Can 0^(ia) Have a Definite Solution for Positive Real a?

[intangible]

Can there be a definite solution for positive real values of a?

My musings so far:

$$0^{ix} = e^{ix\ln0} = (\cos x + i\sin x)^{\ln0} = ( \cos x + i\sin x )^{-\infty} = \left( e^{-\infty} \right)^\ln( \cos x + i\sin x ) } = 0^{\ln( \cos x + i\sin x )}$$

Since 0^0 can be assigned to produce either 1 or 0 depending on the context and for non-zero real values 0^a is always zero, I'm wondering how to deal with this.

Consider the following:

$$0^0 = e^{0\ln 0} = \left( e^0 \right)^{\ln 0} = 1^{-\infty} = 1$$
for $$1^a = 1$$ for all real numbers a

$$0^{a+bi} = e^{(a+bi)\ln0} = e^{a\ln0}(...) = \left (e^{-\infty} \right) ^a (...) = 0^a (...) = 0(...) = 0$$
for $$0^a = 0$$ for a belongs to non-zero real,
hence 0 for all real a,b>0 combinations.

Please share your expertise!

Regards,
intangible

 

[Gib Z]

What are you asking :S? Are you asking why 0^0 is 0 is some cases and 1 in others?

My two cents; In your first line of working, we can get from the first term to the last term in almost one step, knowing exp(ix) = cos x + i sin x. That way, we can avoid dealing with your negative infinities :(

On the second line of working, same thing.

Rather than say $$e^{0 \ln 0$$, which has no meaning, take the limit as x ---> 0, of $$e^{x \ln x}$$. If you have studied about the order of magnitude of functions, you will know that as x --> 0, f(x) = x goes to zero much faster than g(x) = ln x, goes to negative infinity. The limit is 1. And yes, once again, we avoid the infinites =]

I'm not too sure what happened in the third line of working, could you perhaps post that in full? Thanks

 

[HallsofIvy]

First, 0⁰ is not "0 in some cases" and "1 in others". 0⁰ is not defined. Often we are really dealing with limits of the form f(x)^g(x), f and g both having limit 0. In many useful cases, those limits are 0 or 1.

Second, your calculation breaks down as soon as you assert that $ln(0)= -\infty$. That is also not true. ln(0) is not defined- it is no number at all. It is true that in the limit, as x goes to 0, ln(x) "goes to $-\infty$", but you cannot say that it is equal to that. Once you start using "shorthand" as if it were the real thing, it's no surprise you run into difficulties.