# Is zero to the power of an imaginary number still zero?

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.

• 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? Jan 7, 2017 at 6:34
• Ok so for some $e^a = 0$ I can't just say $e^{i*a} = 0^i = 0$? Jan 7, 2017 at 6:36
• @астонвіллаолофмэллбэрг Yeah I put in a background in the comment and answer. Jan 7, 2017 at 6:37
• (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. Jan 7, 2017 at 6:41
• 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. Jan 7, 2017 at 6:44

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.
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.