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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?

--Background--

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? $\endgroup$ – астон вілла олоф мэллбэрг Jan 7 '17 at 6:34
  • $\begingroup$ Ok so for some $e^a = 0$ I can't just say $e^{i*a} = 0^i = 0$? $\endgroup$ – Goldname Jan 7 '17 at 6:36
  • $\begingroup$ @астонвіллаолофмэллбэрг Yeah I put in a background in the comment and answer. $\endgroup$ – Goldname Jan 7 '17 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. $\endgroup$ – астон вілла олоф мэллбэрг Jan 7 '17 at 6:41
  • $\begingroup$ 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 Peterson Jan 7 '17 at 6:44
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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. $\endgroup$ – spaceisdarkgreen Jan 7 '17 at 7:42

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