What is the value of $1^i$? $\,$
Asked
Active
Viewed 5,845 times
3 Answers
75
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.
Carl Mummert
- 81,604
3
$1^{i} = e^{\log(1) i} = e^{0}=1$
Ilmari Karonen
- 26,117
Mykie
- 7,037
-
this sounds good to me too, since 1 is real why isnt $\log(1)$ zero? see also previous comment in other answer – Nikos M. May 27 '14 at 23:19
0
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.
Laurent Lessard
- 2,004
-
4This 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-Álvarez Aug 31 '10 at 02:32
-
-
2+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 09:38
-
Of course the answer will depend on the definition of $1^i$. However, the convention of using the principal branch of the logarithm is not nearly as well established as using the positive square root for $\sqrt{2}$. We usually point it out explicitly when we use the principal branch when doing a computation with complex logarithms. Also, it's a common exercise in complex analysis texts to compute all the possible values of $2^i$ or $i^i$. The case of $1^i$ is completely parallel to those from the viewpoint of the definition of complex exponentiation. – Carl Mummert Aug 31 '10 at 11:30
-
Sure, its a standard exercise to compute sets $\exp(i\cdot exp^{−1}(2))$ and $\exp(i\cdot \exp^{-1}(i))$, involving pre-images of 2 and i under the exponential function. But to pretend that one should routinely consider 'logarithm' functions per se, which aren't monotone functions on the reals --- so that consequently expressions such as $x \mapsto 2^x$ also may not be monotone functions on the reals --- seems quite fanciful. There is more than one way to choose such a 'logarithm' function on ℂ, sure; but for each one, the corresponding definition of $1^x$ yields the same constant function. – Niel de Beaudrap Aug 31 '10 at 12:37
-
4@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
-
4@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 01 '10 at 06:27
-
@Carl: “Complex exponentiation is only ever well-defined relative to a choice of a branch of the logarithm.” – Absolutely true. So we should pick a branch (there is a common way of doing that, so if that choice does what we need, we should), then we can call the logarithm and $(a,b) \mapsto a^b$ a “function” again and continue with more interesting questions. – Christopher Creutzig Sep 01 '10 at 07:42
-
This sounds good to me. Another of these paradoxes (classical) maths are full of – Nikos M. May 27 '14 at 23:18
-
-
-
Dont you miss (and i also) and $i$ in the exponent? So this should be $e^{-2i\pi}$ which is 1? – Nikos M. May 27 '14 at 23:29
-
@MarianoSuárez-Alvarez, sorry got confused. Why since log(1) has real argument, is not always zero? – Nikos M. May 27 '14 at 23:35
-
@Carl Mummert, the example you provided with $1^1/2$ equals -1, is very wrong and i will explain why. First since someone can just interpret it as complex operation or real operation equally, would result in 2 different outcomes, with exactly the same notation and values. This is surely not what you would like to call mathematical clarity? – Nikos M. May 27 '14 at 23:43
-
@Nikos M.: there is nothing that we can do about the fact that the notation $1^{1/2}$ is ambiguous. – Carl Mummert May 28 '14 at 02:37
-
@Carl will have to disagree, yes we can and in fact other answers (although yours is good) have answered this part also – Nikos M. May 28 '14 at 02:39
-
@Carl assuming one to any power to be always one, would it result in some other part of the existing mathematics become scrambled? – Nikos M. May 28 '14 at 02:43
-
I guess if you want to define it differently and make it multivalued, nobody is going to stop you... I just don't think there is any reason for it. The reason log(z) is multivalued is because exp(z) is periodic, and we define $y = \log(z)$ such that $\exp(y) = z$. Making $a^b$ multivalued for real positive $a$ makes no sense.
– Laurent Lessard Sep 01 '10 at 00:40