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$$\lim_{n\to 0} n^{i} = \lim_{n\to 0} e^{i\log(n)} $$ I know that $0^{0}$ is generally undefined, but can equal one in the context of the empty set mapping to itself only one time. I realize that in terms of the equation above, the limit does not exist, but can $0^{i}$ be interpreted in a way to assign it a value? For the curious, I ran in to this when trying to calculate the imaginary-derivative of $\sin(x)$.

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  • $\begingroup$ Well this might be related, but wolframalpha currently gives $\frac{d}{dx} 0^x= 0^x(-\infty)$ $\endgroup$
    – Teoc
    Jan 12, 2015 at 3:48
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    $\begingroup$ In the context of natural numbers and finite combinatorics it is generally safe to adopt a convention that $0^0=1$. Extending this to a complex arithmetic context is fraught with risks, as is the ambition to justify limits of this form generally by analogy to the value of a particular limit of this form. The derivative of the complex-valued sine function is everywhere well-defined. $\endgroup$
    – hardmath
    Jan 12, 2015 at 3:59
  • $\begingroup$ I was not taking the derivative of the complex sine function, but the imaginary-order derivative of sine: $$ \frac{d^{i}}{dx^{i}} \sin(x) $$ $\endgroup$
    – soultrane
    Jan 12, 2015 at 4:15
  • $\begingroup$ @Lubin that's pretty much the answer, why don't you post it as it? $\endgroup$
    – hjhjhj57
    Jan 12, 2015 at 4:16
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    $\begingroup$ @hjhjhj57 the topic starter already indicated that the limit does not exist, so this comment does not add anything. He was asking what value can be assigned to it nevertheless. The mean value of the sequence is $0$ (see my ansswer). $\endgroup$
    – Anixx
    Jan 12, 2015 at 4:31

4 Answers 4

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This was a comment, but @hjhjhj57 suggested that it might serve as an answer.

If you write the right-hand side of your equation as $\lim_{t\to−∞}e^{it}=\lim_{t\to−∞}(e^i)^t$, it’s completely clear that the limit doesn’t exist: you’re taking the number $e^i$, which is on the unit circle, and raising it to a large (but negative) power. You have a point that runs around the unit circle infinitely many times as $t\to−∞$, no limit at all.

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    $\begingroup$ Hope this isn't too late of a response. I was looking at the problem again, and I am now wondering if it is relevant that $|{0^{i}}| = 1$, so the magnitude is convergent, but the phase is not. $\endgroup$
    – soultrane
    Jan 15, 2015 at 3:40
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    $\begingroup$ @soultrane, what you say is surely correct, and maybe we can say that it isolates somewhat the reason for nonconvergence. $\endgroup$
    – Lubin
    Feb 14, 2016 at 21:27
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It is possible to interpret such expressions in many ways that can make sense. The question is, what properties do we want such an interpretation to have?

$0^i = 0$ is a good choice, and maybe the only choice that makes concrete sense, since it follows the convention $0^x = 0$. On the other hand, $0^{-1} = 0$ is clearly false (well, almost—see the discussion on goblin's answer), and $0^0=0$ is questionable, so this convention could be unwise when $x$ is not a positive real.


Digging deeper: One generally defines complex exponentiation as a multi-valued function: if $e^c = a$, then we can define $a^x = e^{cx}$. This is not unique, since it depends on the choice of $c$, but it's a good way to think about quantities like $i^i$ (this is sometimes claimed to be $e^{-\pi/2}$, but it can be interpreted as $e^{-\pi/2 + 2\pi n}$ for any $n\in\mathbb{Z}$)

This approach breaks down for $0^i$, because $0$ has no natural logarithm in the complex numbers. However, if we're comfortable calling $-\infty$ (or $-\infty + 2\pi i n$) a natural log of $0$, then we can say that $0^x = e^{-\infty \cdot x} = 0$ when $x$ has positive real part.

When $x$ has negative real part, this leads us to regard $0^x$ as a quantity with infinite magnitude and undefined argument. When $x$ is imaginary, the argument is still undefined, but the magnitude is multi-valued rather than infinite.


My conclusion is that we should avoid assigning meaning to $0^i$.

Writing $|0^i| = 1$ may be sensible, however, under some circumstances.

