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.

  • $\begingroup$ Not on my wolfram it doesn't. $\endgroup$ Aug 8, 2017 at 13:09
  • 1
    $\begingroup$ @uniquesolution $-(-1)^{1/4} = (-1)(-1)^{1/4} = (-1)^{5/4} = (\cos \pi + i \sin \pi)^{5/4} = e^{5i\pi/4}$ $\endgroup$ Aug 8, 2017 at 13:13
  • $\begingroup$ @uniquesolution see Alternative form in Wolfram $\endgroup$
    – MAN-MADE
    Aug 8, 2017 at 13:18

2 Answers 2


The polynomial $p(x)=x^4+1$ splits into four linear factors over the algebraicely complete field $\mathbb{C}$:


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.


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 $m=2$

  • $\begingroup$ I understand why this is correct but what did I do wrong in my calculation? $\endgroup$
    – green frog
    Aug 8, 2017 at 13:09
  • $\begingroup$ You should check for exponential rule, especially power rule. It does not work for arbitrary number. $\endgroup$
    – 04170706
    Aug 8, 2017 at 13:14
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    $\begingroup$ 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
  • $\begingroup$ I feel like I've been abusing power rule then... $\endgroup$
    – green frog
    Aug 8, 2017 at 13:14
  • $\begingroup$ @BoazMoerman I see! Thanks. $\endgroup$
    – green frog
    Aug 8, 2017 at 13:15

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