I understand that in complex analysis $\arg(z) = \operatorname{Arg}(z) + 2k\pi i$.

In some texts about complex analysis I read things like $\arg_{\tau}(z)$. What does $\tau$ mean? In addition, what does $\operatorname{Arg}(z) = \arg_{−\pi}(z)$ mean? What is $\arg_{0}$?

What are principal values? Is $\operatorname{Arg}(z)$ the principal value of $\arg(z)$?

How do I solve problems like: "Determine a branch of $f(z)=\log(z^3-2)$ that is analytic in $z_0$"?

Thanks in advance.

  • $\begingroup$ Ok, I understand now that $\tau$ means that the function arg(z) jumps to $\tau$ when arg(z)$ = \tau +2\pi$. $\endgroup$ Dec 17, 2012 at 19:12

1 Answer 1


As you said the symbol $\arg_r$ denotes the branch of the argument function that has image $(r,2\pi+r]$. Now for your second question remember that the principal $\log$ (defined for $\arg_{-\pi}$) is complex differentiable everywhere but $S=\left\{z\in \mathbb{C}:z\leq 0\right\}$.

If $z_0\notin S$ choose the principal log.

If $z_0\in S$, choose the branch of $\log$ defined for $\arg_{0}$ that is complex differentiable everywhere but $P=\left\{z\in \mathbb{C}:z\ge 0\right\}$.

That way, $\log z$ will be holomorphic at $z_0$ everytime (unless of course $z_0=0$ where no $\log$ can even be defined). I think you can know work your problem

  • $\begingroup$ You mean $\arg_{-\pi}$, not $\arg_{-\pi/2}$. $\endgroup$
    – mrf
    Dec 18, 2012 at 12:26
  • $\begingroup$ @mrf Yes I do mean that $\endgroup$
    – Nameless
    Dec 18, 2012 at 14:04

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