Prove that the non zero roots of the equation $\tan x =x$ are transcedental.
Solution
We have that:
$$\tan x = \frac{e^{ix}- e^{-ix}}{i \left ( e^{ix} +e^{-ix} \right )}= -i \frac{e^{2ix}- 1}{e^{2ix}+1} \Rightarrow e^{2ix} = \frac{1+i\tan x}{1-i \tan x}$$
If $x$ is an algebraic number, then $2ix$ is also algebraic hence from Lindemann's theorem $e^{2ix}$ is transcedental (for $x\neq 0$). Hence $\tan x$ must also be transcedental.
Solution
We have that:
$$\tan x = \frac{e^{ix}- e^{-ix}}{i \left ( e^{ix} +e^{-ix} \right )}= -i \frac{e^{2ix}- 1}{e^{2ix}+1} \Rightarrow e^{2ix} = \frac{1+i\tan x}{1-i \tan x}$$
If $x$ is an algebraic number, then $2ix$ is also algebraic hence from Lindemann's theorem $e^{2ix}$ is transcedental (for $x\neq 0$). Hence $\tan x$ must also be transcedental.
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