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Thoughtexperiment: 0^(ia)

  1. Jan 24, 2008 #1
    Can there be a definite solution for positive real values of a?

    My musings so far:

    [tex]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 )} [/tex]

    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:

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

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

    Please share your expertise!

  2. jcsd
  3. Jan 24, 2008 #2

    Gib Z

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    Homework Helper

    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 [tex]e^{0 \ln 0[/tex], which has no meaning, take the limit as x ---> 0, of [tex]e^{x \ln x}[/tex]. 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
  4. Jan 25, 2008 #3


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    Science Advisor

    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 [itex]ln(0)= -\infty[/itex]. 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 [itex]-\infty[/itex]", 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.
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