Is 0^i undefined?

[Time Commander Bob]

0^x equals 0, except when x=0 or if x is negative. However, I think I may have found another exception to this rule: 0ⁱ, that is 0 raised to the imaginary unit.
Since:
xⁿⁱ=cos(lnxⁿ)+isin(lnxⁿ)
Then if one substitutes x=0 and n=1, one will get:
0ⁱ=cos(ln0¹)+isin(ln0¹)
The problem is that ln0 is in itself undefined, since e^x cannot equal 0 no matter what finite value of x is used. So does this prove that 0ⁱ is undefined, or have I made some sort of mistake/presumption somewhere?

[Simetrical]

In complex analysis, typically one defines a^z = e^(z log(a)), for some specific value of log(a) that's specified in advance (since the complex log is multivalued). However, obviously this doesn't work at all if a = 0. That's the one complex number where log isn't defined (since e^z is never 0). In that case we just define 0^z = 0, except perhaps for z = 0.

In that light, your paradox doesn't require z = i. It will work for any z at all, except 0. It even works for real z, if your definition of 0^z is e^(z log(0)). So for 0 we just use a different definition.

[Time Commander Bob]

In complex analysis, typically one defines a^z = e^(z log(a)), for some specific value of log(a) that's specified in advance (since the complex log is multivalued). However, obviously this doesn't work at all if a = 0. That's the one complex number where log isn't defined (since e^z is never 0). In that case we just define 0^z = 0, except perhaps for z = 0.

In that light, your paradox doesn't require z = i. It will work for any z at all, except 0. It even works for real z, if your definition of 0^z is e^(z log(0)). So for 0 we just use a different definition.

So why is 0^z defined as 0? Is there some sort of proof you could show me for this? (not questioning the veracity of what you state, merely that I like to know why things are the way they are in mathematics)
Presumably then, 0^-z is undefined.

[smithy17]

I think 0^z is defined as 0 because otherwise we wouldn't be able to use powers like in the real numbers. If 0^i was not zero then (0^i)^-i wouldn't be 0 either, but that is the same as 0^1 which is zero.

[Simetrical]

So why is 0^z defined as 0? Is there some sort of proof you could show me for this? (not questioning the veracity of what you state, merely that I like to know why things are the way they are in mathematics)
Presumably then, 0^-z is undefined.

Well, there's no question of proof. It's just a definition. Thinking about it, though, I'm surely wrong. Powers of zero don't come up much, but 0^z = 0 would be inconsistent. In particular, you'd have 0^(-1) = 0 when you really want 0^(-1) = infinity. I'm not sure what the convention is, in that case. I suppose you're probably right, 0^z is most likely undefined for z not real. Non-real powers of zero don't come up much. Usually it's only integer powers of zero.

(Saying 0^z = 0 for all nonzero complex numbers z, but 0^(-z) is undefined, doesn't make sense. -z is a nonzero complex number too, after all, so then 0^(-z) would have to be 0.)

I think 0^z is defined as 0 because otherwise we wouldn't be able to use powers like in the real numbers. If 0^i was not zero then (0^i)^-i wouldn't be 0 either, but that is the same as 0^1 which is zero.

On the other hand, if 0^i = 0, then (0^i)^i = 0^i = 0, but 0^(i^2) = 0^(-1) = 1/0 is certainly not 0. So it's not consistent either way. Exponentiation in the complex numbers is messy. I think the conventional answer here is probably as Bob originally said, that 0^z is only defined for z a nonnegative real.