Let $f(z)$ be the analytic branch of $\log(z)$ satisfying $f(1)=-2\pi i $. Where is this branch defined?
If $Log(z)$ represents the principal part of $\log$, defined for $\mathbb{C}-(-\infty,0]$, it seems that $f(z)=Log(z)-2\pi i$. But does taking the difference affect where is the branch defined?
Since $f(z)=Log(z)-2\pi i$, I would say that the new branch should be defined in $\mathbb{C}-(-\infty,0]$. On the other hand, given the value of $f$ at $z=1$, I think $f$ should be defined in $\mathbb{C}-\{ iy| y\leq 0 \}$. Which one is the correct one? Will be happy if you will be able to elaborate.
Thank you
EDIT: my misunderstanding is part of the following question:
Specify the value of g(ei), where $g$ is the analytic branch of $z^i$, defined in $\{ \mathbb{C}−{iy|y≤0}\} $ , where $g(1)=e^{2π}$.
As $g(z)=e^{i\log(z)}$, we should choose that branch $f$ of $\log$ the I defined above. But since $f(z)=Log(z)-2\pi i$, why it is defined in $\{ \mathbb{C}−{iy|y≤0}\} $ rather than in $\mathbb{C}-(-\infty,0]$?
My misunderstanding here comes from some basic properties of analytic branches that I can't figure out. Hope you'll be able to help me out with this.
Best Answer
Note that branches of the logarithm (which is the same as of the argument) are defined on the plane excluding any Jordan arc connecting zero to infinity, so what you need is the argument of $ei$ hence of $i$ when the argument of $1$ is $-2\pi$ and you avoid the negative imaginary line, so go counterclockwise.
Now $\arg i -\arg 1=\pi/2$ if you go counterclockwise from to $1$ to $i$ in your domain and do not need to cross the real axis again (if instead of a simple ray like here, you would exclude a spiral for example, you may need to cross the real axis a few times before you get to $i$ from $1$) so $\arg i = -3\pi/2$ here hence $$z^{ei}=e^{i\log(ei)}=e^{i(\log |ei|+i\arg {ei})}=e^{i(1-3i\pi/2)}=e^{3\pi/2}(\cos 1+i \sin 1)$$