What is the value of $1^i$?

Question

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

Answer 1

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.

Answer 2

$1^{i} = e^{\log(1) i} = e^{0}=1$

Answer 3

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.