Semi-discrete Calculus in Complex Analysis

Can we apply the detachment operator in the Complex plane? As in Real Analysis, a theory built on the derivative sign would go through rate calculations, and wherever the derivative is nonzero it would add nothing. But what if we introduce a theory of "complex trends" that computes them directly, without going through the derivative? Would it apply to non-differentiable and even discontinuous functions, as it does on the real line? And would it capture trends in the function's image more reliably than the derivative sign?

Complex Analysis is full of concepts that generalize their Real Analysis counterparts, such as continuity and the derivative. Inspired by the operator's applications in AI, the natural sciences and Real Analysis, let us define a complex version of a function's detachment and study its basic properties. We start from the real line, because the sharper we are there, the clearer it becomes which parts survive in the plane.

From the Real Line

For $f:\mathbb{R}\to\mathbb{R}$, the one-sided detachments [5] at a point $x$ are

$$f_{\pm}^{;}\left(x\right)\equiv\pm\lim_{h\to0^{\pm}}\operatorname{sgn}\left[f\left(x+h\right)-f\left(x\right)\right],$$

and $f$ is detachable at $x$ when both limits exist. The $\pm$ coefficient makes both of them read from left to right: $f_{+}^{;}=+1$ says the values rise as we leave $x$ to the right, and $f_{-}^{;}=+1$ says they rise as we arrive at $x$ from the left. Four facts about this pair shape everything below.

First, the limit is a statement about eventual signs. The sign function takes only the values $-1,0,+1$, so its limit exists exactly when it is eventually constant. Thus $f_{+}^{;}(x)=+1$ means $f(x+h)>f(x)$ for every small enough $h>0$, and $f_{+}^{;}(x)=0$ means a genuine plateau, $f(x+h)=f(x)$ near $x$ on the right. Nothing is divided and no magnitude is measured. The operator is also partial: for $x^3\sin(1/x)$ (with value $0$ at $0$) the sign of $f(h)-f(0)$ keeps changing as $h\to0$, so neither one-sided detachment exists there, although the derivative does and equals $0$.

Second, away from a zero derivative the detachment is the derivative sign. Suppose the one-sided derivative $f_{+}'(x)$ exists and is a nonzero real number. Close to $x$ the quotient $\left(f(x+h)-f(x)\right)/h$ has the sign of $f_{+}'(x)$, and because $h>0$ the increment $f(x+h)-f(x)$ has that sign as well. Hence $f_{+}^{;}(x)=\operatorname{sgn}f_{+}'(x)$, and likewise on the left. Where $f'(x)\neq0$ the detachment repeats $\operatorname{sgn}f'(x)$ on both sides. Where $f'(x)=0$ it can say anything: at the origin, $x^3$, $-x^3$, $x^2$ and the zero function all have derivative $0$, while their pairs $(f_{-}^{;},f_{+}^{;})$ are $(+1,+1)$, $(-1,-1)$, $(-1,+1)$ and $(0,0)$.

Third, a trend at a point is weaker than monotonicity near it. Take $f(x)=x+2x^2\sin(1/x)$ with $f(0)=0$, a classical counterexample [3]. Its difference quotient at the origin is

$$\frac{f(h)-f(0)}{h}=1+2h\sin\left(\frac{1}{h}\right)\ge1-2\left|h\right|>0\qquad\left(0<\left|h\right|<\tfrac{1}{2}\right),$$

so $f'(0)=1$ and $f_{-}^{;}(0)=f_{+}^{;}(0)=+1$: just left of the origin $f$ lies below $f(0)$, and just right of it $f$ lies above. Yet away from the origin

$$f'(x)=1+4x\sin\left(\frac{1}{x}\right)-2\cos\left(\frac{1}{x}\right),$$

which equals $-1$ at $x=1/(2\pi n)$ and $3$ at $x=1/((2n+1)\pi)$. Both sequences tend to $0$, so $f$ is monotone on no interval around the origin. The detachment records only how each side stands relative to the anchor $f(0)$, and says nothing about the order of two values on one side.

Fourth, the detachment survives where every rate is silent. Let $b(x)=e^{-1/x^2}$ with $b(0)=0$. All derivatives of $b$ vanish at the origin, so no Taylor coefficient can tell what happens there. Still, $b(h)>b(0)$ for every $h\neq0$, so $(b_{-}^{;}(0),b_{+}^{;}(0))=(-1,+1)$, which says at once that the origin is a strict local minimum. In general, $(f_{-}^{;},f_{+}^{;})=(-1,+1)$ at an interior point holds if and only if the point is a strict local minimum, with no continuity or differentiability assumed. This works because the detachment uses only comparisons of values. Replacing $f$ by $\phi\circ f$ for a strictly increasing $\phi$ changes no detachment, while the derivative can change completely: with $f(x)=x$ and $\phi(y)=y^3$, the derivative at the origin drops from $1$ to $0$.

The complex plane has no order, so the last fact cannot carry over as it stands. What does carry over is direction. Write a one-sided increment near a point as $c\,t^{\nu}+o(t^{\nu})$ with $c\neq0$. The pair $(\operatorname{sgn}c,\nu)$ says which way the function moves and how fast the change vanishes; in tropical geometry this pair is called the signed valuation [4]. The detachment keeps the first coordinate and drops the second, which is where the rate lives.

