

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

Of course, 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 of all them is assuming already some knowledge from students.

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

Here comes the two promised example. The 2nd one was reminded by several answers and comments about relations with trigonometric functions (but also by notification "The bounty on your question Trigonometry related to Rogers--Ramanujan identities expires within three days"; it seems to be harder than I expect).

Example 1. Find the Fourier expansion of the (unbounded) periodic function $$f(x)=\ln\Bigl|\sin\frac x2\Bigr|.$$

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 $$\Bigl|\sin\frac x2\Bigr|=\sqrt{\frac{1-\cos x}2}$$ and $\operatorname{Re}\ln w=\ln|w|$, where we choose $w=\frac12(1-z)$, we deduce that $$\operatorname{Re}\Bigl(\ln\frac{1-z}2\Bigr)=\ln\sqrt{\frac{1-\cos x}2} =\ln\Bigl|\sin\frac x2\Bigr|.$$ Thus, $$\ln\Bigl|\sin\frac x2\Bigr| =-\ln2-\operatorname{Re}\sum_{n=1}^\infty\frac{z^n}n =-\ln2-\sum_{n=1}^\infty\frac{\cos nx}n.$$ 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. For an integer $a$ relatively prime to $p$, the Legendre symbol $\bigl(\frac ap\bigr)$ is $+1$ or $-1$ depending on whether the congruence $x^2\equiv a\pmod{p}$ is solvable or not. One of elementary consequences of (elementary) Fermat's little theorem is $$\biggl(\frac ap\biggr)\equiv a^{(p-1)/2}\pmod p. \qquad\qquad\qquad {(*)}$$ Show that $$\biggl(\frac2p\biggr)=(-1)^{(p^2-1)/8}.$$

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 $$2^{(p-1)/2}\equiv\sqrt2\cos\frac{\pi p}4\pmod p.$$ It remains to apply ($*$): $$\biggl(\frac2p\biggr) \equiv2^{(p-1)/2} \equiv\sqrt2\cos\frac{\pi p}4 =\begin{cases} 1 & \text{if } p\equiv\pm1\pmod8, \cr -1 & \text{if } p\equiv\pm3\pmod8, \end{cases}$$ which is exactly the required formula.

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Maybe an option is to have them understand that real numbers also do not belong to the real world, that all sort of numbers are simply abstractions. –  Mariano Suárez-Alvarez Jul 1 '10 at 14:50
Probably your electrical engineering students understand better than you do that complex numbers (in polar form) are used to represent amplitude and frequency in their area of study. –  Gerald Edgar Jul 1 '10 at 15:36
Not an answer, but some suggestions: try reading the beginning of Needham's Visual Complex Analysis (usf.usfca.edu/vca/) and the end of Levi's The Mathematical Mechanic (amazon.com/Mathematical-Mechanic-Physical-Reasoning-Problems/dp/…). –  Qiaochu Yuan Jul 1 '10 at 17:05
Your example has a hidden assumption that a student actually admits the importance of calculating F.S. of $\ln\left|\sin{x\over 2}\right|$, which I find dubious. The examples with an oscillator's ODE is more convincing, IMO. –  Paul Yuryev Jul 2 '10 at 3:02
@Mariano, Gerald and Qiaochu: Thanks for the ideas! Visual Complex Analysis sounds indeed great, and I'll follow Levi's book as soon as I reach the uni library. @Paul: I give the example (which I personally like) and explain that I do not consider it elementary enough for the students. It's a matter of taste! I've never used Fourier series in my own research but it doesn't imply that I doubt of their importance. We all (including students) have different criteria for measuring such things. –  Wadim Zudilin Jul 2 '10 at 5:06

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.

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Here is a kind of mime show.

Silently write the formulas for $\cos(2x)$ and $\sin(2x)$ lined up on the board, something like this: $$\cos(2x) = \cos^2(x) \hphantom{+ 2 \cos(x) \sin(x)} - \sin^2(x)$$ $$\sin(2x) = \hphantom{\cos^2(x)} + 2 \cos(x) \sin(x) \hphantom{- \sin^2(x)}$$

Do the same for the formulas for $\cos(3x)$ and $\sin(3x)$, and however far you want to go: $$\cos(3x) = \cos^3(x) \hphantom{+ 3 \cos^2(x) \sin(x)} - 3 \cos(x) \sin^2(x) \hphantom{- \sin^3(x)}$$ $$\sin(3x) = \hphantom{\cos^3(x)} + 3 \cos^2(x) \sin(x) \hphantom{- 3 \cos(x) \sin^2(x)} - \sin^3(x)$$

Maybe then let out a loud noise like "hmmmmmmmmm... I recognize those numbers..."

Then, on a parallel board, write out Pascal's triangle, and parallel to that write the application of Pascal's triangle to the binomial expansions $(x+y)^n$. Make some more puzzling sounds regarding those pesky plus and minus signs.

Then maybe it's time to actually say something: "Eureka! We can tie this all together by use of an imaginary number $i = \sqrt{-1}$". Then write out the binomial expansion of $$(\cos(x) + i\,\sin(x))^n$$ break it into its real and imaginary parts, and demonstrate equality with $$\cos(nx) + i\, \sin(nx).$$

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