Altheman wrote: I would appriciate it if you guys could give me some hints to get started, instead of giving a full solution. Here are some of my thoughts. One of the things that I don't understand is that $ g(z)$ does not seem to restrict the equation at all. Saying the sum of 2 functions($ f(z), f(\omega z + a)$) is a function ($ g(z)$) works for all functions. Also it is always going to be complex input to output because even if the imaginary parts canceled, real numbers are still in the complex set of numbers. One thing that I have been looking at though is that with the omega, we get a cyclic function. Say $ j(z) = \omega z$, then we have $ j^3(z) = x$ I tried using that... $ f(z) + f(\omega z + a) = g(z)$ $ f(\omega z) + f(\omega^2 z + a) = g(\omega z)$ $ f(\omega^2 z) + f(z + a) = g(\omega^2z)$ say $ h(z) = f(z) + f(z + a)$, then we have $ h(z) + h(\omega z) + h(\omega^2 z) = g(z) + g(\omega z) + g(\omega^2 z) = i(z)$ then we have $ i(z) = i(\omega z) = i(\omega^2 z) = ... = i(\omega^n z)$ it seems that $ i(z)$ should be $ c*z^{3n}$ for $ n \in \mathbb{Z}$, but that seems to be a pretty large assumption Altheman wrote: One thing that I have been looking at though is that with the omega, we get a cyclic function. Say $ j(z) = \omega z$, then we have $ j^3(z) = x$ You are almost there! Now note that, even more general, if we take $ j\left(z\right) = \omega z + a$, then $ j^3\left(z\right) = z$ for every $ z\in\mathbb{C}$. Now, the functional equation $ f\left(z\right) + f\left(\omega z + a\right) = g\left(z\right)$ becomes $ f\left(z\right) + f\left(j\left(z\right)\right) = g\left(z\right)$. If we apply this equation not only to z, but also to j(z) and $ j^2\left(z\right)$, then we get $ f\left(j\left(z\right)\right) + f\left(j^2\left(z\right)\right) = g\left(j\left(z\right)\right)$; $ f\left(j^2\left(z\right)\right) + f\left(\underbrace{j^3\left(z\right)}_{ = z}\right) = g\left(j^2\left(z\right)\right)$. Now, we have three linear equations for the three unknowns f(z), f(j(z)), $ f\left(j^2\left(z\right)\right)$. And now, it cannot be that hard anymore... Darij

Solution: Note that, since $\omega$ is a third root of unity not equal to $1$, we have that \[ \omega^2 + \omega + 1 = 0. \]Substitute $ z \rightarrow \omega z + a $ in the original equation to get the following transformation: \[ f(z) + f(\omega z + a) = g(z) \implies f(\omega z + a) + f(\omega^2 z + \omega a + a) = g(\omega z + a). \]We make the same substitution again: \[ f(\omega^2 z + \omega a + a) + f (\omega^3 z + \omega^2 a + \omega a + a) = g(\omega^2 z + \omega a + a). \]Recall that $ \omega^2 + \omega + 1 = 0 $ and $ \omega^3 = 1 $, so we obtain: \[ f(\omega^2 z + \omega a + a) + f(z) = g(\omega^2 z + \omega a + a). \]Adding the three recent equations together, we get: \[ f(z) + f(\omega z + a) + f(\omega^2 z + \omega a + a) = \dfrac {g(z) + g(\omega z + a) + g(\omega^2 z + \omega a + a}{2}. \]We subtract from the second equation and we get: \[ f(z) = \dfrac {g(z) - g(\omega z + a) + g (\omega^2 z + \omega a + a)}{2}. \]We see that there is a one-to-one bijection between $f$ and $g$ here, in the sense that for every $g$, there is a unique $f$. That is, there is exactly one function $f(z)$, given $g(z)$, which satisfies the original functional equation. This was exactly what we wanted to show. $ \blacksquare $