Defining the Complex Detachment

Recall that the signum of a complex number $z$ is defined as:

$$\operatorname{sgn}\left(z\right)\equiv\frac{z}{\left|z\right|}=e^{i\arg\left(z\right)}\quad\left(z\neq0\right),\qquad\operatorname{sgn}\left(0\right)\equiv0,$$

where $\arg\left(\cdot\right)$ is the complex argument function. Geometrically, the sign function maps every nonzero point of the complex plane to its closest point on the unit circle:

On the real line, the detachment was softened relative to the derivative by allowing one-sided limits and by not requiring them to agree. In higher dimensions, one-sided derivatives extend naturally to directional derivatives. The Complex Analysis literature focuses on differentiable and, more specifically, analytic functions. Still, complex partial derivatives, known as Wirtinger derivatives, are common, and they serve to define complex directional derivatives. These appear in advanced research (see [2], p. 722) and as a teaching and auxiliary tool (see [1], p. 19). Here is the analogous way to define a directional detachment.

Definition 1. The $\varphi$-detachment of a function $f:\mathbb{C}\longrightarrow\mathbb{C}$ at a point $z_{0}\in\mathbb{C}$ is

$$f_{\varphi}^{;}\left(z_{0}\right)\equiv e^{-i\varphi}\lim_{t\to0^{+}}\operatorname{sgn}\left[f\left(z_{0}+te^{i\varphi}\right)-f\left(z_{0}\right)\right],\tag{1}$$

so that $f_{\varphi}^{;}$ takes values in $S^{1}\cup\{0\}$, where $S^{1}$ is the unit circle in the complex plane. The limit runs along the ray that leaves $z_{0}$ in the direction $\varphi$, that is, over the points $z$ with $\arg\left(\Delta z\right)=\varphi$ as $\Delta z\to0$. We say that $f$ is $\varphi$-detachable at $z_{0}$ if the limit exists. Since $\left|\operatorname{sgn}w\right|=1$ for every $w\neq0$, the value $0$ occurs exactly when $f$ is constant on a short segment of that ray.

The $e^{-i\varphi}$ coefficient generalizes the $\pm1$ coefficient of the one-sided detachments, and it does so literally. If $f$ is real on the real line, then $\varphi=0$ gives $f_{+}^{;}(x)$, and $\varphi=\pi$ gives $f_{-}^{;}(x)$, because $e^{-i\pi}=-1$ and the ray in direction $\pi$ is the left side of $x$. Both coefficients make the detachment consistent with the derivative sign, as Claim 1 shows.

The definition subtracts $f(z_0)$ for a reason. If $f$ is continuous at $z_{0}$ and $f(z_0)\neq0$, then $\operatorname{sgn}f(z_{0}+te^{i\varphi})$ simply tends to $\operatorname{sgn}f(z_0)$, which reports where the value is and nothing about how it changes. The trend is carried by the increment.

The diagram below compares two ways of turning the increment $\Delta f=f(z)-f(z_0)$ into local information. Applying the limit directly to $\Delta f$ would give $0$ at every point of continuity, so both operators first rescale it:

The following claim shows that the detachment adds information beyond the derivative only where the derivative vanishes.

Claim 1. Let $f:\mathbb{C}\longrightarrow\mathbb{C}$, and suppose the directional difference quotient

$$\frac{\partial f}{\partial\varphi}\left(z_{0}\right)\equiv\lim_{t\to0^{+}}\frac{f\left(z_{0}+te^{i\varphi}\right)-f\left(z_{0}\right)}{te^{i\varphi}}$$

exists and is nonzero. Then $f$ is $\varphi$-detachable at $z_{0}$, and

$$f_{\varphi}^{;}\left(z_{0}\right)=\operatorname{sgn}\left[\frac{\partial f}{\partial\varphi}\left(z_{0}\right)\right].$$

In particular, if $f$ is differentiable at $z_{0}$ with $f'(z_0)\neq0$, then $f_{\varphi}^{;}(z_0)=\operatorname{sgn}f'(z_0)$ for every direction $\varphi$.

Proof. Write $\Delta f=f\left(z_{0}+te^{i\varphi}\right)-f\left(z_{0}\right)$. Since $\operatorname{sgn}$ is multiplicative and $\operatorname{sgn}\left(te^{i\varphi}\right)=e^{i\varphi}$ for $t>0$,

$$\begin{aligned} e^{-i\varphi}\operatorname{sgn}\left(\Delta f\right)&=\frac{\operatorname{sgn}\left(\Delta f\right)}{\operatorname{sgn}\left(te^{i\varphi}\right)}=\operatorname{sgn}\left[\frac{\Delta f}{te^{i\varphi}}\right]\\ &\xrightarrow[t\to0^{+}]{}\operatorname{sgn}\left[\frac{\partial f}{\partial\varphi}\left(z_{0}\right)\right], \end{aligned}$$

where the limit passes inside the sign because the sign function is continuous on $\mathbb{C}\backslash\left\{ 0\right\}$ and the quotient tends to a nonzero number. For differentiable $f$ the quotient tends to $f'(z_0)$ along every ray.$\,\,\,\,\blacksquare$

Detachability did not have to be assumed; it follows. For brevity, we will denote the $\varphi$-detachment at $z_{0}$ by $f^{;}$ rather than $f_{\varphi}^{;}\left(z_{0}\right)$.

