Mathematical contradiction.

[Sentendence]

\(\displaystyle e^{\pi}=(e^{i\pi})^{-i}=(-1)^{-i}=\frac{1}{(-1)^{i}}\)

Implies that

\(\displaystyle e^{\pi} = \frac{1}{(-1)^{i}}\)

\(\displaystyle e^{\pi}(-1)^{i}=1\)

\(\displaystyle e^{\pi}(-1)^{i}=(-1)(-1)\)

\(\displaystyle e^{\pi}(-1)^{i}=e^{i\pi}(-1)\)

\(\displaystyle \ln (e^{\pi}(-1)^{i})=\ln(e^{i\pi}(-1))\)

\(\displaystyle \pi\ln(e)+i\ln(-1)=i\pi \ln(e)+\ln(-1)\)

\(\displaystyle \pi\ln(e)[1-i]=\ln(-1)[1-i]\)

\(\displaystyle \pi\ln(e)[1-i]-\ln(-1)[1-i]=0\)

\(\displaystyle [1-i][\pi\ln(e)-\ln(-1)]=0\)

\(\displaystyle [1-i][\pi-i\pi]=0\\\)

\(\displaystyle \pi[1-i][1-i]=0\\\)

Implies that i = 1

P.S How do you change this to display LaTeX code ?

 

[DrPhil]

\(\displaystyle e^{\pi}=(e^{i\pi})^{-i}=(-1)^{-i}=\frac{1}{(-1)^{i}}\)

Implies that

\(\displaystyle e^{\pi} = \frac{1}{(-1)^{i}}\)

\(\displaystyle e^{\pi}(-1)^{i}=1\)

\(\displaystyle e^{\pi}(-1)^{i}=(-1)(-1)\)

This is far enough to disclose the contradiction:

\(\displaystyle 1 = (-1)(-1)\)

\(\displaystyle \ln(1) =\ln(-1) + \ln(-1)\)

\(\displaystyle 0 = (\pi\ i) + (\pi\ i) = 2\ \pi \ i\)

The flaw is that the logarithm of a complex number is not single-valued. You have to choose what sheet you are on. You can always add or subtract a multiple of \(\displaystyle 2\pi i\).

\(\displaystyle \displaystyle \ln(-1) = (2n + 1)\pi\ i\)

Thus after you take logarithms, your equalities are all modulo \(\displaystyle 2\pi\).