The book I'm using is Complex Variables and Applications, 9th Ed by Brown and Churchill. I'm confused about exercise 14.1.b (pg 43):
For each of the functions below, describe the domain of definition that is understood:
$$ \text{(b)} \qquad f(z) = \operatorname{Arg}\left(\dfrac{1}{z}\right) $$
The answer given is $\mathbb{C}$ with $\operatorname{Re}(z) \neq 0$. I don't understand why it's not simply $z \neq 0$.
I've had a look at other MathSE answers and they mention something about branch cuts and well-definedness. However, the book doesn't mention "branch cuts" until much later. Here's how the book defines $\operatorname{Arg}$ (pg 17):
The principal value of $\arg(z)$, denoted by $\operatorname{Arg}(z)$, is the unique value $\Theta$ such that $-\pi < \Theta \leq \pi$.
And this is how they define the domain of definition (pg 37):
Let $S$ be a set of complex numbers. A function $f$ defined on $S$ is a rule that assigns to each $z$ in $S$ a complex number $w$. The number $w$ is called the value of $f$ at $z$ and is denoted by $f(z)$, so that $w = f(z)$. The set $S$ is called the domain of definition of $f$.
It must be emphasized that both a domain of definition and a rule are needed in order for a function to be well defined. When the domain of definition is not mentioned, we agree that the largest possible set is to be taken. Also, it is not always convenient to use notation that distinguishes between a given function and its values.
I don't see how you can conclude that $\operatorname{Re}(z) \neq 0$ just from this, so I must be missing something; to me,
$$ \operatorname{Arg}\left(\dfrac{1}{i}\right) = \operatorname{Arg}(-i) = -\frac{\pi}{2} $$
seems fine.
Best Answer
You are correct; $\operatorname{Arg}$ is defined for all non-zero complex numbers, including the imaginary axis. The book must have made a typo there.