Geometric Intuition

Suppose that $f$ is $\varphi$-detachable at $z_{0}$, and let $\epsilon>0$. By the definition of the limit, for every $z=z_{0}+te^{i\varphi}$ with $t>0$ small enough,

$$\left|\operatorname{sgn}\left[f\left(z\right)-f\left(z_{0}\right)\right]-f^{;}e^{i\varphi}\right|<\epsilon.$$

So $f(z)$ lies in a thin wedge with its vertex at $f(z_0)$, opening around the ray from $f(z_0)$ in the direction $f^{;}e^{i\varphi}$. The chord-angle analysis below gives the half-angle of the wedge.

Recall that for real functions the detachment is equivalent to the following statement, because the real sign function is discontinuous at zero and takes only three values:

$$\exists\dot{B}_{\pm}\left(x_{0}\right):\forall x\in\dot{B}_{\pm}\left(x_{0}\right):\,\,\,\operatorname{sgn}\left[f\left(x\right)-f\left(x_{0}\right)\right]=\pm f_{\pm}^{;}\left(x_{0}\right).$$

The analogous statement in the complex domain does not necessarily hold, and it is stricter than the existence of the limit in equation $(1)$. It would force the image of a short segment of the ray to lie on a single half-line from $f(z_0)$. The complex sign is continuous away from zero and takes a whole circle of values, so a limit of signs can be approached without being attained. For example, $f(z)=z+iz^{2}$ at $z_0=0$ in the direction $\varphi=0$ has $\Delta f=t+it^{2}$, whose sign tends to $1$ and never equals it.

We know that $\lim_{z\to0}\operatorname{sgn}\left(z\right)$ does not exist, because its value depends on the path along which $z$ approaches zero. Fixing the direction of approach is what turns this limit into something we can compute. The detachability of a real function at a point is synonymous with the function having a "local trend" there. The detachability of a complex function points to a local trend in a broader sense: a $\varphi$-detachable function maps the points of a short segment leaving $z_{0}$ in direction $\varphi$ to points approaching a single ray from $f(z_0)$.

The differentiability of a function at a point lets us compute its values with the linear approximation formula. The function's value is linear in the difference between the points, up to the function $\rho$:

$$f\left(z\right)=f\left(z_{0}\right)+f'\left(z_{0}\right)\left(z-z_{0}\right)+\rho\left(z\right)\left(z-z_{0}\right),$$

where $\lim_{z\to z_{0}}\rho\left(z\right)=0$ and $z\neq z_{0}$.

The detachment suggests a different kind of approximate linear relation for the image, one whose formula carries no information from the domain. Suppose $f$ is $\varphi$-detachable and let $\epsilon>0$. For $z$ on the ray and close enough to $z_{0}$,

$$\left|\operatorname{sgn}\left[f\left(z\right)-f\left(z_{0}\right)\right]-f^{;}e^{i\varphi}\right|<\epsilon.$$

Since $f^{;}e^{i\varphi}$ is a limit of points on the (closed) unit circle, it lies on the circle as well, and the chord between it and $\operatorname{sgn}\left[f\left(z\right)-f\left(z_{0}\right)\right]$ has length less than $\epsilon$. Let us assume without loss of generality that $0<\arg\left(f^{;}e^{i\varphi}\right)<\frac{\pi}{2}$, which amounts to rotating the target plane, a step that leaves chord lengths unchanged. The chord-angle argument assumes $f^{;}\neq0$ and $\epsilon<2$.

Let $\theta_{f}$ be the angle between $\operatorname{sgn}\left[f\left(z\right)-f\left(z_{0}\right)\right]$ and $f^{;}e^{i\varphi}$. On the unit circle, a chord that subtends the angle $\theta_f$ has length $2\sin\left(\left|\theta_{f}\right|/2\right)$, so

$$\left|\operatorname{sgn}\left(\Delta f\right)-f^{;}e^{i\varphi}\right|=2\sin\frac{\left|\theta_{f}\right|}{2}<\epsilon,$$

and therefore

$$\left|\theta_{f}\right|<2\arcsin\left(\frac{\epsilon}{2}\right).$$

We have shown that near $z_{0}$ the argument of the increment of $f$ stays within a fixed angle of the detachment's direction:

$$\left|\arg\left(\Delta f\right)-\arg\left(f^{;}e^{i\varphi}\right)\right|<2\arcsin\left(\frac{\epsilon}{2}\right).\tag{2}$$

We will refer to this analysis, which turns a bound on a chord's length into a bound on its angle, as the "chord-angle" analysis.

Computability and Consistency

As in Real Analysis, detachability at a point does not imply continuity or differentiability, and vice versa. One rule computes most of the examples below at once. It is the complex form of the signed valuation from the first section.

Claim 2 (leading-term rule). Suppose that along the ray in direction $\varphi$,

$$f\left(z_{0}+te^{i\varphi}\right)-f\left(z_{0}\right)=c\,t^{\nu}+o\left(t^{\nu}\right)\qquad\left(t\to0^{+}\right)$$

for some $c\in\mathbb{C}\backslash\{0\}$ and some real $\nu$. Then $f_{\varphi}^{;}\left(z_{0}\right)=e^{-i\varphi}\operatorname{sgn}c$.

Proof. Dividing by $t^{\nu}>0$ does not change the sign, so the sign of the increment equals $\operatorname{sgn}\left(c+o(1)\right)$, which tends to $\operatorname{sgn}c$ by the continuity of the sign function away from $0$.$\,\,\,\,\blacksquare$

The exponent $\nu$ never enters the answer. Two consequences follow directly.

