What Are the Possible Values of 2^(-i)?

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    Expression Imaginary
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Discussion Overview

The discussion revolves around finding all possible values of the expression \(2^{-i}\), exploring its implications in the context of complex numbers and logarithms. Participants engage with the mathematical properties and multivalued nature of logarithmic functions as they relate to this expression.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Technical explanation

Main Points Raised

  • Some participants propose using the relationship \(2^{-i} = e^{-i\log(2)}\) to express the value in terms of exponential functions.
  • Others mention that the logarithm is a multivalued function, suggesting that \( \log(a) = \ln|a| + i\arg(a) \) applies, which introduces the concept of multiple solutions.
  • There is a discussion about whether to apply the complex logarithm or the natural logarithm, with some arguing that since 2 is a complex number, the multivalued nature of the logarithm should be considered.
  • One participant notes that if \(z\) is an integer, the function \(f(z) = 2^z\) has only one solution, while for rational numbers, it has a finite number of solutions, and for other complex numbers, it has finitely many solutions.

Areas of Agreement / Disagreement

Participants express differing views on the application of logarithmic functions and the nature of solutions for \(2^{-i}\). There is no consensus on the best approach to find all possible values, and the discussion remains unresolved regarding the implications of using different logarithmic forms.

Contextual Notes

The discussion highlights the dependence on definitions of logarithmic functions and the implications of treating numbers as complex. The multivalued nature of the logarithm introduces complexity in determining all possible values.

shen07
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Find all possible Values of:

2(-i)
 
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Re: Exercise

shen07 said:
Find all possible Values of:

2(-i)
What have you tried so far?

-Dan
 
Re: Exercise

well i have tried using

uv=evln(u)

and get cos(ln(2))-isin(ln(2))

but is it the solution because i am asked to find all the possible values
 
Re: Exercise

shen07 said:
well i have tried using

uv=evln(u)

and get cos(ln(2))-isin(ln(2))

but is it the solution because i am asked to find all the possible values

The logarithm is a multivalued function so

$$\log(a) = \ln|a|+iarg(a) $$
 
Re: Exercise

Yes i know that but here we have ln(2) and we can't write this in the above form.
 
Re: Exercise

shen07 said:
Yes i know that but here we have ln(2) and we can't write this in the above form.

$$2^{-i}=e^{-i\log(2)}=e^{-i(\ln|2|+2k\pi i)}$$
 
Re: Exercise

ZaidAlyafey said:
$$2^{-i}=e^{-i\log(2)}=e^{-i(\ln|2|+2k\pi i)}$$

but should we not apply complex logarithm ,i.e LOG to complex numbers only, here we should have use LN as far as i have understand. Correct me if i am wrong
 
Re: Exercise

shen07 said:
but should we not apply complex logarithm ,i.e LOG to complex numbers only, here we should have use LN as far as i have understand. Correct me if i am wrong

But 2 is still a complex number .

The function

$$e^{\log(2^{-i})}$$

is a multivalued function so it has infinite solutions .
 
Re: Exercise

ZaidAlyafey said:
But 2 is still a complex number .

The function

$$e^{\log(2^{-i})}$$

is a multivalued function so it has infinite solutions .

ok i understand what you are trying to say. 2 is also found in the complex plane, so we can apply this rule to it, hence obtaining several values with this arbitrary k.

Thanks
 
  • #10
Re: Exercise

For more information consider the following

$$f(z)=2^{z}$$ where $z$ is any complex number .

  1. if $z$ is an integer then the function has only one solution
  2. if $z$ is a rational number then it has finite number of solutions .
  3. If $z$ is any other complex number then it has finitely many solutions .
 
  • #11
Re: Exercise

ZaidAlyafey said:
For more information consider the following

$$f(z)=2^{z}$$ where $z$ is any complex number .

  1. if $z$ is an integer then the function has only one solution
  2. if $z$ is a rational number then it has finite number of solutions .
  3. If $z$ is any other complex number then it has finitely many solutions .
Thanks a Lot for this Idea..Will remember it..:D
 

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