[SOLVED]Evaluating imaginary expression

shen07

Member
Find all possible Values of:

2(-i)

topsquark

Well-known member
MHB Math Helper
Re: Exercise

Find all possible Values of:

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

-Dan

shen07

Member
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

ZaidAlyafey

Well-known member
MHB Math Helper
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
The logarithm is a multivalued function so

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

shen07

Member
Re: Exercise

Yes i know that but here we have ln(2) and we cant write this in the above form.

ZaidAlyafey

Well-known member
MHB Math Helper
Re: Exercise

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

shen07

Member
Re: Exercise

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

ZaidAlyafey

Well-known member
MHB Math Helper
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

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

is a multivalued function so it has infinite solutions .

shen07

Member
Re: Exercise

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

ZaidAlyafey

Well-known member
MHB Math Helper
Re: Exercise

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

shen07

Member
Re: Exercise

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