Question on the $arg$ function

Question

We recently came across $arg$ in our lectures:

$\arg\colon \mathbb{C}\setminus\{0\} \to \mathopen] -\pi, \pi]$,

$ z \to \arg(z)$, so $z=|z|e^{i{\arg(z)}}$

First question: why is the function bounded by $\mathopen]-\pi,\pi]$, surely it could also be bounded by $\mathopen]0,2\pi]$ or even by $[0,2\pi\mathclose[$. I do not understand the significance of its bound as it is periodical in any case?

Second question: I've had a look at the following problem: Solve $z^3=-i$, where $z \in \mathbb{C}$.

Since $\arg(-i) \in \mathopen]-\pi,\pi]$ we are required to say $-i=e^{-i\frac{\pi}{2}+2ki\pi}=z^3$, $k\in \mathbb{Z}$.

Therefore, $z_{k}=e^{\frac{-i\frac{\pi}{2}+2ki\pi}{3}}$, $k=0,1,2$

Using this we get:

$z_{0}=e^{\frac{-i\pi}{6}}$, $\arg(z_{0})=\frac{-\pi}{6}$, which lies in $\mathopen]-\pi,\pi]$

$z_{1}=e^{i\frac{\pi}{2}}$, which is correct as arg also lies in the range

Now the problem arrives, as:

$z_{2}= e^{i\frac{7}{6}\pi}$ and $\arg(z_{2})=\frac{7}{6}\pi$, which does not lie in the range $\mathopen]-\pi,\pi]$. Does this mean $z_{2}$ is not a solution to the problem? Or, is $z_{2}$ still as solution, as it is periodical and $\arg(z_{2})=\arg(z_{2})+2(-1)\pi=-\frac{5}{6}\pi \in \mathopen]-\pi,\pi]$. Am I missing something, or just leading myself astray? Thank you for your help.

Answer 1

As to your first question -- this is simply a matter of convention (and I do not think that the convention you gave is universal). As you noted, the periodicity would allow arg to be defined to have values on any half-open interval of length $2\pi$ -- in order for arg to be a well-defined function, we must merely choose one such interval arbitrarily.

As to your second question -- if we take $z = e^{-\frac56\pi}$, $z^3 = (e^{-\frac56\pi})^3 = e^{-\frac52\pi} = e^{-2\pi}e^{-\frac12\pi} = e^{-\frac12\pi} = -i$. Since $z = e^{-\frac56\pi}$ satisfies the equation, I do not think we should discount it. So yes, sometimes we will need to take advantage of the periodicity to adjust our argument so it falls in the desired range.

Answer 2

Answer to question 1:

Yes, there is nothing particularly special about the choice $(-\pi,\pi]$, and in many conventions, $[0,2\pi)$ is used for the range of arguments.

Answer to question 2:

$z_2=\exp\left(i\frac76\pi\right)$ is a solution, since $$e^{i\frac76\pi}\equiv e^{i\frac76\pi-2\pi i}=e^{-i\frac56\pi}$$So the answer that you'd be expected to give is this one, since its argument lies in the correct range. However, this does not mean that the solution with $\arg=\frac76\pi$ is not a solution to the equation - it is, and it is the exact same number as the one with $\arg=-\frac56\pi$. You can see this by computing $$\cos\left(\frac76\pi\right)+i\sin\left(\frac76\pi\right)=\cos\left(-\frac56\pi\right)+i\sin\left(-\frac56\pi\right)$$

Answer 3

The convention that $\arg z$ has range $]-\pi,\pi]$ corresponds to the convention that the cut to form the Riemann surface for $\sqrt{z}$ is along the negative real axis; and the convention that the cut to form the Riemann surfsce for $\log z$ is along the negative real axis.

With these matching conventions, we have $$ \sqrt{z} = \sqrt{|z|}\;e^{i(\arg z)/2} \\ \log z = \log |z| + i \arg z $$