soft question - 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.

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
$$
f(x)=lnBigl|sinfrac x2Bigr|.
$$

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

Consider the function on the interval $xin[varepsilon,2pi-varepsilon]$.
The series
$$
sum_{n=1}^inftyfrac{z^n}n, qquad z=e^{ix},
$$
converges for all values $x$ from the interval.
Since
$$
Bigl|sinfrac x2Bigr|=sqrt{frac{1-cos x}2}
$$
and $operatorname{Re}ln w=ln|w|$, where we choose $w=frac12(1-z)$,
we deduce that
$$
operatorname{Re}Bigl(lnfrac{1-z}2Bigr)=lnsqrt{frac{1-cos x}2}
=lnBigl|sinfrac x2Bigr|.
$$
Thus,
$$
lnBigl|sinfrac x2Bigr|
=-ln2-operatorname{Re}sum_{n=1}^inftyfrac{z^n}n
=-ln2-sum_{n=1}^inftyfrac{cos nx}n.
$$
As $varepsilon>0$ can be taken arbitrarily small,
the result remains valid for all $xne2pi k$.

Example 2.
Let $p$ be an odd prime number.
For an integer $a$ relatively prime to $p$,
the Legendre symbol $bigl(frac apbigr)$ is $+1$ or $-1$
depending on whether the congruence
$x^2equiv apmod{p}$ is solvable or not.
One of elementary consequences of (elementary) Fermat's little theorem is
$$
biggl(frac apbiggr)equiv a^{(p-1)/2}pmod p.
qquadqquadqquad {(*)}
$$
Show that
$$
biggl(frac2pbiggr)=(-1)^{(p^2-1)/8}.
$$

Solution.
In the ring $mathbb Z+mathbb Zi=Bbb Z[i]$, the binomial formula implies
$$
(1+i)^pequiv1+i^ppmod p.
$$
On the other hand,
$$
(1+i)^p
=bigl(sqrt2e^{pi i/4}bigr)^p
=2^{p/2}biggl(cosfrac{pi p}4+isinfrac{pi p}4biggr)
$$
and
$$
1+i^p
=1+(e^{pi i/2})^p
=1+cosfrac{pi p}2+isinfrac{pi p}2
=1+isinfrac{pi p}2.
$$
Comparing the real parts implies that
$$
2^{p/2}cosfrac{pi p}4equiv1pmod p,
$$
hence from $sqrt2cos(pi p/4)in{pm1}$ we conclude that
$$
2^{(p-1)/2}equivsqrt2cosfrac{pi p}4pmod p.
$$
It remains to apply ($*$):
$$
biggl(frac2pbiggr)
equiv2^{(p-1)/2}
equivsqrt2cosfrac{pi p}4
=begin{cases}
1 & text{if } pequivpm1pmod8, cr
-1 & text{if } pequivpm3pmod8,
end{cases}
$$
which is exactly the required formula.