e, i, and π (Posted on 2004-03-02)

	Submitted by Tristan
Rating: 2.8571 (7 votes)
Solution:	(Hide)
e^(iπ) = -1 (e^(iπ))^i = (-1)^i e^(i²π) = (-1)^i e^(-π) = (-1)^i (e^(-π))*(e^i) = ((-1)^i)*(e^i) e^(i-π) = (-e)^i (e^(i-π))^(π) = ((-e)^i)^(π) e^(iπ-π²) = (-e)^(iπ) e^(iπ)*(e^(-π²) = (-e)^(iπ) -(e^(-π²)) = (-e)^(iπ) So, (-e)^(iπ) = -(e^(-π²))

	Subject	Author	Date
	Answer	K Sengupta	2008-03-21 12:14:30
	Solution	Praneeth	2007-08-20 01:17:33
	Anotehr elegant method...	Kenny M	2006-02-07 21:16:33
	re: solution - text problems	Tristan	2004-03-02 20:14:10
	Notes:	Gamer	2004-03-02 14:50:27
	re: too easy for this site	Charlie	2004-03-02 13:30:16
	re(2): solution	Charlie	2004-03-02 13:20:42
	re(2): solution	Richard	2004-03-02 13:14:05
	re: solution	Richard	2004-03-02 13:10:47
	re: too easy for this site	SilverKnight	2004-03-02 12:42:12
	too easy for this site	red_sox_fan_032003	2004-03-02 12:35:58
	solution	Charlie	2004-03-02 09:08:39