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

[shen07]

Find all possible Values of:

2^(-i)

 

[topsquark]

Re: Exercise

Find all possible Values of:

2^(-i)

What have you tried so far?

-Dan

 

[shen07]

Re: Exercise

well 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

 

[alyafey22]

Re: Exercise

well 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

The logarithm is a multivalued function so

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

 

[shen07]

Re: Exercise

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

 

[alyafey22]

Re: Exercise

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

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

 

[shen07]

Re: Exercise

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

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

 

[alyafey22]

Re: Exercise

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 .

 

[shen07]

Re: Exercise

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

 

[alyafey22]

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 .

 

[shen07]

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 .

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