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(-#)^(pi) = NonReal Answer

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ok well the topic kind of explains my questoin but why is this?? dose this have to do with my calculator or dose it have to do with pi itself its been bothering me and this seemed like the right forum to ask even though it isnt related to games (well it could be as almost anything can) but thought id as thanks[grin]

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Taking the non-integer power of a negative number is usually basically the same as taking the square root of negative one. You get an imaginary (and thus non-real) number. The square root of -1 is the same as (-1)1/2. I don't know the formalities behind it, but it appears that all rational numbers with an even denominator (when the fraction is fully reduced), and all irrational numbers produce this behavior.

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Agony - Hrm. No. This is nonsense.

e = cos θ + i.sin θ  [by definition]

e = -1

(-x)y = x.(e)y = x.eiyπ = x.cos yπ + ix.sin iπ

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if y is rational, it can be written as:

y = p/q, q>0

=> (-x)^(p/q)

with a^(1/b) = b-th sqrt of a:

=> (-x)^(p/q) = q-th sqrt of (-x)^p

(-x)^p is a negative number for all uneven p.

=> for all uneven p there is no answer.

That should be the formalities for rational numbers, i hope ;)

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It comes from how powers is defined. As Agony said, (-1)^0.5 is imaginary number.

How power is defined on complex numbers:
Let we have complex number c that angle between +x axis and c is q and |c|=r
(Note that for positive numbers, q=0 and for negative, q=pi , and for imaginary, q=pi/2 )

and if we have complex number ck with angle q' and absolute value r' ,
r'=rk (r is always positive)
q'=(q+n*2*pi)*k where n is integer.

Note that in general, ck have infinitely many values. But in some cases, most of 'em is equal.

For positive c with zero imaginary part, there's only one purely real ck. For negative c, there may be real values and may not, depending to k... If k is irrational, there's infinitely many complex values of ck.

On computers, usually, only one specific value is choosen...

that is, c=r*eiq

Fruny: i think if op would know what's eiq is all about, there would be no question...

edit: looked up font table for theta, replaced alpha by theta for standartness, typos.
edit2: damn,i see theta as exactly like 8.
edit3:some HTML masochism.
edit4: damn, WTF font size="+1" and bold is so big?

[Edited by - Dmytry on November 10, 2004 3:24:44 PM]

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so really my questoin had nothing to do with pi it has to do with no int powers being represnted as fraction like so:

woa turn down your font ok i get it chill out geeze

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for any integer k:
(-1)^(2k) = 1
(-1)^(2k+1) = -1

and inversely : x^(-y) = x^(1/y)

-1 = (-1)^(1/(2k+1))
such as (-1)^(1/3) = third root of -1. which asks: what number, times itself, and times itself again equals -1? answer: -1.
-1*-1*-1 = -1


(-1)^(1/2k) such as (-1)^(1/2) = sqr(-1). which asks the question: what number, times itself, equals -1? try all day, but you will never find a real solution. so, the representation we have for this is the imaginary number i = sqr(-1).

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