I feel very dumb asking this. I'm trying to calculate $e^{\pi i n/4}$ for odd $n.$ I say the following: $e^{\pi i n/4} = (e^{\pi i n})^{1/4} = (-1)^{1/4}.$ However, Wolfram Alpha says that for $n = 5$ we have $-(-1)^{1/4}.$ I am confused.
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$\begingroup$ Not on my wolfram it doesn't. $\endgroup$– uniquesolutionAug 8, 2017 at 13:09
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1$\begingroup$ @uniquesolution $-(-1)^{1/4} = (-1)(-1)^{1/4} = (-1)^{5/4} = (\cos \pi + i \sin \pi)^{5/4} = e^{5i\pi/4}$ $\endgroup$– Daniel XiangAug 8, 2017 at 13:13
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$\begingroup$ @uniquesolution see Alternative form in Wolfram $\endgroup$– MAN-MADEAug 8, 2017 at 13:18
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2 Answers
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The polynomial $p(x)=x^4+1$ splits into four linear factors over the algebraicely complete field $\mathbb{C}$:
$p(x)=(x-e^{i\frac{\pi}{4}})(x-e^{-i\frac{\pi}{4}})(x-e^{i\frac{5\pi}{4}})(x-e^{-i\frac{5\pi}{4}})$
and each of the corresponding roots could be taken as $``(-1)^{\frac{1}{4}}"$.
Clearly $e^{i\frac{5\pi}{4}}=-e^{i\frac{\pi}{4}}$ (This might explain why Wolfram found "the negative of Your root"). However since $\mathbb{C}$ is no ordered field there is no unique fourth root.
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Let $n=2m+1$
$e^{i\pi (2m+1)/4}=e^{i\pi m/2}e^{i\pi /4}=\Big(\dfrac{1+i}{\sqrt 2}\Big)e^{\dfrac{i\pi m}{2}}$
Now put $m=2$
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$\begingroup$ I understand why this is correct but what did I do wrong in my calculation? $\endgroup$ Aug 8, 2017 at 13:09
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$\begingroup$ You should check for exponential rule, especially power rule. It does not work for arbitrary number. $\endgroup$– 04170706Aug 8, 2017 at 13:14
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5$\begingroup$ You used the identity $(e^z)^w=e^{(zw)}$ which is true for the positive real numbers, but not for complex numbers in general. Instead the multi-valued relation $(e^z)^w=e^{(z+2\pi n)w}$ holds where $n \in \mathbb{Z}$. $\endgroup$ Aug 8, 2017 at 13:14
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$\begingroup$ I feel like I've been abusing power rule then... $\endgroup$ Aug 8, 2017 at 13:14
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