I have this problem right here:
Find a branch of $\log{(z^2+1)}$ that is analytic as $z=0$ and takes the value $2\pi i$ there.
If I just plug in $z=0$ and use the principal branch I would just get $0$, $\log{1}$ is $0$ and the argument is $0$? So what do i do? Can i just cut the plane at the negative real axis and define the branch as $\pi \leq \arg{z} < {3\pi}$ ? If so how do i state that mathematically?
Best Answer
Well,let's proceed by definition of $\log{w}$:
$\log_{\alpha}{w}=\log|w| +i\theta$;$\alpha\le\theta<\alpha+2\pi$
Put $w=z^2+1$ and you get;$\log_{\alpha}{(z^2+1)}=\log|z^2+1| +i\theta$;$\alpha\le\theta<\alpha+2\pi$
put $z=0$ and you get:$\log_{\alpha}{(1)}= iarg(1)$;$\alpha\le arg(1)<\alpha+2\pi$
choose the branch:$(\pi,3\pi)$ so that $log_\pi{1}=2i\pi$[Note the branch :$(\pi,3 \pi)$,all the inequalities are strict because we want the function to be analytic.]
PS:$\log_\alpha{z}$ represents that for $\log$ function branch is $[\alpha,\alpha+2\pi)$