12/19/2007, 04:03 PM
(1-i)^i=e
ln(1-i)=1/i=-i is this true?
if we expand ln(1-i) = -i -i^2/2+i^3/3-i^4/4........since i is infinitisemal, i^2, i^3, i^4 etc vanishes in summation compared to -i, if we stay in the same scale with our calculations so :
ln(1-i)=-i=1/i and (1-i)^i=e
Main idea was that +i is in fact, infinitesimal of any scale provided we do calculations while staying in that scale. Then -i is infinity of the same grade.
ln(1-i)=1/i=-i is this true?
if we expand ln(1-i) = -i -i^2/2+i^3/3-i^4/4........since i is infinitisemal, i^2, i^3, i^4 etc vanishes in summation compared to -i, if we stay in the same scale with our calculations so :
ln(1-i)=-i=1/i and (1-i)^i=e
Main idea was that +i is in fact, infinitesimal of any scale provided we do calculations while staying in that scale. Then -i is infinity of the same grade.