～～虛數變有理數？？～～

Firstly, I should say sorry to you that I have posted a wrong answer to you at the link below:
http://hk.knowledge.yahoo.com/question/?qid=7006111803054

Therefore, you should be more careful to my answer: :>

In fact, we have much more stronger result by the theorem in the reference below.

i^i is TRANSCENDENTAL and hence this is NOT a rational number.

A complex number is transcendental if it is not algebraic.
If an number is algebraic, then it is a root of non-zero irreducible polynomial with rational (or integral ) coefficient(s).
Hence, every rational number is algebraic.

The following link has more about transcendental number.
http://hk.knowledge.yahoo.com/question/?qid=7006112902966&others=1

I end up at here with the link of the reference book below.
http://books.google.com/books?vid=ISBN052139791X&id=SmsCqiQMvvgC&printsec=frontcover&dq=Transcendental+and+algebraic+numbers#PRA1-PA11,M1

2006-12-17 06:57:11 補充：
i is an algebraic number since i^2 +1=0

2006-12-22 01:20:57 補充：
Actually, the answer of 無知少年
is exactly correct. However, if we restricted the argument, then there is no more problem.Then, it remains to identify whether e^(-pi/2) is transcendental or not. If we just look at this form, we almost have no any information.

2006-12-22 01:30:14 補充：
However, if we switch the number into another form, i^i , this is totally different picture. The complex number i, is a root of the polynomial equation x^2 +1=0This means that i is algebraic since all the coefficients are integers.

2006-12-22 01:33:00 補充：
Now, by Baker Theorem , i^i is transcendental since, if we consider the power term,1 and i are linear independent over the field of rational numbers.You see!! The key is the POWER term!! This must be an algebraic number other than rationals.

2006-12-22 01:37:18 補充：
It is obvious that the rational power of algebraic number is , of course, algebraic. However, if it is not rational, then Baker Theorem tells us that it is transcendental if the power term is algebraic other rational.

2006-12-22 01:38:18 補充：
SORRY............term is algebraic other than rational.

2006-12-22 21:04:47 補充：
SORRYIt is sufficient to use Gelfond–Schneider theorem only.Baker Theorem is a more generalized theorem.I am sorry that I used to refer Baker Theorem.

2006-12-22 21:05:12 補充：
http://en.wikipedia.org/wiki/Gelfond-Schneider_theorem

係就係實數,但你要知道佢係唔確定的,
因為:
e^[i*(4k+1)π/2]=cos[(4k+1)pi/2]+isin[(4k+1)pi/2]=i
k為整數
所以:
i^i=e^[-(4k+1)π/2],隨k的不同而不同,為一單對多的映射,所以要睇你點去定義佢.

Sorry, I can't reply that question with a definite proof right now, my bet is that most likely it is NOT, by my instinct . But I would really like to ask the previous person who replied this question as claiming e^(-pi/2) as rational (i.e. can be written as p/q where p, q are both integer), is it possible for you to provide me the p and q that you have found?

2006-12-20 20:26:55 補充：
貓朋 - 可以解釋清楚少少點解e^(-pi/2)是超越數嗎?