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What have you tried so far?Find all possible Values of:
2(-i)
The logarithm is a multivalued function sowell 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
\(\displaystyle 2^{-i}=e^{-i\log(2)}=e^{-i(\ln|2|+2k\pi i)}\)Yes i know that but here we have ln(2) and we cant write this in the above form.
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\(\displaystyle 2^{-i}=e^{-i\log(2)}=e^{-i(\ln|2|+2k\pi i)}\)
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
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 .
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 .