A Number Raised to the m Power

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Discussion Overview

The discussion revolves around the properties of numbers raised to the mth power, particularly focusing on whether raising a number to an even power results in a real number, with m being a positive even integer. The scope includes mathematical reasoning and exploration of complex numbers.

Discussion Character

  • Exploratory, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant inquires about a proof that a number raised to the mth power is always real when m is a positive even number.
  • Another participant provides an example of a complex number raised to the fourth power, yielding a complex result.
  • A participant expresses doubt about the initial claim and questions if the statement holds true specifically for m equal to 2.
  • Another participant responds with a mathematical expansion of a complex number squared, indicating that the result is not necessarily real unless certain conditions are met.
  • A later reply suggests checking the case for m equal to 4 by further expanding the expression derived from squaring a complex number.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the original claim. There are multiple competing views regarding the conditions under which raising a number to an even power results in a real number, particularly when complex numbers are involved.

Contextual Notes

The discussion highlights limitations related to assumptions about the nature of the numbers being raised to powers, particularly the distinction between real and complex numbers. The mathematical steps and conditions necessary for the results to hold true are not fully resolved.

Bashyboy
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Hello Everyone,

I was wondering, does anyone know of a proof that showed if a number is raised to the mth power, where m is a positive even number, the number is always real?
 
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(1+2i)4 = -7-24i
 
Drats! I was hoping it was true! How about if m were only 2? Would the statement then be true?
 
It isn't: (a + bi)(a + bi) = a^2 + 2abi + b^2i^2 = a^2 + 2abi - b^2
 
Bashyboy said:
Drats! I was hoping it was true! How about if m were only 2? Would the statement then be true?

##(1+i)^2=1+2i-1=2i##

Your claim will only be true for ##m=2## if the number is only real or imaginary. This is clear by expanding a complex number as a binomial.

##(a+bi)^2=a^2-b^2+2abi##

In order for this to be real, either ##a## or ##b## must be zero. You should check the case when ##m=4## by squaring ##(a^2-b^2+abi)## to see what you get.


Edit: I see that you figured it out as I was posting
 
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