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

In summary, the possible values for 2(-i) can be found by using the complex logarithm function, resulting in infinite solutions due to the multivalued nature of the function. The function e^{\log(2^{-i})} can have one solution if z is an integer, finite solutions if z is a rational number, and finitely many solutions for any other complex number.
  • #1
shen07
54
0
Find all possible Values of:

2(-i)
 
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  • #2
Re: Exercise

shen07 said:
Find all possible Values of:

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

-Dan
 
  • #3
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
 
  • #4
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

\(\displaystyle \log(a) = \ln|a|+iarg(a) \)
 
  • #5
Re: Exercise

Yes i know that but here we have ln(2) and we can't write this in the above form.
 
  • #6
Re: Exercise

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

\(\displaystyle 2^{-i}=e^{-i\log(2)}=e^{-i(\ln|2|+2k\pi i)}\)
 
  • #7
Re: Exercise

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

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
 
  • #8
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

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

is a multivalued function so it has infinite solutions .
 
  • #9
Re: Exercise

ZaidAlyafey said:
But 2 is still a complex number .

The function

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

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

\(\displaystyle f(z)=2^{z}\) 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

\(\displaystyle f(z)=2^{z}\) 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
 

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

What is an imaginary expression?

An imaginary expression is a mathematical expression that involves the use of imaginary numbers, which are numbers that can be represented as a multiple of the square root of -1. These numbers are often denoted by the letter "i" and are used to solve equations that cannot be solved using only real numbers.

How do you evaluate an imaginary expression?

To evaluate an imaginary expression, you can use the basic rules of arithmetic, such as the distributive property and the order of operations. It is also important to remember that the square root of -1, denoted by "i", can be simplified to -1 when squared. This can help simplify the expression and make it easier to evaluate.

What is the purpose of evaluating imaginary expressions?

Evaluating imaginary expressions is important in many fields of science and mathematics, such as engineering and physics. It allows us to solve equations and problems that involve complex numbers, which cannot be solved using only real numbers. This helps us better understand and model real-world phenomena.

Can imaginary expressions have real solutions?

Yes, imaginary expressions can have real solutions. This occurs when the imaginary parts of the equation cancel each other out, leaving only a real number as the solution. For example, the equation x^2 + 1 = 0 has the imaginary number i as a solution, but x^2 - 1 = 0 has the real numbers 1 and -1 as solutions.

Are there any limitations or restrictions when evaluating imaginary expressions?

One limitation when evaluating imaginary expressions is that the result will always have a real part and an imaginary part. This means that the final answer will involve both a real number and the imaginary unit i. Additionally, the use of imaginary numbers may not always be applicable or necessary, depending on the context of the problem.

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