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- #1

- Thread starter shen07
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- Thread starter
- #1

- Aug 30, 2012

- 1,198

What have you tried so far?Find all possible Values of:

2^{(-i)}

-Dan

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- #3

- Jan 17, 2013

- 1,667

The logarithm is a multivalued function sowell i have tried using

u^{v}=e^{vln(u)}

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

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

\(\displaystyle \log(a) = \ln|a|+iarg(a) \)

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- #5

- Jan 17, 2013

- 1,667

\(\displaystyle 2^{-i}=e^{-i\log(2)}=e^{-i(\ln|2|+2k\pi i)}\)Yes i know that but here we have ln(2) and we cant write this in the above form.

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- #7

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

- Jan 17, 2013

- 1,667

But 2 is still a complex number .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

The function

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

is a multivalued function so it has infinite solutions .

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- #9

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.But 2 is still a complex number .

The function

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

is a multivalued function so it has infinite solutions .

Thanks

- Jan 17, 2013

- 1,667

For more information consider the following

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

- if $z$ is an integer then the function has only one solution
- if $z$ is a rational number then it has finite number of solutions .
- If $z$ is any other complex number then it has finitely many solutions .

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- #11

For more information consider the following

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

- if $z$ is an integer then the function has only one solution
- if $z$ is a rational number then it has finite number of solutions .
- If $z$ is any other complex number then it has finitely many solutions .

Thanks a Lot for this Idea..Will remember it..