If $f$ is analytic at $z_{0}$ and $f-f(z_0)$ has a zero of order $m\geq1$ there, with leading Taylor coefficient $a_{m}$, then $c=a_{m}e^{im\varphi}$ and

$$f_{\varphi}^{;}\left(z_{0}\right)=\operatorname{sgn}\left(a_{m}\right)e^{i\left(m-1\right)\varphi}.$$

For $m=1$ this is Claim 1 again. For every $m\geq2$ the derivative vanishes, while the detachment, as $\varphi$ goes once around $z_0$, winds $m-1$ times around the unit circle. The detachment shows the order of a critical point through its winding; the value of the derivative cannot.

If $f$ has a pole of order $m$ at $z_{0}$, with leading Laurent coefficient $a_{-m}$ and any finite value assigned at $z_{0}$, then $c=a_{-m}e^{-im\varphi}$ with $\nu=-m$, and

$$f_{\varphi}^{;}\left(z_{0}\right)=\operatorname{sgn}\left(a_{-m}\right)e^{-i\left(m+1\right)\varphi}.$$

Let us now work through some basic examples to gain intuition.

Example 1. Let us consider the following function:

$$f\left(z\right)=\begin{cases} \frac{1}{z}, & z\neq0,\\ 0, & z=0. \end{cases}$$

Let us calculate the $\varphi$-detachments at the origin by the definition:

$$f_{\varphi}^{;}\left(0\right)=e^{-i\varphi}\lim_{t\to0^{+}}\operatorname{sgn}\left[\frac{1}{te^{i\varphi}}\right]=e^{-i\varphi}\cdot e^{-i\varphi}=e^{-2i\varphi},$$

in agreement with the pole case of Claim 2 ($m=1$, $a_{-1}=1$). This function is neither continuous nor differentiable at the origin, and the detachment still provides monotony information there.

Example 2. Let $f\left(z\right)=z^{2}$. The derivative is $f'\left(z_{0}\right)=2z_{0}$. At $z_{0}=0$ the derivative vanishes, so we calculate the detachment from its definition:

$$f_{\varphi}^{;}\left(0\right)=e^{-i\varphi}\lim_{t\to0^{+}}\operatorname{sgn}\left(t^{2}e^{2i\varphi}\right)=e^{-i\varphi}\cdot e^{2i\varphi}=e^{i\varphi}.$$

As in Real Analysis, the directional derivatives provide no monotony information at the stationary point, while the directional detachment does. This is the case $m=2$ of Claim 2: one full turn of $\varphi$ turns the detachment once around the circle.

Example 3. Let $f\left(z\right)=\sqrt{z}$, the principal branch, with $\varphi\in(-\pi,\pi]$. The derivative is $f'\left(z_{0}\right)=\frac{1}{2\sqrt{z_{0}}}$. At $z_{0}=0$ the derivative does not exist, so we calculate the detachment from its definition:

$$f_{\varphi}^{;}\left(0\right)=e^{-i\varphi}\lim_{t\to0^{+}}\operatorname{sgn}\left(\sqrt{t}\,e^{\frac{i\varphi}{2}}\right)=e^{-i\varphi}\cdot e^{\frac{i\varphi}{2}}=e^{-\frac{i\varphi}{2}}.$$

The image of the ray leaves the origin in the direction $e^{i\varphi/2}$, and the detachment records that direction relative to $\varphi$. Here $\nu=\frac{1}{2}$ in Claim 2. As in Real Analysis, the detachment can provide monotony information at the derivative's singularities, for continuous functions as well.

Example 4. Let

$$f\left(z\right)=\begin{cases} z^{2}\sin\left(\frac{1}{\left|z\right|}\right), & z\neq0,\\ 0, & z=0. \end{cases}$$

Since $\left|f(z)\right|\leq\left|z\right|^{2}$, $f$ is differentiable at $z=0$ with $f'(0)=0$. Along every ray, however,

$$\operatorname{sgn}\left[f\left(te^{i\varphi}\right)\right]=e^{2i\varphi}\operatorname{sgn}\left[\sin\left(\frac{1}{t}\right)\right],$$

which keeps switching between $e^{2i\varphi}$ and $-e^{2i\varphi}$ as $t\to0^{+}$. So the directional detachment at the origin does not exist in any direction. As in Real Analysis, the complex detachment may provide no monotony information at points of infinite oscillation, even where the derivative exists.

Example 5. Let $f\left(z\right)=\Re\left(z\right)$, whose image is the real line. This function is nowhere complex differentiable, but its directional difference quotients exist. Along the ray, $\Delta f=t\cos\varphi$, so the sign of its directional quotient from Claim 1 is

$$\operatorname{sgn}\left[\frac{\partial f}{\partial\varphi}\left(z_{0}\right)\right]=\operatorname{sgn}\left[\frac{t\cos\varphi}{te^{i\varphi}}\right]=\operatorname{sgn}\left[\cos\left(\varphi\right)e^{-i\varphi}\right]=\begin{cases} e^{-i\varphi}, & \left|\varphi\right|<\frac{\pi}{2},\\ 0, & \left|\varphi\right|=\frac{\pi}{2},\\ -e^{-i\varphi}, & \text{else}. \end{cases}$$

Since the quotient may vanish, let us calculate the $\varphi$-detachment directly from the definition:

$$f_{\varphi}^{;}\left(z_{0}\right)=e^{-i\varphi}\lim_{t\to0^{+}}\operatorname{sgn}\left[t\cos\varphi\right]=\begin{cases} e^{-i\varphi}, & \left|\varphi\right|<\frac{\pi}{2},\\ 0, & \left|\varphi\right|=\frac{\pi}{2},\\ -e^{-i\varphi}, & \text{else}. \end{cases}$$

Here the detachment agrees with the sign of the directional quotient even where that quotient vanishes ($\left|\varphi\right|=\frac{\pi}{2}$). Along those rays $f$ is constant, which is exactly what the value $0$ means.

