Is {(-2)^5}^(1/5) a complex number ?

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Discussion Overview

The discussion centers around the expression E = {(-2)^5}^(1/5) and whether it is a complex number. Participants explore the implications of calculating this expression, including the different roots that can arise from it.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether E = {(-2)^5}^(1/5) is a complex number, noting their calculation yields E = -2, while another source provides a complex result.
  • Another participant points out that the source assumes the principal root and suggests using the real-valued root option to obtain -2.
  • A participant identifies a potential typo in the expression and clarifies that E = {(-2)^5}^(1/5) simplifies to (-32)^(1/5), which has five solutions, including one real number and four complex numbers.
  • It is noted that Wolfram Alpha defaults to providing the principal root, and participants discuss how to obtain all five solutions, including the real and complex roots.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the roots of the expression, with some asserting that -2 is a valid solution while others emphasize the existence of multiple complex solutions. The discussion remains unresolved regarding the preferred interpretation of the roots.

Contextual Notes

Limitations include the dependence on the choice of root (principal vs. real-valued) and the potential for confusion regarding the representation of complex numbers in this context.

phydis
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Notice how the alpha page says
Assuming the principal root | Use the real‐valued root instead
If you select the real valued option you'll get -2.
 
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You have a typo
E = {(-2)^5}^(1/5)=(-32)^(1/5)
There are five numbers such that x^5=-32
(-32)^(1/5) should be one of them
they are
1.61803398874989+1.17557050458495 i
-0.61803398874989+1.90211303259031 i
-2
-0.61803398874989-1.90211303259031 i
1.61803398874989-1.17557050458495 i

We chose 1.61803398874989+1.17557050458495 i as it is first on the list and the most reasonable choice. Some people inexplicably pick -2.
 
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There are *five* solutions to x5=(-2)5. One of them is the real number -2. The other four are complex numbers. Wolfram Alpha gives you the principal root by default. You can force it to yield the real-valued root by clicking on "Use the real‐valued root instead".

You can see all five solutions if you instead ask WA to solve x5=(-2)5.
 
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got it. Thanks everyone!
 

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