# Thread: C(omplex) puzzle

1. ## C(omplex) puzzle

What's wrong with this?
$\displaystyle {\displaystyle e^{i2\pi n}=1 \text{ ; for } n=0,\pm 1, \pm 2, \hdots }$

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

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

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

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

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

$\displaystyle 1=e^{-4\pi^2n^2}$, which is true only for $\displaystyle n=0$.

2. The third line, where for some reason you put that:

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

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

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

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

This, of course, is only true when n = 0.
This line is correct ($\displaystyle e=e^{1+i2\pi n} \Rightarrow e^{1+i2\pi n}=\left(e^{1+i2\pi n}\right)^{1+i2\pi n}$) . The incorrect line is the one after the line you wrote - $\displaystyle \left(e^a\right)^b$ does not necessarily equal $\displaystyle e^{ab}$ when $\displaystyle b$ is not real (and the above argument just shows this leads to a contradiction).