Auxiliary Terminology

Assumptions and abbreviations. We will use the following conventions below:

• We focus on directional detachments in a given angle $\varphi$. Therefore, by a $\delta$-"environment" of $z_{0}$ we mean the segment of length $\delta$ that starts at $z_{0}$ and is tilted by $\varphi$.
• When we add the arguments of two unit vectors, we assume without loss of generality that both arguments lie between $0$ and $\pi$. This is possible whenever the two vectors are not opposite: a common rotation brings both into the open upper half-plane, and a rotation changes none of the moduli used below.
• For two points $z_{0}$ and $z$ in the domain of a function $f$, we abbreviate $\Delta f=f\left(z\right)-f\left(z_{0}\right)$.
• An operator applied to a function is evaluated at $z_{0}$ unless we say otherwise. For example, $\operatorname{sgn}\left(f\right)=\operatorname{sgn}\left(f\left(z_{0}\right)\right)$.

We assume detachability, but neither differentiability nor continuity, of the functions in the results below. As a tradeoff, we use the following concepts.

Definition 2. Let $f:\mathbb{C}\longrightarrow\mathbb{C}$ and $z_{0}\in\mathbb{C}$. Then $f$ is sign-continuous at $z_{0}$ if $\operatorname{sgn}\left(f\right)$ is continuous there.

Definition 3. Let $f,g:\mathbb{C}\longrightarrow\mathbb{C}$ and $z_{0}\in\mathbb{C}$. Then $f,g$ are:

• Bimodular at $z_{0}$ if for each $z$ in its environment, $\left|\Delta f\right|=\left|\Delta g\right|$.
• Locally bimodular at $z_{0}$ if for each $z$ in its environment, $\left|f\left(z_{0}\right)\Delta g\right|=\left|g\left(z_{0}\right)\Delta f\right|$.
• Spatially bimodular at $z_{0}$ if for each $z$ in its environment, $\left|f\left(z\right)\Delta g\right|=\left|g\left(z\right)\Delta f\right|$.

We also say that $g$ has constant modulus at $z_{0}$ if $\left|g\left(z\right)\right|=\left|g\left(z_{0}\right)\right|$ for each $z$ in its environment.

Example 6. A function that is continuous at $z_{0}$ with $f(z_0)\neq0$ is sign-continuous there, because the sign function is continuous away from $0$. This includes points of non-differentiability, such as the cusp of $1+\left|z\right|$ at the origin. Sign-continuity can also hold at a discontinuity: the function equal to $1$ at the origin and to $2$ elsewhere has the constant sign $1$.

Example 7. Any function $f:\mathbb{C}\longrightarrow\mathbb{C}$ is everywhere bimodular to its reflections with respect to the axes,

$$g_{1}\left(z\right)=-f\left(z\right),\quad g_{2}\left(z\right)=\overline{f\left(z\right)},\quad g_{3}\left(z\right)=-\overline{f\left(z\right)},$$

and to its rotations by a unit vector, $g_{4}\left(z\right)=e^{i\xi}f\left(z\right)$, since none of these changes $\left|\Delta f\right|$.

Example 8. Let $f,g:\mathbb{C}\longrightarrow\mathbb{C}$ and $z_{0}\in\mathbb{C}$. If $f$ is a constant multiple of $g$ on an environment of $z_{0}$, then they are locally and spatially bimodular there. Indeed, write $f=cg$ on that environment. Then $\Delta f=c\Delta g$, and

$$\left|f\left(z_{0}\right)\Delta g\right|=\left|c\right|\left|g\left(z_{0}\right)\right|\left|\Delta g\right|=\left|g\left(z_{0}\right)\Delta f\right|.$$

The same computation with $z$ in place of $z_{0}$ gives spatial bimodularity. The condition is sufficient but not necessary.

Remark 1. Let $f:\mathbb{C}\longrightarrow\mathbb{C}$ be $\varphi$-detachable at $z_{0}$ and let $\epsilon>0$. The chord-angle analysis showed that detachability implies the bound $(2)$ on the argument of $\Delta f$ on some environment of $z_{0}$. Let us use the following notation for brevity:

$$\theta_{f}\equiv\arg\left(\frac{\Delta f}{f^{;}e^{i\varphi}}\right),\qquad\left|\theta_{f}\right|<2\arcsin\left(\frac{\epsilon}{2}\right),\tag{3}$$

so that

$$\operatorname{sgn}\left(\Delta f\right)=f^{;}e^{i\left(\varphi+\theta_{f}\right)}.\tag{4}$$

Remark 2. Similarly, if $f$ is sign-continuous at $z_{0}$, then given $\epsilon>0$ there is an environment on which $\operatorname{sgn}f(z)$ lies within distance $\frac{\epsilon}{2}$ of $\operatorname{sgn}\left(f\right)$. Let us use the notation

$$\theta\equiv\arg\left(f\left(z\right)\right)-\arg\left(f\left(z_{0}\right)\right),$$

which can be rewritten as

$$\operatorname{sgn}\left(f\left(z\right)\right)=\operatorname{sgn}\left(f\left(z_{0}\right)\right)e^{i\theta}.\tag{5}$$

The following lemma is elementary.

