Why include negative $n$ in a Fourier expression? If we make the Fourier series $$\sum_{n=-\infty}^\infty a_n e^{in\theta},$$
what is the point of explicitly including the negative terms? It seems just using evenness and oddness of cosine and sine we can quickly get rid of them.
 A: The functions $t\mapsto\cos(nt)$ $(n\geq0)$ and $t\mapsto \sin(nt)$ $(n\geq1)$ together form a "basis" for the space of all real- or complex-valued $2\pi$-periodic functions, as do the functions $t\mapsto e^{in t}$ $(n\in{\mathbb Z})$.
Leaving out the $t\mapsto e^{in t}$ with $n<0$ would be devastating, as the remaining set of basis functions is no longer "complete". Not even $t\mapsto\cos t$ could be developed into a "complex Fourier series" then.
When working on a concrete problem with concrete functions having certain symmetries (being real-valued, even, odd, etc.) it is usually preferable to use the  "$\cos$-$\sin$-basis", since the symmetries present in $f$ are visibly reproduced in the coefficients (e.g., all coefficients are real, or all $b_n=0$, etc.).
But for theoretical considerations about convergence of Fourier series in general, or for the proof of formulas about convolution, etc., the "complex basis" has immense advantages. These come mainly from the universal and simple addition formula
$$e^{i(m+n) t}\equiv e^{im t}\cdot e^{in t}\qquad(t\in{\mathbb R})$$
and from the fact that sums of the form $\sum_{k=0}^n \sin(k t)$ and similar are now just geometric series.
A: Christian Blatter's answer is excellent, but I want to add a small observation. The real Fourier series is traditionally introduced as$$\frac12\!a_0+\sum_{n=1}^{\infty}(a_n\cos nt+b_n\sin nt).$$It is not explained why the first term is, exceptionally, multiplied by $\frac12$; it would seem more natural to write simply$$\sum_{n=0}^{\infty}(a_n\cos nt+b_n\sin nt).$$But it turns out that this would lead to having an exceptional factor $2$ introduced when the first term is evaluated---an ugly feature. The real reason for the $\frac12$ is that the complex form$$\sum_{n\in\Bbb Z}c_n\exp \mathrm i nt$$is the fundamental one, and the factor $\frac12$ relates the two series uniformly: $a_n=c_n+c_{-n}$ (and $ b_n=\mathrm i(c_n-c_{-n})$), so that the constant term is $c_0=\frac12\!(c_0+c_{-0})=\frac12\!a_0.$
