Why is the argument function defined in Complex Analysis discontinuous? I've been working through Spivak's Calculus and I'm currently working through Ch. 26 on Complex Functions. One of the functions they define is the argument function. Recall that the argument of a nonzero complex number $z$ is a (real) number $\theta$ such that:
$$z = |z|(\cos \theta + i \sin \theta)$$
There are infinitely many arguments for $z$m but just one which satisfies $0 \leq \theta < 2\pi$. If we call this unique argument $\theta(z)$, then $\theta$ is a (real-valued) function (the "argument function") on $\{z \in \mathbb{C}: z \neq 0\}$.
Now there is a claim in the text that the argument function is discontinuous at all nonnegative real numbers. Attached is two screenshots that correspond to one another.


In the text the example they give is after shifting and redefining the argument function $\theta(z)$. Even after redefining it I still don't see how the function is discontinuous. At the current moment I'm not concerned with the formal proof of how it is discontinuous, but an overall picture of why.
To see what I'm trying to get at, when first reading that the argument function is discontinuous I envisioned what would happen is that when solving for the argument I would end up with an undefined function. So if we attempt to solve for the argument of a complex number $z = x+iy$, recalling the trigonometric definition of a complex number $z$, we arrive at:
$$\arctan\bigg(\frac{y}{x}\bigg) = \theta$$
Restricting $\theta$ to the first quadrant of the complex plane. Using $\arccos$ or $\arcsin$ will have similar requirements for their individual functions.
From here since the claim was that the argument function is discontinuous at every positive real number and keeping in mind that the $x$ or $y$ are defined as real numbers I thought of taking the limit, that would mean something along the lines of if $a,b \in \mathbb{R}$:
$$\lim_{(a,b) \to (x,y)}\arctan\bigg(\frac{b}{a}\bigg) = \theta(x,y) = \arctan\bigg(\frac{y}{x}\bigg)$$
Writing this out now as I think about it and the behaviour of multivariable functions, I believe that the discontinuities can occur because approaching the limiting value (in this case $(x,y)$) happens over an infinite amount of different paths and some of these will possibly have instances where the argument function will produce different values compared to the expected result.
Is this interpretation correct as to why the argument function is discontinuous? at least in an informal way?
 A: By the way, the definition of $\theta$ is NOT $\arctan\left(\frac{y}{x}\right)$. This is only valid if you restrict $\theta$ to the first quadrant of the complex plane. The full correct definition requires a bunch of case work.
Let $a\geq 0$ be any real number. Then, as Spivak mentions, $\theta$ is not continuous at the point $a$; the definition of continuity says that given a small $\epsilon$ ball around the point $\theta(a)$, you must be able to find a small $\delta$ ball around the point $a$, such that $\theta$ maps the $\delta$ ball INTO the $\epsilon$ ball (see page 535, figure 3). The crux of the argument is the following: if $\alpha$ is close to $a$, but lies "above" (i.e has positive imaginary part, for example $\alpha = a+ i\frac{\delta}{2}$) then $\theta(\alpha)$ is a "small" positive number; i.e it is close to $\theta(a)=0$.
On the other hand, a point like $\beta= a-i\frac{\delta}{2}$ is also close to $a$; in the sense that $|\beta-a|<\delta$, however, its argument $\theta(\beta)$ is very close to $2\pi$; i.e $\theta(\beta)$ lies outside the $\epsilon$ ball drawn around $\theta(a)$.
So, to summarize, the trouble is that no matter how small of a ball you draw around the point $a$, you can always find points above and below which map close to $0$ and $2\pi$ respectively. It is this large discrepancy which is the reason for the discontinuity.

Another way of phrasing it (more computational, less geometric) is to say that you have two paths along which we get different values for the limit: if we consider the paths $\gamma_1(t) = a+it$ and $\gamma_2(t) = a-it$, then
\begin{align}
\lim_{t\to 0^+}\theta(\gamma_1(t)) = 0 \quad \text{but} \quad
\lim_{t\to 0^+}\theta(\gamma_2(t)) = 2\pi.
\end{align}
The differences in limits show that $\theta$ is discontinuous at the point $a$. Similar line of reasoning works to show any of the other argument functions you define will also be discontinuous.
A: The argument function must be discontinous along a ray from the origin to the point at infinity.  Here, with the definition you give, the ray is the positive real axis.  Since the argument function is multi-valued, there has to be somewhere where it jumps by $2\pi$.
Let $\epsilon>0$.  Consider $z_1=x_0+i\epsilon$ and $z_2=x_0-i\epsilon$.  Then $\theta(x_0+i\epsilon)=\tan^{-1} \frac{\epsilon}{x_0}$ and $\theta(x_0-i\epsilon)=2\pi-\tan^{-1} \frac{\epsilon}{x_0}$.  Let $\epsilon\to 0$.  We see that the limits $\lim_{\epsilon\to 0} z_1 =\lim_{\epsilon\to 0} z_2 = x_0$ but $\lim \theta(z_1) \ne \lim \theta(z_2).$