Lemma 3. Let $z_{1},z_{2}\in\mathbb{C}$. If $\left|z_{1}\right|=\left|z_{2}\right|$ and $0<\arg\left(z_{1}\right),\arg\left(z_{2}\right)<\pi$, then

$$\arg\left(z_{1}+z_{2}\right)=\frac{\arg\left(z_{1}\right)+\arg\left(z_{2}\right)}{2}.\tag{6}$$

Indeed, $0$, $z_{1}$, $z_{1}+z_{2}$ and $z_{2}$ form a rhombus, and the diagonal of a rhombus bisects its angle.

Lemma 4. Let $f,g:\mathbb{C}\longrightarrow\mathbb{C}$ be $\varphi$-detachable at $z_{0}\in\mathbb{C}$ and let $\epsilon>0$. Then there exists an environment of $z_{0}$ such that for each $z$ there:

$$\left|e^{\frac{i}{2}\left(\theta_{f}+\theta_{g}\right)}-1\right|<\epsilon.$$

Proof. Take the environment on which formula $(3)$ holds for both $\theta_{f}$ and $\theta_{g}$. A chord-angle analysis of the angle $\frac{\theta_{f}+\theta_{g}}{2}$ gives

$$\begin{aligned} \left|e^{\frac{i}{2}\left(\theta_{f}+\theta_{g}\right)}-1\right|&=2\left|\sin\frac{\theta_{f}+\theta_{g}}{4}\right|\leq2\sin\frac{\left|\theta_{f}\right|+\left|\theta_{g}\right|}{4}\\ &<2\sin\left(\arcsin\frac{\epsilon}{2}\right)=\epsilon, \end{aligned}$$

where the strict inequality uses the bound in formula $(3)$ for both angles and the monotonicity of $\sin$ near zero.$\,\,\,\,\blacksquare$

Lemma 5. Let $f,g:\mathbb{C}\longrightarrow\mathbb{C}$ be $\varphi$-detachable at $z_{0}\in\mathbb{C}$, where $f$ is sign-continuous. Let $\epsilon>0$. Then there exists an environment of $z_{0}$ such that for each $z$ there:

$$\left|e^{\frac{i}{2}\left(\theta+\theta_{f}+\theta_{g}\right)}-1\right|<\epsilon.$$

Proof. Apply formula $(3)$ with $\frac{\epsilon}{2}$ in place of $\epsilon$, so that $\left|\theta_{f}\right|,\left|\theta_{g}\right|<2\arcsin\left(\frac{\epsilon}{4}\right)$. By the sign-continuity of $f$, for each $z$ close enough to $z_{0}$,

$$\left|\operatorname{sgn}\left(f\left(z\right)\right)-\operatorname{sgn}\left(f\right)\right|<\frac{\epsilon}{2},$$

and a chord-angle analysis bounds $\theta=\arg\left(f\left(z\right)\right)-\arg\left(f\left(z_{0}\right)\right)$ in absolute value by $2\arcsin\left(\frac{\epsilon}{4}\right)$. Writing $a=\arcsin\left(\frac{\epsilon}{4}\right)$,

$$\begin{aligned} \left|e^{\frac{i}{2}\left(\theta+\theta_{f}+\theta_{g}\right)}-1\right|&=2\left|\sin\frac{\theta+\theta_{f}+\theta_{g}}{4}\right|<2\sin\left(\frac{3}{2}a\right)\\ &<2\sin\left(2a\right)=4\sin\left(a\right)\cos\left(a\right)\\ &\leq4\sin\left(a\right)=\epsilon.\,\,\,\,\blacksquare \end{aligned}$$

Algebraic Rules

The rules for sums, products and quotients below rest on one step. If $u,v\in S^{1}$ are not opposite, then by Lemma 3 (after a common rotation), for all small angles $\alpha,\beta$,

$$\left|\operatorname{sgn}\left(ue^{i\alpha}+ve^{i\beta}\right)-\operatorname{sgn}\left(u+v\right)\right|=\left|e^{\frac{i}{2}\left(\alpha+\beta\right)}-1\right|.\tag{7}$$

Each proof writes the increment of the combined function as a positive multiple of $e^{i\varphi}\left(ue^{i\alpha}+ve^{i\beta}\right)$ and then bounds the right-hand side of $(7)$ with Lemma 4 or Lemma 5.

Claim 6. Constant multiple rule. Let $f:\mathbb{C}\longrightarrow\mathbb{C}$ be $\varphi$-detachable at $z_{0}\in\mathbb{C}$, and let $c\in\mathbb{C}$. Then $cf$ is also $\varphi$-detachable there and:

$$\left(cf\right)^{;}=\operatorname{sgn}\left(cf^{;}\right).$$

Proof. Directly from the definition of the detachment:

$$\begin{aligned} \left(cf\right)^{;}&=e^{-i\varphi}\lim_{t\to0^{+}}\operatorname{sgn}\left[c\Delta f\right]=\operatorname{sgn}\left(c\right)\,e^{-i\varphi}\lim_{t\to0^{+}}\operatorname{sgn}\left(\Delta f\right)\\ &=\operatorname{sgn}\left(c\right)f^{;}=\operatorname{sgn}\left(cf^{;}\right).\,\,\,\,\blacksquare \end{aligned}$$

