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.

Visual Complex Analysis(usf.usfca.edu/vca/) and the end of Levi'sThe Mathematical Mechanic(amazon.com/Mathematical-Mechanic-Physical-Reasoning-Problems/dp/…). $\endgroup$Visual 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$9more comments