Exponentiation with zero base, complex exponent?

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The discussion centers on defining the expression 0^z for complex z, highlighting the complexities that arise when z is not real. It is established that for real z, the behavior is clear: 0^z equals 0 for positive real z, is undefined for negative real z, and is indeterminate for z equal to zero. However, when z is complex, particularly in the form of bi, the expression becomes indeterminate due to the logarithmic properties involved, as ln(0) is undefined. The consensus suggests that 0^z should be considered zero if the real part of z is positive, undefined if negative, and indeterminate when the real part is zero. Overall, the challenges of defining 0^z for complex numbers stem from the inherent limitations of logarithmic functions and the nature of exponentiation with a zero base.
bcrowell
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Is there a good general definition of 0^z, 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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Is it not legitimate to use:

$$0^z=0^{a+bi}=0^a0^{bi}=0\quad\text{for}~~a>0$$
?
 
What does $$0^{bi}$$ mean?
 
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##.
 
johnqwertyful said:
What does $$0^{bi}$$ mean?

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

This is indeterminate with magnitude = 1.
 
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?
 
Amplifying on what mathman said, it seems like any expression of the form 0^{bi} is an indeterminate form. For example, suppose \epsilon>0 and \delta are both real. Then for

\lim_{\epsilon\rightarrow0^+,\ \delta\rightarrow0}|\epsilon^{\delta+i}|=e^{\delta\ln\epsilon}

has a value that depends on how rapidly \epsilon and \delta approach zero compared to each other. So I think 0^z is zero if Re(z)>0, infinite if Re(z)<0, and indeterminate if Re(z)=0.
 
I think if Re(z)<0 then you basically have a divide by 0 going on, and should be undefined and not "infinite".
 
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.
 
  • #10
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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