The Lindemann-Weierstrass theorem tells us for which algebraic $\alpha$ the exponential $e^{\alpha}$ is also algebraic. I wonder what happens when we replace $\alpha$ by $2\pi i \alpha$. In other words, for which algebraic $\alpha$ is $e^{2\pi i\alpha}$ also algebraic ?
My thoughts : obviously, if $\alpha$ is rational then $e^{2\pi i\alpha}$ is a root of unity and so is algebraic.
Most of the idea is discussed in the commented link. To finish this problem, assume on the contrary that $\beta=e^{2\pi i \alpha}$ is algebraic for an algebraic irrational $\alpha$. Then $$ \frac{\log \beta}{\log(-1)} = \frac{2 \pi i\alpha}{\pi i} = 2\alpha. $$
Now, this implies that $\beta = (-1)^{2\alpha}$ with algebraic irrational $\alpha$. This contradicts Gelfond-Schneider.