If e^itπ is algebraic , is $t$ a rational number. [closed]

I have a elementary question:If e^itπ is algebraic , is $t$ a rational number. I do not know whether it is right

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Welcome to MO! This question is not really within the narrow scope of this site. It would seem to fit better on math.stackexchange.com a similar site with a broader scope. Yet in fact something similar (which answers your question) was already asked there math.stackexchange.com/questions/4323/… So there you could find more details on this. –  quid Mar 9 '13 at 9:47
en.wikipedia.org/wiki/Gelfond–Schneider_theorem –  Alain Valette Mar 9 '13 at 10:29
How is Gelfond--Schneider directly relevant here? Without some explanation I have a very hard time seeing this being useful for OP. Actually, I am also a bit dubious what is even meant. –  quid Mar 9 '13 at 13:14
@quid: I admit, I read the question too fast... –  Alain Valette Mar 9 '13 at 14:23
@Alain Valette: Thank you for the reply, this explains it. –  quid Mar 9 '13 at 14:32

closed as off topic by Marc Palm, quid, Alain Valette, Felipe Voloch, Benjamin SteinbergMar 9 '13 at 13:19

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If we take any real algebraic integer $s$ with $0 < s <1,$ then, $t = s + i \sqrt{1-s^{2}}$ is an algebraic integer which lies on the unit circle. Now apply this with $s = \sqrt{2}-1.$ Note that $t$ generates a degree $4$ extension of the rationals. If $t$ were a primitive $m$-th root of unity, we would have $\phi(m) = 4$ so that $m = 8.$ Hence $t$ would be a primitive $8$-th root of unity, but it is not, as each primitive $8$-th root of unity has real part $\pm \frac{1}{\sqrt{2}}.$
Which question is "this question"? For the one asked on this site it might be still simpler to consider $(3 + 4i)/5$; which is mentioned in the question on math.SE. Discussions of questions asked on other sites in answers here seem off-topic. –  quid Mar 9 '13 at 12:37