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Wolfram Alpha returns nothing. I tried working out the algebra on my own and I am stuck. Let $0^i=x$. Then $i=\log_0(x)$. $i=\ln(x)/\ln(0)$. $\ln(0)=-\infty$. Any number $n$ over complex infinity is $0$ so I end up with $i=0$. Or I could multiply both sides by $\ln(0)$ which is $-\infty$. So I end up with $i\cdot(-\infty)=\ln(x)$. Yielding $x=e^{-\infty\cdot i}$. I know I took quite a few liberties there in defining functions but having no answer at all just seems strange. Any insights?

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    $\begingroup$ You are doing supreme algebraic gymnastics that has never been seen before. On a serious note : You also have to be very cautious about which operations you are doing. Some of the operations may not be legal. $\endgroup$
    – Someone
    Dec 22, 2019 at 7:12
  • $\begingroup$ You have done so many "beyond imagination" things., though I think $0^i$ is undefined because $0^0$ is undefined. Have a look: youtube.com/watch?v=12Nae7qYxs4 $\endgroup$ Dec 22, 2019 at 7:19

4 Answers 4

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The standard definition for exponentiation of complex numbers is

$$a^z=\exp(\ln(a^z))=\exp(z\ln(a))$$

Now, this definition works perfectly when $a\in\mathbb{R}$ and $a>0$. It still works when $a$ is a non-zero complex number, but you have to be careful to define which branch of the logarithm you are working with, something that might be above your level of expertise (I base this off your initial question). However, there is no way to define $\ln(0)$ as there is no complex number $z\in\mathbb{C}$ such that

$$e^z=0$$

But you ask, why isn't

$$0^i=\exp(\ln(0^i))=\exp(i\ln(0))=\exp(-i \cdot \infty)=0$$

since $\ln(0)=-\infty$ (I know, infinity isn't a number but it is useful for this demonstration). However, if this were the case, then

$$0=0^i=(0^i)^i=0^{-1}=\infty$$

which is a contradiction. Another way to look at is why isn't

$$\exp(-a\infty)=0$$

Well, if $a$ is a positive real number then it is zero. However, if $a<0$ then

$$\exp(-a\infty)=\exp(-(-|a|)\infty)=\exp(|a|\infty)=\exp(\infty)=\infty$$

That is, it depends on what is "multiplying" the infinity which determines what the final value is. Since $i$ isn't real, there is no way to get this final value.

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Well, exponentials regarding complex numbers are tricky to handle. How would you define an expression such as $a^b$ for complex numbers $a$ and $b$? The usual definition is $$a^b:=\exp(b\log a)$$ where $\exp$ is the exponential function and $\log$ is the complex logarithm---which is multivalued, so in order to make $a^b$ well-defined we must have chosen a branch. Here we run into a problem: there is no way to assign a meaningful value to $\log0$, hence no way to assign value to such an expression as $\exp(i\log0)$. So the expression is indeterminate.

You might ask: how is this consistent with the fact that $0^x=0$ for any real $x>0$? Well, if we think of $0^x=\exp(x\log0)$, along the positive real axis it makes sense to think of $\log0$ as essentially $-\infty$ (more precisely: $\lim_{\epsilon\to 0^+}\log\epsilon=-\infty$), and so $\exp(x\log0)$ can be thought of as essentially $\exp(-\infty)$ which is essentially $0$ (again more precisely, $\lim_{\epsilon\to-\infty}\exp\epsilon=0$). But there is no such thing in the complex plane. If we now treat $\log0$ as "essentially $-\infty$" we will end up with an expression such as $0^i=\exp(-i\infty)$. The only meaningful way to interpret this is as the limit $\exp(-i\infty)=\lim_{z\to\infty}\exp(-iz)$, but this simply doesn't exist.

In summary: $0^a=0$ for real $a>0$ because we can say $0^a=\exp(a\log0)$ which is "essentially" $\exp(-a\infty)$ which is "essentially" $0$ if $a$ is a positive real. In the general case for complex numbers $a$ however, this expression is not well-defined.

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The function $f:\mathbb{R}^+\to\mathbb{C}$ defined by $f(x)=0^x$ for positive real numbers $x$ has a unique analytic continuation to a function $f:\mathbb{C}\to\mathbb{C}$, namely $f(z)=0$ for all $z$. So in that sense you can say that $0^i=0$. But of course you've relinquished some of the other conventions of exponential notation, such as $0^{-1}=1/0=\infty$.

Remark: In going from $0^i=x$ to $i=\log_0(x)$, the OP implicitly assumes the function $0^z$ is invertible. But as this answer shows, that's not the case.

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There are various answers already addressing the issue, but I will try to go into more depth to provide a satisfying answer.

When it comes to $x$ being a real number (or more generally, an element of a monoid in $\cdot,$) defining $x^n$ is very straightforward if $n$ is a natural number (or $0,$ but in higher-level mathematics, $0$ is, more often than not, also treated as a natural number). We say that $$x^n:=\prod_{m\in{n}}[x],$$ where $[x]$ is the constant sequence whose output is $x.$ This can easily be extended to integers $n,$ as long as $x$ is an element of a group. For real numbers, this translates to $x$ being nonzero. However, if we try to go any further and extend this for rational $n$ instead, things begin to fall apart, and we no longer have a good way of giving a definition without achieving some undesirable situations or inconsistencies. Still, when it comes to complex to complex numbers $x,$ we try to make the best of it, and we work in terms of roots of unity and the $n$th root of a complex number. However, it becomes completely hopeless if we try to extend it to real exponents, let alone complex exponents, and let us not mention exponents in other algebraic structures. The only way to make these latter work is if we restrict what $x$ can be to a specific set of values, which is not satisfying, and what that set of values is may not be unique, and even if we do the restriction, there may not be a unique value that exists that one can assign to $x^c$ when $c$ is real or complex, even despite the restrictions. So working with expressions like this is something mathematicians tend to avoid, due to how inconvenient it is.

Nonetheless, you may see that in some circumstances, the definition $z^w=\exp[w\log(z)],$ provided a suitable definition of $\log,$ is quite useful, but this comes at the cost of defining $0^z.$ Here, we arrive at a situation where the definition of exponentiation based on rational or integer exponents is just incompatible with the definition for complex eponents. The expression $0^i$ is right where these incompatible definitions meet each other, and because of this, none of the definitions can possibly work for it. $i$ is a complex number, but $\exp[i\log(0)]$ does not exist, and none of the previous definitions give an unambiguous extension of $0^x$ to $0^i.$ Because of this, the notation is just extremely problematic.

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