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?


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 $[\alpha,\alpha+2\pi)$

  • $\begingroup$ Is $log_{\pi} 1$ or is alpha instead of pi? $\endgroup$ – MonsieurGalois Dec 1 '15 at 14:00
  • $\begingroup$ @MonsieurGalois,it completely depends upon which branch you choose.For example if question demands for principal branch then you take $\alpha$ to be $-\pi$ etc. $\endgroup$ – Suraj_Singh Dec 4 '15 at 12:39

I would consider something like $$ f(z) = 2\pi i + \int_0^z \frac{2w}{1+w^2} \, dw, $$ where the integral is taken along the straight line joining the origin to $z$, provided that $z$ is not on the imaginary axis with $\lvert \Im(z) \rvert \geq 1 $. Excluding that set gives you a naturally-induced branch you can use.

  • $\begingroup$ Im not sure that i understand what you mean with that integral. The solution manual simply says this $\mathcal{L}_{\pi}(z^2+1)$, what does that mean? $\endgroup$ – user269620 Sep 18 '15 at 17:51
  • $\begingroup$ One would hope that that notation is defined somewhere in the book: it's not something I've seen before. My idea is to create an analytic function with the correct derivative and value at a point: by Cauchy's theorem, this is sufficient to produce a unique analytic function, and then it's a matter of specifying a unique continuously varying path for each value of $z$: the places where continuity cannot be maintained can be divided from each other by means of branch cuts. $\endgroup$ – Chappers Sep 18 '15 at 19:10
  • $\begingroup$ @Chappers,at $z=0, log(z^2+1)$ becomes zero and argument of zero is not defined.So what about this? $\endgroup$ – Suraj_Singh Sep 20 '15 at 13:53
  • $\begingroup$ @kilimanjaro It does? I thought $\log{1}=0$. $\endgroup$ – Chappers Sep 20 '15 at 13:57
  • $\begingroup$ @Chappers,I made silly mistake! $\endgroup$ – Suraj_Singh Sep 20 '15 at 14:02

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