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What is the value of $1^i$? $\,$

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up vote 51 down vote accepted

First, a concrete example of things that can happen with complex exponentiation if you aren't careful: $1 = e^{2\pi i}$, so we can naively try to compute $1^i = (e^{2\pi i})^i = e^{(2\pi i)i} = e^{-2\pi}$.

The formal moral of that example is that the value of $1^i$ depends on the branch of the complex logarithm that you use to compute the power. You may already know that $1=e^{0+2ki\pi}$ for every integer $k$, so there are many possible choices for $\log(1)$.

The textbook definition of complex exponentiation states that $1^i = e^{\log(1)i}$ where "log" is a branch of the complex logarithm.

Now $e^{(2ki\pi)i} = e^{-2k\pi}$. If you take $k = 0$, which corresponds to using the principal branch of the logarithm, you get an answer of $1^i = e^0 = 1$. If you take $k = 2$ as in the example above, using the fact that $1=e^{2\pi i}$, you get $1^i = e^{-2\pi}$. There are infinitely many possible values for $1^i$, corresponding to different branches of the complex logarithm.

The confusing point here is that the formula $1^x = 1$ is not part of the definition of complex exponentiation, although it is an immediate consequence of the definition of natural number exponentiation.

A second point that can be confusing is that the function $e^z$ used above is really the complex exponential function $\exp(z)$, which is defined by a power series. Otherwise, the definition of complex exponentiation would be circular.

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Thanks for catching my mistake about the fact that the branches do matter. I deleted my response since it a) it was wrong and b) it's now unneccesary in light of your answer. You may wish you modify your paragraph "As another response has pointed out....", though. Sorry for the inconvenience. –  Jason DeVito Aug 30 '10 at 22:47
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I think this solution is wrong -- the function $a^b$ is single-valued and unambiguously defined whenever $a$ is real and positive. I posted a solution with more detail. –  Laurent Lessard Aug 31 '10 at 8:55
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There's no exception in the definition of complex exponentiation for the case when the base is positive and real, or for the case when the base is 1. For example, it's quite common in complex analysis to distinguish between the single-valued function $\exp(z)$ and the multiple-valued function $e^z=\exp(z\log(e))$. It is true that you can make the exponential single-valued by picking a particular branch of the logarithm, but any such choice is somewhat arbitrary. The complex exponential itself is inherently multivalued, like the complex logarithm. –  Carl Mummert Aug 31 '10 at 11:24
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@LL: I think you may be confusing the multivalued complex function $\exp(z\log(e))$ with the single-valued complex function $\exp(z)$. Both of these functions, unfortunately, can be denoted $e^z$, and the only term I know to refer to either one is "exponential function". In my post of 11:24 I am referring to the multivalued one. Many texts in complex analysis mention this notational issue, usually so that the authors can say that $e^z$ will be used to refer to $\exp(z)$ in their book. But the original question is not about the function $\exp(z)$, it is about $\exp(i\log(1))$. –  Carl Mummert Aug 31 '10 at 19:32
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Every real number is (identified with) a complex number. One could equally well point out that $i$ is not a real number, and so the usual definitions of real-valued exponentiation in elementary calculus cannot handle $1^i$ because they are only defined for real inputs. The complex exponential function $z^\alpha$ is defined for all nonzero complex $z$, even those identified with real numbers. There is no other familiar definition that applies to the power $1^i$. If you had a definition of $z^\alpha$ that distinguishes real and complex inputs, how would you define a Riemann surface for it? –  Carl Mummert Sep 1 '10 at 2:40
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it's 1. 1 to the power of anything is 1.

Edit: I'll elaborate. In defining $a^b$, we know intuitively what to do in certain cases. When $a$ is a positive integer and $b$ is an integer, for example. This definition is easily extended to when $a$ is real and positive, and $b$ is a real number. In the case where $a$ is negative, or complex, we run into trouble... one way to get around this is to define the power in terms of the logarithm, as $a^b = e^{b \log a}$. Then, we have the issue that the logarithm is multivalued, so we could get different answers depending on which branch we choose. This is the approach discussed in Carl's solution.

I do not believe this proposed method applies for the case where $a$ is positive and real, which is the relevant case being discussed. In the case where $a$ is positive and real, $a^b$ is unambiguously defined as $a^b = e^{b \ln a}$, where $\ln a$ is the unique real number $x$ satisfying $e^x = a$. This definition works even in the case where $b$ is complex. So we can apply this to the original problem: $1^i = e^{i\log(1)} = e^0 = 1$ (Myke, you got my +1!). In fact, $1^z = 1$ for any complex $z$. This viewpoint is shared by MathWorld, WolframAlpha, and Wikipedia, for what that's worth.

I guess at some level, it's a question of semantics and preference. Why not use multiple branches and all that jazz for the case where $a>0$? Because (and this is my opinion), it's unnecessary, and doesn't fit with existing definitions. The exponential function $e^z$ is well-defined via a power series and I think everyone would agree that it is single-valued, even when $z$ is complex. It makes no sense to me that $a^z$ should be any different when $a$ is real and positive.

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This is only part of the truth. It is also true that $1^i=e^{-2\pi}$... In particular, your claim «1 to the power of anything is 1» is false—without extra context, at least. –  Mariano Suárez-Alvarez Aug 31 '10 at 2:32
    
I made a massive edit. –  Laurent Lessard Aug 31 '10 at 8:49
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+1. It is true that there are multiple functions which provide inverses for restrictions of the exponential function; but it is not true that "$1^i=1$ and also $1^i=\mathrm e^{-2\pi}$". At best, this is shorthand for the position that the absence of a 'distinguished' inverse function for $\exp(x)$ entails that there is no meaningful convention for interpreting the expression "$1^x$". But then one should also view the expressions $\ln(1)$ and $\sqrt 2$ with equal suspicion. --- In practice, we identify a distinguished inverse by convention; $1^i = 1$ for the same reason that $0^0 = 1$. –  Niel de Beaudrap Aug 31 '10 at 9:38
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@Niel de Beaudrap: Complex analysis book routinely consider arbitrary branches of the logarithm. I disagree with your last sentence, even when $x$ is real. If we take a branch of the logarithm where $\log(1) = 2\pi i$ then $1^{1/2} = \exp(\pi i) = -1$ which is not equal to $1^1 = \exp(2\pi i) = 1$. Complex exponentiation is not the same function as real exponentiation; they only agree for appropriate branches of the logarithm, and then that agreement has to be proved, it is not a definition. Complex exponentiation is only ever well-defined relative to a choice of a branch of the logarithm. –  Carl Mummert Aug 31 '10 at 13:51
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@Carl Mummert: Well, you've violated the (perhaps not clearly articulated) precondition of that sentence: selecting a log function which is monotone on the reals --- in particular, real on the reals. (As you show, violating this allowed $\sqrt 1$ to be negative, violating a convention which you seem to suggest as being independent, or "better established".) --- I doubt that this debate is resolvable; but I'm interested in notation with utility, which requires consistency with other utilitarian choices. If I wanted more branches, I'd refer to $\exp^{-1}$, which is what is really meant anyway. –  Niel de Beaudrap Sep 1 '10 at 6:27
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$1^{i} = e^{\log(1) i} = e^{0}=1$

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