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I looked up $i^i$ and it is said that it is equal to $\exp(-\pi/2)\approx 0.20787$.

But I tried the following: $$i^i=\exp[\ln{(i^i)}]=\exp[i\ln(i)]=\exp(i\ln(e^{i(\pi/2+2\pi k)}))$$ $$=\exp(i\cdot i(\pi/2+2\pi k))=\exp(-\pi/2-2\pi k)$$

I also get $\exp(-\pi/2)$ as a solution but also get infinite many solutions involving the parameter $k$. I know if you restrict your domain on $0$ to $2\pi$ the only solution is $\exp(-\pi/2)$, but mathematically it doesn't make sense to me to restrict yourself only to get a unique solution. Where is my mistake in my reasoning?

EDIT: Such an expression should only have one "true" answer. Restricting the domain to smaller regions seems to me as the statement which is less "true". So I would like to hear arguments from you, why the one or the other is the "true" expression.

$i^i=\exp(-\pi/2)$ is not true, because as $i^i=\exp(-pi/2+2\pi k)$. It is not true because it is not a complete description. To make my point clear. Lets consider $x^2-1=0$. If you say the solution to this equaiton is $x=1$ then this is wrong, as $x=-1$ is also a solution.

Beeing able to define functions is not an argument in my opinion, as I am only trying to calculate a value. We also accept $x^2+y^2=r^2$ as a representation for the circle.

Uniqueness is also not a proper argument in my opinion, as $\sqrt{a^2}$ is also not unique.

I am looking forward for some insightfull answers.

MrYouMath
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    Possible duplicate of Prove that $i^i$ is a real number – J. M. ain't a mathematician Jun 03 '16 at 11:52
  • I don't want to prove it. I want to know why my answer is different. – MrYouMath Jun 03 '16 at 11:54
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    The fact of the matter is simple: there are infinitely many possible values for $i^i$, and none of them is any more correct than any other one. There is no "true" value of $i^i$, really. Actually, the same issue also arises with $1^i$, as I wrote about here: http://math.stackexchange.com/questions/3668/what-is-the-value-of-1i/3674#3674 – Carl Mummert Jun 03 '16 at 12:21
  • Thank you alot. But by "true" I dont mean values. If you have cubic polynomial, you look for all solutions not only one solution. And using $\exp(-pi/2-2\pi k)$ determins all possible solutions. – MrYouMath Jun 03 '16 at 12:24
  • Your insistence that there must be only one "true" answer is simply misguided. The expression $i^i$ represents a countable collection of distinct values, and it's up to you -- or, more frequently, the mathematical community -- to pick a principal value. There's no reason whatsoever that we should choose $2$ rather than $-2$ as the value for $\sqrt{2}$ other than the fact that it is accepted convention. Same thing. It may seem more "true" to most people, but that's because we have a learned bias toward choosing a positive number over a negative number for the principal value. – MPW Jun 03 '16 at 12:27
  • @MPW: please read my previous comment. – MrYouMath Jun 03 '16 at 12:29
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    I think it isn't clear what your question is. The expression $i^i$ represents each of the values $${\cdots,e^{-\frac92\pi},e^{-\frac52\pi},e^{-\frac12\pi}, e^{\frac32\pi}, e^{\frac{7}{2}\pi}, \cdots }$$ – MPW Jun 03 '16 at 12:35
  • @MPW: You are right it is hard to describe :). I think I will delete this question, because I don't think that I will get an appropriate answer for this question. But I will wait 1 or 2 hours, maybe I am lucky :D. – MrYouMath Jun 03 '16 at 12:36
  • Leave the question, don't delete it. It doesn't matter if you think there is an appropriate answer, it is instructive for people to read the question and the comments and answers. – MPW Jun 03 '16 at 12:39

1 Answers1

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You must fix a determination of your argument if you want a unique answer. For example choose the principal determination which is $\arg(z) \in (-\pi,\pi].$ It will determine a unique logarithm and hence a unique "power function".

With that determination, $$i^i = e^{i\ln(i)}=e^{i(\ln|i|+i\arg(i))}=e^{-\pi/2},$$ since the principal argument of $i$ is $\pi/2$.

I would add that precisely it makes sens to restrict yourself yo get a unique solution. The concept of multi-valued function is mathematically dangerous and leads to a lot of confusion. (Unless we use more advanced tools like Riemann surfaces.)

C. Dubussy
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  • -1: But I dont want a unique answer, why should I force an expression to be unique? There is no obvious reason for me to do this. I allready said, that I could restrict my domain to get a solution. – MrYouMath Jun 03 '16 at 12:00
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    Because if you want a function, then you must have a unique answer. Of course if you absolutely want to use a multi-valued function, then your answer is true. – C. Dubussy Jun 03 '16 at 12:02
  • There should only be one "true" solution to this problem. And I want to know your arguments why $\exp(-\pi/2)$ deserves to be the "true" solution. – MrYouMath Jun 03 '16 at 12:05
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    The correct statement is "there is a true solution for each determination of the argument". $\exp(-\pi/2)$ is the true solution (by definition of the power) if you choose the principal determination. Because it is "the most common used", generally we consider that it is the true solution. It is just an agreement. – C. Dubussy Jun 03 '16 at 12:09
  • I dont consider "Agreements" and "most common used" as mathematical arguments. – MrYouMath Jun 03 '16 at 12:14
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    This is related to those questions we get from time to time: how come my calculator says the cube root of $-1$ is a complex number and not $-1$? – GEdgar Jun 03 '16 at 12:20
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    @MrYouMath : Of course you are free to use the mathematical definition you want. I just gave you the most common one and explain why you will find the value $i^i = e^{-\pi/2}$ in most of books. If you prefer to define $i^i$ as the set ${\exp(-\pi/2+2\pi k) : k \in \mathbb{Z}}$ that's your right. But in general it is convenient to take the definitions used by most mathematicians. The point is "You can make mathematical rigourous arguments once you have fixed the definitions and this part remains arbitrary". – C. Dubussy Jun 03 '16 at 12:24