Clearly, one has $\sin(\pi x)\neq0$ for all $x\in\mathbb{R}\backslash\mathbb{Z}$. But, one also has $\sin(\pi x)=\frac{e^{\imath\pi x}-e^{-\imath\pi x}}{2\imath}=\frac{(e^{\imath\pi})^x-(e^{-\imath\pi})^x}{2\imath}=\frac{(\cos(\pi)+\imath\sin(\pi))^x-(\cos(-\pi)+\imath\sin(-\pi))^x}{2\imath}=\frac{(-1+\imath0)^x-(-1-\imath0)^x}{2\imath}=\frac{(-1)^x-(-1)^x}{2\imath}=0$ seemingly for all $x\in\mathbb{R}$ where $(\pm\imath)^2=-1$. With $x=0.5$, for example, one would then have $\sin(0.5\pi)=\frac{(-1)^{0.5}-(-1)^{0.5}}{2\imath}=\frac{\sqrt{\imath^2}-\sqrt{\imath^2}}{2\imath}=0$ rather than $\sin(0.5\pi)=\frac{\sqrt{-1} - \sqrt{-1}}{2\imath}=\frac{\sqrt{\imath^2} - \sqrt{(-\imath)^2}}{2\imath}=1$ (which is correct) while $\sin(0.5\pi)=\frac{\sqrt{-1} - \sqrt{-1}}{2\imath}=\frac{\sqrt{(-\imath)^2} - \sqrt{\imath^2}}{2\imath}=-1$. But how to decide when to take $-1=\imath^2$ or $-1=(-\imath)^2$, how do the sign of the arguments of $\cos$ and/or $\sin$ (in Euler's identity) relate to $(\pm\imath)^2=-1$ and how about $0\neq x\in\mathbb{C}$?
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2$e^{i\pi x} = \left(e^{i\pi}\right)^x$ is not valid for real $x$, only integer. – Ninad Munshi Mar 23 '21 at 18:27
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3It is NOT true in general that $e^{z_1z_2}=(e^{z_1})^{z_2}$ – Mark Viola Mar 23 '21 at 18:27
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This question states basically the exact same problem but in an easier way. As the helpful comments already wrote $e^{i\pi x}=(e^{i\pi})^x$ only if $x$ is an integer. This Wikipedia article about "De Moivre's formula" might be interesting.
De Moivre's formula does not hold for non-integer powers. The derivation of de Moivre's formula above involves a complex number raised to the integer power $n$. If a complex number is raised to a non-integer power, the result is multiple-valued.
This is because the complex logarithm is not a bijective funtion. An example: $e^{2i\pi}$ and $e^{4i\pi}$ are equal. Taking $\log$'s on both expressions yields $2\pi i=4\pi i$, which is false. See here for more detail.
vitamin d
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Side note that DeMoivre's Theorem may be alternatively derived by showing that $e^{i\alpha} \times e^{i\beta} = e^{i(\alpha + \beta)}.$ However, your analysis was on point, and the side note is not. – user2661923 Mar 23 '21 at 18:47
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First two words in my comment. That although there is an alternative derivation of DeMoivre's Theorem, this alternate proof is not relevant to the point that you were making in your answer. – user2661923 Mar 23 '21 at 18:53
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What point was I trying to make in my answer? I only said that the OP should look at that Wikipedia article, and then I quoted something, (without any derivations). Are we talking about the same thing? – vitamin d Mar 23 '21 at 18:56
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Your point that $e^{i\pi x} = \left(e^{i\pi}\right)^x$ only if $x$ is an integer. – user2661923 Mar 23 '21 at 19:00
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I wasn't criticisizing your answer at all. I was simply indicating that your analysis focused on the derivation of DeMoivre's Theorem, as it pertains to raising $n$ to a positive integer. So, your reasoning was certainly logical. And so, not that it was relevant to the point that you were making, DeMoivre's Theorem can be derived in an alternative manner. – user2661923 Mar 23 '21 at 19:04
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