Claim 7. Sum and difference rules. Let $f,g:\mathbb{C}\longrightarrow\mathbb{C}$ be $\varphi$-detachable and bimodular at $z_{0}\in\mathbb{C}$, with $f^{;}\pm g^{;}\neq0$. Then $f\pm g$ is also $\varphi$-detachable there and:

$$\left(f\pm g\right)^{;}=\operatorname{sgn}\left(f^{;}\pm g^{;}\right).$$

Proof. We prove the formula for the sum. The difference follows by applying it to $f$ and $-g$, which is bimodular to $f$ as well (Example 7) and has $(-g)^{;}=-g^{;}$ by Claim 6. The detachments $f^{;},g^{;}$ are nonzero: if one of them were $0$, that function's increment would vanish on the environment, bimodularity would make the other increment vanish too, and then $f^{;}+g^{;}=0$. Let $\epsilon>0$. There exist $\delta$-environments of $z_{0}$ on which the following conditions hold:

1. On the $\delta_{1}$-environment, the bound of Lemma 4 holds, since $f$ and $g$ are detachable.
2. On the $\delta_{2}$-environment, $\left|\Delta f\right|=\left|\Delta g\right|$, since $f,g$ are bimodular.

For each $z$ in the $\min\left\{ \delta_{1},\delta_{2}\right\}$-environment, formula $(4)$ for $f$ and $g$ and the common modulus give

$$\Delta f+\Delta g=\left|\Delta f\right|e^{i\varphi}\left(f^{;}e^{i\theta_{f}}+g^{;}e^{i\theta_{g}}\right),$$

so with $u=f^{;}$ and $v=g^{;}$ in formula $(7)$,

$$\begin{aligned} &\left|\operatorname{sgn}\left[\Delta\left(f+g\right)\right]-\operatorname{sgn}\left(f^{;}+g^{;}\right)e^{i\varphi}\right|\\ &=\left|\operatorname{sgn}\left(f^{;}e^{i\theta_{f}}+g^{;}e^{i\theta_{g}}\right)-\operatorname{sgn}\left(f^{;}+g^{;}\right)\right|\\ &=\left|e^{\frac{i}{2}\left(\theta_{f}+\theta_{g}\right)}-1\right|<\epsilon, \end{aligned}$$

where the inequality is the first condition.$\,\,\,\,\blacksquare$

The condition $f^{;}+g^{;}\neq0$ cannot be dropped. When the two detachments are opposite, the increments nearly cancel, and the direction of what remains depends on details that neither detachment records.

Claim 8. Product rule. Let $f,g:\mathbb{C}\longrightarrow\mathbb{C}$ be $\varphi$-detachable and spatially bimodular at $z_{0}\in\mathbb{C}$, with $f(z_0),g(z_0)\neq0$. Assume that $f$ is sign-continuous and $g$ has constant modulus at $z_{0}$, or vice versa, and that the bracket on the right-hand side below is nonzero. Then $fg$ is also $\varphi$-detachable at $z_{0}$, and:

$$\left(fg\right)^{;}=\operatorname{sgn}\left[f^{;}\operatorname{sgn}\left(g\right)+g^{;}\operatorname{sgn}\left(f\right)\right].$$

Proof. Let $\epsilon>0$, and assume without loss of generality that $f$ is sign-continuous and $g$ has constant modulus at $z_{0}$. There exist $\delta$-environments of $z_{0}$ on which the following conditions hold:

1. On the $\delta_{1}$-environment, the bound of Lemma 5 holds, since $f,g$ are detachable and $f$ is sign-continuous.
2. On the $\delta_{2}$-environment, $\left|g\left(z\right)\right|=\left|g\left(z_{0}\right)\right|$, since $g$ has constant modulus.
3. On the $\delta_{3}$-environment, $\left|f(z)\Delta g\right|=\left|g(z)\Delta f\right|$, since $f,g$ are spatially bimodular.

Let $z$ be in the $\min\left\{ \delta_{1},\delta_{2},\delta_{3}\right\}$-environment. Adding and subtracting $f\left(z\right)g\left(z_{0}\right)$,

$$\left(fg\right)\left(z\right)-\left(fg\right)\left(z_{0}\right)=f\left(z\right)\Delta g+g\left(z_{0}\right)\Delta f.$$

By the second and third conditions, both terms have the same modulus $M=\left|f(z)\Delta g\right|=\left|g(z_0)\Delta f\right|$. Formulas $(4)$ and $(5)$ then give

$$\Delta\left(fg\right)=Me^{i\varphi}\left[\operatorname{sgn}\left(f\right)g^{;}e^{i\left(\theta+\theta_{g}\right)}+\operatorname{sgn}\left(g\right)f^{;}e^{i\theta_{f}}\right].$$

Let $u=\operatorname{sgn}\left(f\right)g^{;}$ and $v=\operatorname{sgn}\left(g\right)f^{;}$. These are unit vectors: if $f^{;}$ or $g^{;}$ were $0$, the conditions above would make both increments vanish and the bracket in the claim would be $0$. By formula $(7)$,

$$\begin{aligned} &\left|\operatorname{sgn}\left[\Delta\left(fg\right)\right]-\operatorname{sgn}\left(u+v\right)e^{i\varphi}\right|\\ &=\left|\operatorname{sgn}\left(ue^{i\left(\theta+\theta_{g}\right)}+ve^{i\theta_{f}}\right)-\operatorname{sgn}\left(u+v\right)\right|\\ &=\left|e^{\frac{i}{2}\left(\theta+\theta_{f}+\theta_{g}\right)}-1\right|<\epsilon, \end{aligned}$$

where the inequality is due to Lemma 5. Since $u+v=f^{;}\operatorname{sgn}\left(g\right)+g^{;}\operatorname{sgn}\left(f\right)$, this is the claim.$\,\,\,\,\blacksquare$

