1

I have $$ e^{1+2 \pi i}= e^{1} e^{2 \pi i} = e,$$ then $$e=e^{1+2 \pi i} = ( e^{1+ 2 \pi i})^{1+2 \pi i} = e^{(1+2 \pi i) (1+2 \pi i)}= e^{1+4 \pi i - 4 \pi^2}= ee^{4 \pi i}e^{-4 \pi^2}= ee^{-4 \pi^2}.$$ Thus $ e^{-4 \pi^2} = 1$

Can anybody help me?

Beth
  • 389
  • 3
    I guess what it amounts to is that, for positive real numbers and certain other patterns, we can take $(a^b)^c = a^{bc},$ but it can fail in other cases. This failure is interesting, – Will Jagy Oct 15 '18 at 00:27
  • 1
    You may want to check the answers from this question https://math.stackexchange.com/questions/2399675/two-contradictory-ways-to-calculate-e2-pi-ii –  Oct 15 '18 at 02:28

0 Answers0