Linkedin Envelope

Semi-discrete Calculus in Complex Analysis

Can we apply the detachment operator in the Complex plane? As in Real Analysis, a theory based on the derivative sign would go through rates calculations and be trivial. But what if we can introduce a theory about the properties of “complex trends” while calculating them directly without going through the derivative? Would this theory apply to non-differentiable and even discontinuous functions, similar to the case in Real Analysis? And would it capture trends in the function’s image more reliably and coherently than the derivative sign?

Complex Analysis is abundant with concepts that generalize their Real Analysis counterparts, such as continuity and the derivative. Inspired by its AI, natural sciences, and Real Analysis applications, let us consider a complex version of a function’s detachment and analyze its basic mathematical properties.

Defining the Complex Detachment

Recall that the signum of a complex number $z$ is defined as: $$sgn\left(z\right)\equiv\frac{z}{\left|z\right|}=e^{iarg\left(z\right)},$$ where $arg\left(\cdot\right)$ is the complex argument function. Geometrically, this means that the sign function maps any point in the complex plane to its closest counterpart on the unit circle:

Note that in Real Analysis, we suggested softening the definition of the detachment with respect to the derivative by considering merely one-sided detachments, defined by one-sided limits, and didn’t require them to agree. We naturally extend the notion of one-sided derivatives in higher dimensions with directional derivatives. The Complex Analysis literature conventionally focuses on differentiable and, more specifically, analytic functions. However, complex partial derivatives, also known as Wirtinger derivatives, are prevalent. They particularly serve to define the complex directional derivatives, which are useful as well, depending on the context. For example, they are applied in advanced research (see [2], p. 722), and also serve as a pedagogical and auxiliary thought tool (see [1], p. 19). Let us suggest an analogous natural way to define the directional detachment.

Definition 1. Let us define the $\varphi$-detachment of a function $f:\mathbb{C}\longrightarrow\mathbb{C}$ at a point $z_{0}\in\mathbb{C}$ as follows: $$\begin{array}{ccc} & f_{\varphi}^{;}:\mathbb{C}\longrightarrow S^{1}\bigcup\left\{ 0\right\} \\ & f_{\varphi}^{;}\left(z_{0}\right)\equiv e^{-i\varphi}\underset{\begin{array}{c} \left|z\right|\to\left|z_{0}\right|\\ \arg\left(\Delta z\right)=\varphi \end{array}}{\lim}sgn\left[f\left(z\right)-f\left(z_{0}\right)\right],\label{detachment_definition}\tag{1} \end{array}$$ where $S^{1}$ is the unit circle in the complex plane. We will say that $f$ is $\varphi$-detachable at $z_{0}$ if the limit exists. We may think of the $e^{-i\varphi}$ coefficient as a generalization of the $\pm 1$ coefficient from the one-sided detachment in Real Analysis. Both coefficients render the detachment more consistent with the derivative sign, as shown in the following statement.

Feel free to gain intuition by hovering over the following diagram:

$\Delta f=f(z)-f(z_0)$

The following claim shows that the detachment adds a calculation value with respect to the derivative only in cases where the derivative vanishes.

Claim 1. If $f:\mathbb{C}\longrightarrow\mathbb{C}$ is both differentiable and $\varphi$-detachable at the point $z_{0},$ and $f’\left(z_{0}\right)\neq0,$ then: $$f_{\varphi}^{;}\left(z_{0}\right)=sgn\left[\frac{\partial f}{\partial\varphi}\left(z_{0}\right)\right].$$
Proof. According to the definition: $$\begin{align*}f_{\varphi}^{;}\left(z_{0}\right) &\equiv e^{-i\varphi}\underset{\begin{array}{c} \left|z\right|\to\left|z_{0}\right|\\ \arg\left(\Delta z\right)=\varphi \end{array}}{\lim}sgn\left[f\left(z\right)-f\left(z_{0}\right)\right]=\underset{\begin{array}{c} \left|z\right|\to\left|z_{0}\right|\\ \arg\left(\Delta z\right)=\varphi \end{array}}{\lim}\frac{sgn\left[f\left(z\right)-f\left(z_{0}\right)\right]}{e^{i\varphi}}=\underset{\begin{array}{c} \left|z\right|\to\left|z_{0}\right|\\ \arg\left(\Delta z\right)=\varphi \end{array}}{\lim}\frac{sgn\left[f\left(z\right)-f\left(z_{0}\right)\right]}{sgn\left(z-z_{0}\right)} \\ &=\underset{\begin{array}{c} \left|z\right|\to\left|z_{0}\right|\\ \arg\left(\Delta z\right)=\varphi \end{array}}{\lim}sgn\left[\frac{f\left(z\right)-f\left(z_{0}\right)}{z-z_{0}}\right]=sgn\underset{\begin{array}{c} \left|z\right|\to\left|z_{0}\right|\\ \arg\left(\Delta z\right)=\varphi \end{array}}{\lim}\left[\frac{f\left(z\right)-f\left(z_{0}\right)}{z-z_{0}}\right]=sgn\left[\frac{\partial f}{\partial\varphi}\left(z_{0}\right)\right], \end{align*}$$ where the fifth transition is because the sign function is continuous in $\mathbb{C}\backslash\left\{ 0\right\}$.
For brevity, we will denote the $\varphi$-detachment at the point $z_{0}$ by $f^{;},$ rather than $f_{\varphi}^{;}\left(z_{0}\right).$

