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I was wondering if someone could possibly help me figure out how to show $e^{2 \pi i \theta}$ is not algebraic when $\theta$ is irrational.

Thanks!

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2 Answers 2

up vote 6 down vote accepted

Note that as stated the result is not correct. For example, there is a real number $t$ such that $\cos(2\pi t)$ is equal to $\dfrac{3}{5}$, and it can be shown that $t$ is not rational. But if $\theta$ is an irrational algebraic number, then indeed $e^{2\pi i\theta}$ is transcendental.

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Oops. My apologies. Thank you! –  JKasahara Feb 17 '13 at 20:26
    
How does this help one understand the assertion? Could you shed some more light upon the question? Thanks for the attention in any case. –  awllower Feb 18 '13 at 3:33
    
@awllower: There are many irrationals $\theta$ such that $e^{2\pi i \theta}$ is rational, and in particular algebraic. The correct result is that if $\theta$ is an algebraic irrational then $e^{2\pi i \theta}$ is transcendental. That is straight Lindemann-Weierstrass. –  André Nicolas Feb 18 '13 at 3:53

Use Lindemann–Weierstrass theorem

http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem

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Thank you very much! –  JKasahara Feb 17 '13 at 20:39

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