# C(omplex) puzzle

• Oct 24th 2010, 10:31 AM
courteous
C(omplex) puzzle
What's wrong with this?
${\displaystyle e^{i2\pi n}=1 \text{ ; for } n=0,\pm 1, \pm 2, \hdots }$

$\displaystyle {ee^{i2\pi n}=e=e^{1+i2\pi n} }$

$e^{1+i2\pi n}=\left\{ e^{1+i2\pi n} \right\}^{(1+i2\pi n)}$

$e=e^{(1+i2\pi n)^2}=e^{1+i4\pi n-4\pi^2n^2}=e^{1+i4\pi n}e^{-4\pi^2n^2}$

$e^{1+i4\pi n}=e$

$e=ee^{-4\pi^2n^2}$

$1=e^{-4\pi^2n^2}$, which is true only for $n=0$.
• Oct 24th 2010, 11:03 AM
Aryth
The third line, where for some reason you put that:

$e^{1 + i2\pi n} = \left\{e^{1 + i2\pi n}\right\}^{(1 + i2\pi n)}$

This, of course, is only true when n = 0.
• Oct 24th 2010, 11:19 AM
Unbeatable0
Quote:

Originally Posted by Aryth
The third line, where for some reason you put that:

$e^{1 + i2\pi n} = \left\{e^{1 + i2\pi n}\right\}^{(1 + i2\pi n)}$

This, of course, is only true when n = 0.

This line is correct ( $e=e^{1+i2\pi n} \Rightarrow e^{1+i2\pi n}=\left(e^{1+i2\pi n}\right)^{1+i2\pi n}$) . The incorrect line is the one after the line you wrote - $\left(e^a\right)^b$ does not necessarily equal $e^{ab}$ when $b$ is not real (and the above argument just shows this leads to a contradiction).