In a general setting, I would be comfortable saying that $|0^i| = e^{2\pi n}$, for any $n\in\mathbb{Z}$.

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    $\begingroup$ "follows the convention $0^x=0$" - where u got this "convention" from? "and $0^0=0$ is questionable" - it is not questionable, it it plainly wrong, it is usually define to be 1 or left undefined. $\endgroup$
    – Anixx
    Jan 12, 2015 at 7:28
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    $\begingroup$ @Anixx $0^0 = 1$ is also plainly wrong, but may be used in some circumstances (like power series evaluation) as a consequence of the formal identity $a^0 = 1$, which is valid for $a\neq 0$. By the same token, $0^x = 0$, which is valid for $x>0$, may be used in circumstances where it is regarded as a formal identity. $\endgroup$ Jan 12, 2015 at 7:38
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    $\begingroup$ No, it is not wrong, it is widely accepted defginition. Prove it is wrong. $\endgroup$
    – Anixx
    Jan 12, 2015 at 7:48
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    $\begingroup$ @Anixx Where to begin? Most introductory calculus texts have a section on the problems with $0^0$, because it leads to some of the more common mistakes (from my experiences teaching calculus, anyway). For example, $\lim_{x\to 0^+} e^{-1/x} = 0$ and $\lim_{x\to 0^+} x = 0$, but $\lim_{x\to 0^+} (e^{-1/x})^x \neq 1$. $\endgroup$ Jan 12, 2015 at 10:57
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    $\begingroup$ @Anixx Of course, $0^0 = 1$ is quite useful when doing "standard" algebra, e.g. power series, the binomial theorem, etc. But these things all have their limitations. In the comments to goblin's answer, there is even a link to a paper with $0^{-1} = 0$, which turns out to be a completely reasonable definition in some scenarios of interest. $\endgroup$ Jan 12, 2015 at 11:05
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Wait.

The first reasonable way to understand if a good definition of $0^i$ can be given, is to define $$ 0^i:=\lim_{z\to0}z^i=\lim_{r\to0}(re^{i\theta})^i=e^{-\theta}\lim_{r\to0}r^i $$ But now $r^i=e^{i\log r}=\cos(\log r)+i\sin(\log r)$, and being $\log r\stackrel{r\to0}{\longrightarrow}-\infty$, we conclude the limit $\lim_{z\to0}z^i$ can't exist, thus $0^i$ can't be defined.

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We have:

$$\lim_{x\to0} e^{i \log x}=\lim_{x\to\infty} e^{-ix}=\lim_{x\to\infty} (-1)^{-x}=\lim_{x\to\infty} (-1)^{x}$$

The limit as you noticed, does not exist... But if you want to assign a value nevertheless... well, the mean value of $(-1)^n$ will be zero:

$$\lim _{x\to\infty} \frac 1{x}\int_0^x (-1)^x =0$$

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  • $\begingroup$ Why the downvote? $\endgroup$
    – Anixx
    Jan 12, 2015 at 4:06
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    $\begingroup$ In that case $$ e^{-i\infty} = e^{-i\infty \frac{2\pi}{5}} = \left(\cos \left(-\frac{2\pi}{5} \right) + i\sin \left(-\frac{2\pi}{5} \right)\right)^\infty . $$ $\endgroup$
    – hjhjhj57
    Jan 12, 2015 at 4:20
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    $\begingroup$ By writing $e^{-i\infty}$ you assume that the object $e^{-i\infty}$ exists. Then you assume that De Moivre's formula can be used when $x=\infty$ twice. It's not a very clear answer. By rephrasing it as a limit then all of these objections goes away. $\endgroup$
    – Winther
    Jan 12, 2015 at 4:24
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    $\begingroup$ That expression isn't even defined, it's plain and simple abuse of notation. We all know the dangers of handling limits as ordinary numbers. $\endgroup$
    – hjhjhj57
    Jan 12, 2015 at 4:26
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    $\begingroup$ Because, even if your conclusion is the same (or similar), and you've edited your post: 1.- I still don't like the "$(-1)$" thing. 2.- He was very careful (or careful enough) when dealing with infinity, and the idea he's defending is $0^i=0$, which is nowhere mentioned in your answer. $\endgroup$
    – hjhjhj57
    Jan 12, 2015 at 14:53

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