0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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

4
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $m=2$

$\endgroup$
5
  • $\begingroup$ 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
  • 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
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.