What is the definition of a^i?

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Discussion Overview

The discussion centers around the definition and implications of raising a real number to the power of the imaginary unit \(i\), specifically \(a^i\), and whether this results in a complex number. Additionally, a secondary question regarding the proof of the fundamental theorem of algebra is briefly mentioned.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant defines \(a^i\) as \(e^{i(\log(a))} = \cos(\log(a)) + i\sin(\log(a))\).
  • Another participant confirms that \(i^i = e^{[(i)(\pi/2)(i)]} = e^{(-\pi/2)}\), suggesting this is correct.
  • A later reply notes that \(a^i\) is not single-valued, indicating complexity in the function's behavior.
  • There is a mention of the difficulty of proving that an \(n\)-degree polynomial has \(n\) roots, with a reference to the fundamental theorem of algebra.

Areas of Agreement / Disagreement

Participants express differing views on the implications of \(a^i\) and its nature, particularly regarding its multi-valued characteristic. The discussion on the polynomial roots remains less developed, with no consensus reached on the proof's accessibility or complexity.

Contextual Notes

The discussion does not resolve the implications of multi-valued functions in the context of complex numbers, nor does it clarify the assumptions underlying the definitions and proofs mentioned.

3.14lwy
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(1)

what is the definition of a^i ? (where a is a real number , i = (-1)^(1/2) )

is this still a complex number ?




(2)

where can I find the prove of "a n degree polynomial has n roots" ??
 
Mathematics news on Phys.org
1) a^i =e^(i(log(a)))=cos(log(a)) + isin(log(a))

2) A proof of the fundamental theorem of algebra, first formulated by Gauss is very difficult. A simpler proof can be found by the time you get into the study of Complex Functions, generally a Junior college study.
 
thank you!

then , is i^i = e^[(i)*(pi/2)*(i)] = e^(-pi/2) ??
 
Yes, it is.
 
With the proviso that ai is, like most complex functions, not single valued.
 

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