請大家幫忙 寫出一個函數F(X) 令到F^-1(X

Sub X = F(a), we have
F^-1(F(a)) = 1/F(F(a))
F(F(a)) = 1/a

Consider F(a) = a^i, where i is the imaginary unit (i.e. i = sqrt(-1))

F(F(a)) = (a^i)^i = a^-1 = 1/a

F^-1(a) = a^(-i) = 1/F(a)

The link defines clearly what does (to the power i) means.

2010-07-06 16:57:22 補充：
Thanks for the comment, after reading the wiki I think I was wrong for some part.
Can we constraint the domain of the function to make this work

e.g. F(x) is define over all X such that (x ^ i) ^i = x^-1?

這題是很難估計出答案的，因此要計算：
F^(-1)(X) = 1/F(X)
F(1/F(X)) = X
F(F(X)) = 1/X

屬於http://eqworld.ipmnet.ru/en/solutions/fe/fe2315.pdf這類型

2010-07-06 12:45:55 補充：
但(a^i)^i真是= a^(i × i)嗎？請留意http://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities，不要搞錯概念啊！

2010-08-02 00:16:35 補充：
但是wolfram alpha卻真是告訴我們(x^i)^i = x^-1（http://www.wolframalpha.com/input/?i=%28x%5Ei%29%5Ei）。到底wolfram alpha是否「走火」？