I just want to make sure that $0^i = 0$, but for some reason I couldn't find anything about this online.

Is this true?


I'm trying to prove that some exponent is zero. I thought I'd raise each side to the power of $i$ so that I could use Euler's formula.

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    $\begingroup$ I do not think $0^i$ is even defined. You can complex exponentiate only positive real numbers, as far as I know. It would help if you added context to your question i.e. what were you working upon when you hit this question? $\endgroup$ Jan 7, 2017 at 6:34
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    $\begingroup$ Ok so for some $e^a = 0$ I can't just say $e^{i*a} = 0^i = 0$? $\endgroup$ Jan 7, 2017 at 6:36
  • $\begingroup$ @астонвіллаолофмэллбэрг Yeah I put in a background in the comment and answer. $\endgroup$ Jan 7, 2017 at 6:37
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    $\begingroup$ (Though I honestly wished you could) you can't. This is because, the function $e^z$ does not have zero in it's range! The fallacy is in the fact that there is not even a complex $a$ such that $e^a=0$. The rest is absolutely fine, but then take a minute: what would be the nature of the exponential function if $e^a=0$ for some $a$ (which would be non-zero, hence invertible)? Then, $e = e^1 = {e^a}^{\frac 1a} = 0^ {\frac 1a} = 0$, which can't happen! So this is the discrepancy. Hence, $0^i$ can't actually be defined properly. $+1$ for adding context to your question, though. $\endgroup$ Jan 7, 2017 at 6:41
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    $\begingroup$ I believe the standard is that $0^k$ is "undefined" unless $\Re(k)$ is positive. $0^{a+bi}=0^a0^{bi}$. If $a$ is negative, $0^a$ is not defined. If $a=0$ there is more than one interpretation but it is commonly taken to be undefined. $\endgroup$
    – David P
    Jan 7, 2017 at 6:44

2 Answers 2


Standard definition of a complex exponent $w^z$ is $$ w^z=e^{z\log(w)} = e^{a\log(w)}e^{bi\log(w)}$$

where $z = a +bi.$ For $w = re^{i\theta},$ this gives $$ w^z = e^{a\ln(r)}e^{ai\theta}e^{bi\ln(r)}e^{-b\theta}. $$ For $w = 0,$ we have $r=0$ and $\theta =$ whatever. $\ln(0)$ is undefined but we can take the limit $r\downarrow0$ so that $\ln(r)\downarrow-\infty.$ The limit of the expression for $w^z$ is only zero if $a >0,$ otherwise it blows up or oscillates. In other words if $\Re(z)>0$ then $0^z=0.$ Otherwise it's infinite or undefined.

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    $\begingroup$ Initially made a stupid mistake that made me think I had more to say than what was covered in the comments. Turns out i didn't. $\endgroup$ Jan 7, 2017 at 7:42

Let’s consider the general expression $x^i$. We can say that: $$ x^i = e^{i \ln(x)} $$

and, according to Euler’s formula $e^{i \theta} = cos(\theta) + i \sin(\theta)$, we can say that: $$ x^i = e^{i \ln(x)} = \cos(\ln(x)) + i \sin(\ln(x)) $$ Now, if $x = 0$, we have: $$ 0^i = \cos(\ln(0)) + i \sin(\ln(0)) $$

$ln(0)$ doesn’t exist, but, if we take the limit, we can say that $ln(0) \to - \infty$ Therefore: $$ 0^i = \cos(-\infty) + i \sin(-\infty) $$ But the functions $\sin(x)$ and $\cos(x)$ don’t have limits when $x$ approaches to infinite, since they oscillate. Therefore, $\cos(-\infty) + i \sin(-\infty)$ is undefined, hence $0^i$ is undefined and it’s not equal to 0. For the same reason $\infty^i$ and $(-\infty)^i$ are undefined too.


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