I am following Stein and Shakarchi on Complex analysis. On page 98 they define a branch of the logarithm to be a choice of domain for the logarithm. I suspect they mean a choice of interval for the complex argument $\theta$. I cite:

Theorem 6.1 Suppose $\Omega$ is simply connected with $1 \in \Omega$ and $0 \not \in \Omega$. Then in $\Omega$, there is a branch of the logarithm $F(z) = \log_\Omega (z)$ such that (i) $F$ is holomarphic in $\Omega$, (ii) $e^{F(z)} = z$ for all $z \in \Omega$ and (iii) $F(r) = \log(r)$ for real numbers $r$ sufficiently close to $1$.

[Proof omitted]

For example in the slit plane $\mathbb{C} \setminus (- \infty,0]$ we have the principal branch of the logarithm [with the formula] $\log (z) = \log r + i \theta$ for $\theta \in (- \pi, \pi)$

So the authors have found

(a) an interval for $\theta$ and (b) a formula for the principal branch of the logarithm.

How did the authors do this (in particular the interval for $\theta$)?

If $a \in \mathbb{R}$, to me it seems that $(a\pi,(a+2)\pi)$ would have been an equally good choice and that this choice does not depend on theorem 6.1.

Can you guide me on how to work out the other branches of the logarithm and also the appropriate formulas for them?

  • 1
    $\begingroup$ If $\Omega \subset \Bbb{C}^*$ is a simply connected domain fix $a$ and $e^b = a$ and $k$ then $L(z) = b+2i \pi k+\int_a^z\frac1s ds$ is an analytic branch of $\log $ and all the analytic branches arise this way for some $k$. Simply connectedness is to ensure that $\int_a^z\frac1s ds$ doesn't depend on the chosen path $a \to z$. $\endgroup$
    – reuns
    Jun 27 '19 at 20:43
  • $\begingroup$ A general statement is $\operatorname{Log}(z) = \log(|z|) + i \operatorname{arg}_\alpha(z)$. We merely have to choose some branch on the Riemann Surface (you can imagine it as an infinitely branched spiral) on which we get a unique logarithm; we merely choose our argument to lie in any interval $[\alpha, \alpha + 2\pi]$ (where we must delete one of the endpoints for uniqueness). In your case, simply think of think as whenever $\theta \in (\alpha, \alpha + 2\pi]$ we can take a straight branch cut at angle $\alpha$ (of course we must be more careful with curved branch cuts - see reun's comment) $\endgroup$ Jun 27 '19 at 20:48
  • $\begingroup$ It is not that the authors "have found" and interval for $\theta$, but that they define the principal branch with $|\theta|<\pi$; how they derive the formula $\log z=\log r+i\theta$ is already in the book (page 99). $\endgroup$
    – user9464
    Jun 27 '19 at 22:18

In short, the reason for this particular choice of an interval is condition (iii). What you said about taking any $(a\pi,(a+2)\pi)$ is not entirely correct (at least if you want the principal branch of the logarithm), because if you take any $a$ that is not an odd integer, your interval will contain a multiple of $\pi$ of the form $(2n+1)\pi$ with $n\in \mathbb{Z}.$ But such an argument would mean that the associated complex number lies on the negative real axis which we removed from the domain $\Omega.$ Hence, the only options for the interval are $((2k+1)\pi,(2k+3)\pi)$ with $k\in\mathbb{Z}.$

In principle any $k$ is sufficient to fulfill conditions (i) and (ii), but let us now also consider condition (iii). Take any $k$ other than $-1,$ then the interval for $\theta$ will not contain $0.$ Therefore, the corresponding branch of the logarithm can never take real values so it cannot fulfill condition (iii). We are therefore left with only this particular choice of interval.

You can apply similar reasoning to any domain $\Omega$ of the form $$\mathbb{C}\setminus \{re^{i\varphi}:r\in \mathbb{R},r\geq 0\}$$ where you take the entire complex plane and remove just the ray starting at the origin with angle $\varphi\in\mathbb{R}.$ As an exercise, you can think about which choices for $\phi$ are allowed to fulfill the conditions of Theorem 6.1. You could also remove any other curve starting at the origin and going to infinity (like an Archimedian or logarithmic spiral) but then your argument in the formula of the logarithm will also depend on $r.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.