Real or Complex Number?
Is the value of i^i a Real Number or Complex Number?
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answers:
Φ² = Φ+1 say: i = e^(iπ(1/2 + 2k)) ∀k∈ℤ
iⁱ = (e^(iπ(1/2 + 2k)))^i ∀k∈ℤ
iⁱ = e^(-(4k+1)π/2) ∀k∈ℤ

iⁱ = {..., e^(7π/2), e^(3π/2), e^(-π/2), e^(-5π/2), e^(-9π/2), ...}
iⁱ ≈ {..., 59609.74, 111.3178, 0.2078796, 0.0003882032, 0.00000007249473, ...}
Real numbers
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sepia say: i^i
= 0.20787957635076190854695561983497877003...
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Jeffrey say: i^i = exp(-((pi/2) + 2*pi*n)); where "n" is an element of the integer set.

The value will always be real but no specific value
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oyubir say: i^i is not defined. It has no value. You cannot give unambiguous value to it.

Those who say exp(-pi/2) or approx. 0.20788 are implicitly defining a logarithm for complex number, which cannot exist (unambiguously) for the same reason.

It is tempting to say i=exp(i*pi/2) (that is exact. And non ambiguous. No reasoning can give you another value for exp(i*pi/2). i is the only coherent value. So it cost nothing to define exponential of complex numbers)
and therefore
i^i = exp(i*pi/2)^i = exp(i² * pi/2) = exp(-pi/2)

But then you have to keep in mind that exp(5pi/2) is also i.
Therefore i^i should also be exp(-5pi/2) = approx. 0.000388


and what about i^i * i^i then ?
It is (i^i)^2 or (i^2)^i or i^(2i)
That is 0.2079^2 or (-1)^i or i^(2i)
So, (-1)^i is 0.043214 then?

But (-1)^i * (-1)^i is also 1^i.
So 1^i = 0.00187?
Probem is, we already defined 1^i. It is 1. Because it is quite comfortable to have 1^anything=1. And because it is exp(i*ln(1)) = exp(0) = 1.


You have to keep in mind that all this is a matter of definition. There is no strict truth. No absolute answer obtained only by pure logic from the definition of complex.
So if you wish to say that i^i is 0.2079, you can. You can say, in any maths paper, say at the begining, "I define i^i as exp(-pi/2)".
But then, you also have to set a set of artificial, complicated rule, to lift the ambiguities I have given here (and those were only examples).
You could just use normal computation rules from there and get to a conclusion in your paper. Because unless you check every rule, your conclusion will be stained by the ambiguity of your definition, which means that it is possible to get different results from the same computation.

Reason why the only reasonnable thing to do is to say "we don't define i^i. i^i has no value"
Thanks to the fact that i^i has no value, we can blindly do any calculus, knowing that we still can trust well known rules such as a^b^c=a^(b*c) and a^c * b^c = (a*b)^c.
If you give a value to i^i, then you have also to decide a bunch of exceptions for those rules.



Short answer: there is no i^i
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Morningfox say: i^i is a real number, approximately 0.20787957635...
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Captain Matticus, LandPiratesInc say: It's real and equal to e^(-pi/2)

i =>
0 + i =>
cos(pi/2) + i * sin(pi/2) =>
e^((pi/2) * i)

i^i =>
(e^((pi/2) * i))^i =>
e^((pi/2) * i * i) =>
e^((pi/2) * (-1)) =>
e^(-pi/2)
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