What combinations satisfy this complex exponentiation property?

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Discussion Overview

The discussion revolves around the conditions under which the property of complex exponentiation holds for the product of two complex numbers, specifically in the form of \([( \pm ia)( \pm ib)]^{-\alpha} = (\pm ia)^{-\alpha} (\pm ib)^{-\alpha}\), where \(a, b, \alpha > 0\). The focus is on identifying which combinations of signs satisfy this property.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Post 1 introduces the main question regarding the combinations of signs that satisfy the exponentiation property.
  • Post 2 suggests using the definition of exponentiation and logarithms to analyze the problem, specifically questioning when the logarithm of the product equals the sum of the logarithms.
  • Post 3 claims that the property holds in all cases except for the combination (-,-), providing a specific example to illustrate this point.
  • Post 4 expresses agreement with the previous post, affirming the conclusion reached about the (-,-) case.

Areas of Agreement / Disagreement

Participants generally agree that the property holds for all combinations except for the (-,-) case, which is specifically noted as an exception. However, the reasoning behind this conclusion is not fully explored, leaving some aspects of the discussion unresolved.

Contextual Notes

The discussion relies on the definitions of exponentiation and logarithms, and the implications of sign combinations in complex exponentiation are not fully detailed. There may be additional assumptions or mathematical steps that are not explicitly stated.

bruno67
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Suppose we have the product

[(\pm ia) (\pm ib)]^{-\alpha}
wherea, b, \alpha >0. For which of the combinations (+,+), (+,-), (-,+), and (-,-) is the following property satisfied?

[(\pm ia) (\pm ib)]^{-\alpha}=(\pm ia)^{-\alpha} (\pm ib)^{-\alpha}
 
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Hi Bruno67! :smile:

Using the definition of exponentiation we have that

[(\pm ia)(pm ib)]^{-\alpha}=e^{-\alpha Log((\pm ia)(\pm ib))}

So the question becomes when

Log((\pm ia)(\pm ib))=Log(\pm ia)+Log(\pm ib)

Solve this using the definition of the logarithm.
 
Thanks, so it holds in all cases except the (-,-) one. In that case we have

[(-ia) (-ib)]^\alpha = (-ia)^\alpha (-ib)^\alpha (-1)^{2\alpha}.
 
Indeed!:smile:
 

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