Is (-i)^(-m) Equal to cos((m*pi)/2)+i*sin((m*pi)/2) in Complex Analysis?

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Discussion Overview

The discussion revolves around the relationship between the expression (-i)^(-m) and the trigonometric form cos((m*pi)/2) + i*sin((m*pi)/2) within the context of complex analysis. Participants explore whether this relationship holds true and seek methods of proof.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant questions the validity of the relationship and seeks proof.
  • Another suggests that demonstrating the periodicity of the left-hand side could simplify the proof process by limiting the cases to consider.
  • A third participant provides a transformation of -i using exponential notation, stating that -i = exp(-i(pi/2)), and derives (-i)^(-m) = exp(i(m*pi/2)), proposing to apply Euler's identity to further analyze the expression.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the truth of the relationship, and multiple viewpoints regarding the proof and approach remain present.

Contextual Notes

There are unresolved assumptions regarding the periodicity and the application of Euler's identity, as well as the implications of the transformation provided.

dado033
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is this relashion true? or false?
if it is true how can I proof it?
(-i)^(-m) = cos((m*pi)/2)+i*sin((m*pi)/2)
 
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Perhaps try showing the left hand side is periodic. Then I think this would mean that you would only have to work out a finite number of cases.
 
dado033 said:
is this relashion true? or false?
if it is true how can I proof it?
(-i)^(-m) = cos((m*pi)/2)+i*sin((m*pi)/2)


-i=exp(-i(pi/2))
Therefore (-i)^(-m)=exp(i(m*pi/2))
Now apply Euler's identity.
 
thank u very much
 

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