, 2014

Solution:

Answer: $\boxed{\frac{1}{3}}$

Let $r = |z|$, $\theta = \arg z$, and $C = \frac{\Re z}{|z|} = \cos \theta = \cos \frac{2\pi j}{k}$ for some $j$ with $\gcd(j, k) = 1$. The condition of the four consecutive points lying on a parabola is equivalent to having the finite difference
$$f(3) - 3 f(2) + 3 f(1) - f(0) = 0$$
This implies
$$\begin{aligned} f(3) - f(0) & = 3 [f(2) - f(1)] \\ \Longleftrightarrow r^{3} \cos (3 \theta) - 1 & = 3 \left(r^{2} \cos (2 \theta) - r \cos (\theta)\right) \\ \Longleftrightarrow r^{3} \left(4 C^{3} - 3 C\right) - 1 & = 3 \left(r^{2} (2 C^{2} - 1) - r C\right) \end{aligned}$$
Now we simply test the first few possible values of $k$.

$k = 1$ implies $C = 1$, which gives $r^{3} - 1 = 3(r^{2} - r) \Longrightarrow (r - 1)^{3} = 0 \Longrightarrow r = 1$. This is not allowed since $r = 1$ implies a periodic function.

$k = 2$ implies $C = -1$, which gives $-r^{3} - 1 = 3 r^{2} + r \Longrightarrow (r + 1)^{3} = 0$, again not allowed since $r > 0$.

$k = 3$ implies $C = -\frac{1}{2}$. This gives $r^{3} - 1 = \frac{-3}{2}(r^{2} - r) \Longrightarrow (r - 1)\left(r + \frac{1}{2}\right)(r + 2) = 0$. These roots are either negative or $1$, again not allowed.

$k = 4$ implies $C = 0$. This gives $-1 = -3 r^{2} \Longrightarrow r = \pm \frac{1}{\sqrt{3}}$. $r = \frac{1}{\sqrt{3}}$ is allowed, so this will generate our answer.

Again by finite differences (or by any other method of interpolating with a quadratic), we get $2a = f(0) + f(2) - 2 f(1) = \frac{2}{3}$, so $a = \frac{1}{3}$.