 Solution

Put $q=10/Re>0$. The frequency formula becomesFor large positive Reynolds number, $\sqrt{q-q^2/4}=\sqrt{q}[1-q/8+O(q^2)]$. ThereforeMore precisely the real correction is $-5/Re$, and the next imaginary correction is $O(R{e}^{-3/2})$. The growing branch has a rate tending to zero as $Re\to \infty$, despite being unstable at every sufficiently large finite $Re$.