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Exponentiation with zero base, complex exponent?

  1. Aug 7, 2014 #1

    bcrowell

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    Is there a good general definition of [itex]0^z[/itex], where z may be complex? The cases where z is real (and positive, negative, or zero) are straightforward, but what if z isn't real? Are there arbitrary branch cuts involved, or is there some universal definition?
     
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  3. Aug 7, 2014 #2

    Matterwave

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    Is it not legitimate to use:

    $$0^z=0^{a+bi}=0^a0^{bi}=0\quad\text{for}~~a>0$$
    ?
     
  4. Aug 7, 2014 #3
    What does $$0^{bi}$$ mean?
     
  5. Aug 7, 2014 #4

    Matterwave

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    Hmm, I just figured either the answer is indeterminant for all non-real ##z## or it would be ##0## for all ##z|\text{Re}(z)>0##.
     
  6. Aug 8, 2014 #5

    mathman

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    $$0^{bi}=e^{biln0}$$

    This is indeterminate with magnitude = 1.
     
  7. Aug 8, 2014 #6

    Matterwave

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    As long as the magnitude = 1, even if the phase is indeterminant, can we say that the product, and therefore final answer is 0? Or do we just say the function is indeterminant for all non real z?
     
  8. Aug 8, 2014 #7

    bcrowell

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    Amplifying on what mathman said, it seems like any expression of the form [itex]0^{bi}[/itex] is an indeterminate form. For example, suppose [itex]\epsilon>0[/itex] and [itex]\delta[/itex] are both real. Then for

    [tex]\lim_{\epsilon\rightarrow0^+,\ \delta\rightarrow0}|\epsilon^{\delta+i}|=e^{\delta\ln\epsilon}[/tex]

    has a value that depends on how rapidly [itex]\epsilon[/itex] and [itex]\delta[/itex] approach zero compared to each other. So I think [itex]0^z[/itex] is zero if Re(z)>0, infinite if Re(z)<0, and indeterminate if Re(z)=0.
     
  9. Aug 8, 2014 #8

    Matterwave

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    I think if Re(z)<0 then you basically have a divide by 0 going on, and should be undefined and not "infinite".
     
  10. Aug 8, 2014 #9

    dextercioby

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    I don't think 0 can be a valid base for exponentials. Think of the defining properties. 0^(-n) = 1/0^n which doesn't exist for n in N. 0^(ix) is ill-defined, because complex exponentiation goes through natural logarithm for which ln 0 doesn't exist.
     
  11. Aug 8, 2014 #10

    WWGD

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    If you want to define z^a for a non-integer (irrational, actually), and z complex, you have to set up a branch of complex log and then define ##z^a := e^{(alogz)}## to have an unambiguous definition ( and single-valued function). But then you have the problem that log0 =ln|0|+iarg0 . The argument is variable, depending on the branch, but ln0 cannot be defined.

    And, from another perspective, a log z will be a local inverse for expz , which has no global inverse, since it is not 1-1 ( it is infinite-to-1, actually) , but has local inverses, e.g., by the inverse function theorem, and these local inverses are precisely the branches of logz. But exp z is never 0.
     
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