Kloosterman sum

Remarks

The value does not depend on the choice of representatives or of $\bar{x}$, since the phase is defined modulo $1$. Example: $m = 5$, $a = b = 1$. The pairs $(x, \bar{x})$ are $(1, 1), (2, 3), (3, 2), (4, 4)$, so the phases are $2/5, 1, 1, 8/5$ and $$K(1, 1; 5) = e(2/5) + 2 + e(3/5) = 2 + 2\cos(4\pi/5) \approx 0.382.$$ Non-example: the sum $\sum_{x = 0}^{m-1} e((a x + b \bar{x})/m)$ over all residues is not defined, since $\bar{x}$ does not exist when $\gcd(x, m) > 1$. The sum is real, because replacing $x$ by $-x$ conjugates each term. If $\gcd(a, m) = 1$, then $K(a, b; m) = K(1, ab; m)$. For $b = 0$, the sum $K(a, 0; m)$ is the Ramanujan sum $c_m(a)$. It is an exponential sum in which the phase function is $x \mapsto (ax + b\bar{x})/m$.