Show that if $z\in \mathbb{C}$ is a non-rational solution to $\sqrt{3}(1+z)=\tan\left(z\pi/2\right)$, then $z$ is transcendental.

Question

This is exercise 1 from chapter 3 of Pillars of Transcendental Number Theory (Natarajan, Thangadurai), which is on the Gelfond-Schneider Theorem. There are a few clear ways to use the Gelfond-Schneider Theorem; if I can write $z=\alpha^{\beta}$ with $\alpha$ algebraic, not equal to $0$ or $1$ and $\beta$ irrational, than I am done. This seems like a big wish though, I've not been able to do this.

Another idea is to have some kind of thing like $\alpha^{z}=\gamma$, where $\gamma$ is algebraic, as this would mean $z$ has to be transcendental (we're already told $z$ is irrational so this seems to me like a good way to get to the contradiction). However, I've not been able to get this to work yet either.

I may also be going completely the wrong way about it completely; any hints would be very helpful!

Answer 1

Assume $z$ is algebraic, so $\sqrt{3}(1+z)$ is algebraic. Then, consider that $\tan(\pi x)=i\frac{e^{-i\pi x}-e^{i\pi x}}{e^{-i\pi x}+e^{i\pi x}}$, which is an algebraic function of $e^{i\pi x}=(-1)^x$, which is transcendental for irrational algebraic $x=z/2$ by the Gelfond-Schneider Theorem. Thus, we reach a contradiction, so non-rational solutions $z$ must be transcendental.