$$ f(x) = \begin{cases} 0 & x \not\equiv 4 \pmod 5 \\ 1 & x \equiv 4 \pmod 5 \end{cases} $$
and $$ f(x) = \frac{1}{5} \sum_{k=0}^4 \cos\left(\frac{2 \pi}{5} k (x-4) \right) $$
Can someone help me understand where this formula comes from and what's really happening here? I know it comes from the real part of $$ f(x) = \frac{1}{5} \sum_{k=0}^4 \zeta^{k(x-4)} $$ but I don't understand why the real part of this is zero only on $4$.
I can see that it works but I don't understand the motivation behind what the person did to come up with this.
P.S. I don't know any complex analysis :(
Let $\zeta = e^{2\pi i/5}$ be a primitive fifth root of unity. That is to say, $\zeta^5 = 1$ and $\zeta \ne 1$. So the set $\{\zeta^k\}_{k=0}^4 = \{1, \zeta, \zeta^2, \zeta^3, \zeta^4\}$ are the distinct roots of the quintic polynomial $$z^5 - 1 = 0.$$ As such, $z^5 - 1$ admits a factorization of the form
$$z^5 - 1 = (z - 1)(z - \zeta)(z - \zeta^2)(z - \zeta^3)(z - \zeta^4).$$
Expanding out the factorization on the RHS and comparing coefficients, we must then have
$$0 = 1 + \zeta + \zeta^2 + \zeta^3 + \zeta^4.$$
Since the real and imaginary parts of the RHS do not interact, this implies that
$$0 = 1 + \Re(\zeta) + \Re(\zeta^2) + \Re(\zeta^3) + \Re(\zeta^4).$$
And since $\Re(\zeta^k ) = \cos \frac{2\pi}{5} k$, we obtain $$0 = \sum_{k=0}^4 \cos \frac{2\pi}{5} k.$$
Now let $x \in \mathbb Z$ and consider the sum $$1 + \zeta^x + \zeta^{2x} + \zeta^{3x} + \zeta^{4x}.$$ Notice how when $x = 1$, this becomes equivalent to our previous situation. But when $x = 0$, the terms of this sum all equal $1$. In fact, if we substitute some integer values of $x$, and use $\zeta^{5} = 1$ to reduce the exponent, we see that:
$$\begin{array}{c|l} x & 1 + \zeta^x + \zeta^{2x} + \zeta^{3x} + \zeta^{4x} \\ \hline 0 & 1 + 1 + 1 + 1 + 1 = 5 \\ 1 & 1 + \zeta^1 + \zeta^2 + \zeta^3 + \zeta^4 = 0 \\ 2 & 1 + \zeta^2 + \zeta^4 + \zeta^1 + \zeta^3 = 0 \\ 3 & 1 + \zeta^3 + \zeta^1 + \zeta^4 + \zeta^2 = 0 \\ 4 & 1 + \zeta^4 + \zeta^3 + \zeta^2 + \zeta^1 = 0 \\ \hline 5 & 1 + 1 + 1 + 1 + 1 = 5 \\ \vdots & \vdots \end{array}$$ It looks like the roots just permute whenever $x \not\equiv 0 \pmod 5$. That is not a coincidence: in general, whenever $x$ is relatively prime to $5$, the set $\{x, 2x, 3x, 4x\} \pmod 5$ will be some permutation of $\{1,2,3,4\}$. This explains the relationship to the behavior of the function $f$: all that remains is to modify $x$ to $x-4$, and we now see that the sum $$\sum_{k=0}^4 \cos \frac{2\pi}{5}k(x-4)$$ will be equal to $5$ if and only if $x \equiv 4 \pmod 5$, in which case each term will be $\cos 0 = 1$. Otherwise, if $x \not\equiv 4 \pmod 5$, the terms will simply rearrange themselves and will sum to $0$. I leave a formal proof of this as an exercise for the reader.
As a further exercise, what would happen if we were looking at the function $$g(x) = \sum_{k=0}^7 \cos \frac{2\pi}{8} kx?$$ For which values of $x$ would this sum not equal $0$? And more generally, what about
$$g_n(x) = \sum_{k=0}^{n-1} \cos \frac{2\pi}{n} k x$$
for a positive integer $n$?