Is is true that $e^{i\pi\alpha}$ is transcendental whenever $\alpha$ is irrational? And if so is there a simple proof?
Transcendence of $e^{i\pi\alpha}$ when $\alpha$ is irrational
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0No, $e^{i\pi}=-1$, therefore $e^{i\pi\alpha}=(e^{i\pi})^{\alpha}=(-1)^{\alpha}$. – 2017-01-15
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1@barakmanos $(e^{i\pi})^{\alpha} \neq e^{i\pi \alpha}$ in general. Even so, then what? $(-1)^{\alpha} :=\exp(\alpha \log (-1)) = e^{i \pi \alpha}$, back to the starting point. – 2017-01-15
1 Answers
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No: try $\alpha = \dfrac{\tan^{-1} 0.75}{\pi} \approx 0.204832764699\ldots$.
Then $e^{i\pi\alpha}=0.8 + 0.6i$