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A necessary condition for a nonzero solution is $u(z) u(a-z) = 1$. If that is true and $u$ is entire, it has an entire square root. Now $\sqrt{u(z)} \sqrt{u(a-z)} = u(a/2) = \pm 1$: if it is $1$, then $f(z) = \sqrt{u(z)}$ is a solution. If it is $-1$, then $f(z) = (z - a/2) \sqrt{u(z)}$ is a solution.