Claim 9. Quotient rule. Let $f,g:\mathbb{C}\longrightarrow\mathbb{C}$ be $\varphi$-detachable and locally bimodular at $z_{0}\in\mathbb{C}$, with $f(z_0)\neq0$ and $g\neq0$ near $z_0$, where $g$ is sign-continuous at $z_{0}$. Assume also that $\operatorname{sgn}\left(g\right)f^{;}\neq\operatorname{sgn}\left(f\right)g^{;}$. Then $\frac{f}{g}$ is also $\varphi$-detachable there, and:

$$\left(\frac{f}{g}\right)^{;}=\operatorname{sgn}\left[\frac{\operatorname{sgn}\left(g\right)f^{;}-\operatorname{sgn}\left(f\right)g^{;}}{g^{2}}\right].$$

Proof. First, let us develop the expression by the definition of the limit. Over a common denominator,

$$\left(\frac{f}{g}\right)\left(z\right)-\left(\frac{f}{g}\right)\left(z_{0}\right)=\frac{g\left(z_{0}\right)\Delta f-f\left(z_{0}\right)\Delta g}{g\left(z\right)g\left(z_{0}\right)},$$

so

$$\begin{aligned} \left(\frac{f}{g}\right)^{;}&=\lim_{t\to0^{+}}\frac{e^{-i\varphi}\operatorname{sgn}\left[g\left(z_{0}\right)\Delta f-f\left(z_{0}\right)\Delta g\right]}{\operatorname{sgn}\left(g\left(z\right)\right)\operatorname{sgn}\left(g\right)}\\ &=\frac{1}{\operatorname{sgn}\left(g\right)^{2}}\lim_{t\to0^{+}}e^{-i\varphi}\operatorname{sgn}\left[g\left(z_{0}\right)\Delta f-f\left(z_{0}\right)\Delta g\right], \end{aligned}$$

where the second transition is because $g$ is sign-continuous at $z_{0}$, provided the last limit exists. By local bimodularity, both terms inside the bracket have the same modulus $M=\left|g\left(z_0\right)\Delta f\right|=\left|f\left(z_0\right)\Delta g\right|$, and formula $(4)$ gives

$$g\left(z_{0}\right)\Delta f-f\left(z_{0}\right)\Delta g=Me^{i\varphi}\left[\operatorname{sgn}\left(g\right)f^{;}e^{i\theta_{f}}-\operatorname{sgn}\left(f\right)g^{;}e^{i\theta_{g}}\right].$$

Let $u=\operatorname{sgn}\left(g\right)f^{;}$ and $v=-\operatorname{sgn}\left(f\right)g^{;}$. Both are unit vectors, since by local bimodularity one increment vanishes exactly when the other does, which would give $u+v=0$; and they are not opposite by assumption. Next, let us show with an $\epsilon$-$\delta$ analysis that the sign of the bracket tends to $\operatorname{sgn}\left(u+v\right)$. Let $\epsilon>0$. There exist $\delta$-environments of $z_{0}$ on which the following conditions hold:

1. On the $\delta_{1}$-environment, the bound of Lemma 4 holds, since $f$ and $g$ are detachable.
2. On the $\delta_{2}$-environment, $\left|f(z_0)\Delta g\right|=\left|g(z_0)\Delta f\right|$, since $f,g$ are locally bimodular.

For each $z$ in the $\min\left\{ \delta_{1},\delta_{2}\right\}$-environment, formula $(7)$ gives

$$\begin{aligned} &\left|\operatorname{sgn}\left(ue^{i\theta_{f}}+ve^{i\theta_{g}}\right)-\operatorname{sgn}\left(u+v\right)\right|\\ &=\left|e^{\frac{i}{2}\left(\theta_{f}+\theta_{g}\right)}-1\right|<\epsilon, \end{aligned}$$

where the inequality is due to Lemma 4. Hence the last limit exists and equals $\operatorname{sgn}\left(u+v\right)$. Finally, $\operatorname{sgn}\left(g\right)^{2}=\operatorname{sgn}\left(g^{2}\right)$ and $\operatorname{sgn}\left(w\right)/\operatorname{sgn}\left(q\right)=\operatorname{sgn}\left(w/q\right)$ give the formula.$\,\,\,\,\blacksquare$

References

[1] Shabat, B.V., 2003. Introduction to Complex Analysis-excerpts.

[2] Druţu, C. and Kapovich, M., 2018. Geometric group theory (Vol. 63). American Mathematical Soc..

[3] Gelbaum, B.R. and Olmsted, J.M.H., 1964. Counterexamples in Analysis. Holden-Day (Dover reprint, 2003).

[4] Allamigeon, X., Gaubert, S. and Skomra, M., 2020. Tropical spectrahedra. Discrete & Computational Geometry, 63, pp. 507–548.

[5] Shachar, A., 2010. Introduction to Semi-discrete Calculus. arXiv:1012.5751.