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.

Thank you in advance!

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.

Visual Complex Analysis(usf.usfca.edu/vca/) and the end of Levi'sThe Mathematical Mechanic(amazon.com/Mathematical-Mechanic-Physical-Reasoning-Problems/dp/…). $\endgroup$ – Qiaochu Yuan Jul 1 '10 at 17:05Visual Complex Analysissounds 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. $\endgroup$ – Wadim Zudilin Jul 2 '10 at 5:065more comments