Geometric Intuition

Suppose that a function is $\varphi$-detachable at $z_{0}.$ Then by the limit definition, given $\epsilon>0,$ for any sufficiently small environment of $z_{0},$ the function’s values at a point $z$ there, satisfy that: $$\left|sgn\left[f\left(z\right)-f\left(z_{0}\right)\right]-f^{;}\right|<\epsilon.$$ This means that $f\left(z\right)$ is bound between two lines intersecting at $z_{0}.$ The angle between the spanning vector of each line and $f^{;},$ is $\epsilon.$
Recall that in the realm of the real functions, the detachment definition is equivalent to the following statement: $$\exists\dot{B}_{\pm}\left(x_{0}\right):\forall x\in\dot{B}_{\pm}\left(x_{0}\right):\,\,\,\,\,sgn\left[f\left(x\right)-f\left(x_{0}\right)\right]=f_{\pm}^{;}\left(x_{0}\right),$$ due to the discontinuity of the sign function at zero. However, an analogous statement in the complex domain doesn’t necessarily hold. Such a statement is more strict than the existence of the limit stated in equation $(\ref{detachment_definition})$. Geometrically, the former is defined merely for functions whose image near a point is on a straight line. The reason lies in the different geometric interpretations of the discontinuity of the sign function in one vs. two dimensional functions.

We know that the limit $\underset{z\to0}{\lim}sgn\left(z\right)$ doesn’t exist because it depends on the different paths by which we let $z$ approach zero. Can this intuition help us establish a geometric interpretation for the detachment?

The detachability of a real valued function at a point is a synonymous with the function having a “local trend” there. The detachability of a complex function points out the existence of a “local trend” in a broader sense.

Intuitively, a $\theta$-detachable function maps points along a line in the environment of the point, to points “approaching” a single line.

The differentiability of a function at a point allows us to calculate its values with the linear approximation formula. It states that 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 $\underset{z\to z_{0}}{\lim}\rho\left(z\right)=0$ and $z\neq z_{0}.$

In contrast, the detachment suggests a different kind of approximated linear relation of the function’s image, where the formula itself does not involve information from the domain. Suppose $f$ is $\varphi$-detachable, and let $\epsilon>0.$ According to the limit definition, we have that for $z$ in a sufficiently small environment of $z_{0}$ such that $\arg\left(z-z_{0}\right)=\varphi$:

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

Since $f^{;}$ is the limit of points on the (closed) unit circle, it has to be on the circle as well. The size of the chord between $sgn\left[f\left(z\right)-f\left(z_{0}\right)\right]$ and $f^{;}e^{i\varphi}$ is at most $\epsilon.$ Let us assume without loss of generality that $0<\arg\left(f^{;}e^{i\varphi}\right)<\frac{\pi}{2}.$ Let $\theta_{f}$ be the angle between $sgn\left[f\left(z\right)-f\left(z_{0}\right)\right]$ and $f^{;}e^{i\varphi}.$ The formula for the size of a chord given the angle $\theta_{f}$ and radius $1$ yields: $$\left|sgn\left[f\left(z\right)-f\left(z_{0}\right)\right]-f^{;}e^{i\varphi}\right|=2\cdot1\cdot\sin\left(\frac{\theta_{f}}{2}\right)\Longrightarrow\theta_{f}=2\arcsin\left(\frac{\left|sgn\left[f\left(z\right)-f\left(z_{0}\right)\right]-f^{;}e^{i\varphi}\right|}{2}\right)<2\arcsin\left(\frac{\epsilon}{2}\right).$$
Therefore, we showed that the argument of the difference between the function’s image at points in the environment of $z_{0}$ and its image at $z_{0}$ satisfies: $$arg\left(f\left(z\right)-f\left(z_{0}\right)\right)\in\left(\arg\left(f^{;}e^{i\varphi}\right)-2\arcsin\left(\frac{\epsilon}{2}\right),\arg\left(f^{;}e^{i\varphi}\right)+2\arcsin\left(\frac{\epsilon}{2}\right)\right).\label{arg_delta_f_bounds}\tag{2}$$ We will refer the above analysis that associates the bound on an the chord length with that of the angle, as the “chord-angle” analysis.

Computability and Consistency

Similar to the case in Real Analysis, detachability at a point does not necessarily imply continuity or differentiability – and vice versa. Let us illustrate 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 by the definition: $$f_{\varphi}^{;}\left(0\right)\equiv e^{-i\varphi}\underset{\begin{array}{c} \left|z\right|\to\left|z_{0}\right|\\ \arg\left(\Delta z\right)=\varphi \end{array}}{\lim}sgn\left[f\left(z\right)-f\left(z_{0}\right)\right]=e^{-i\varphi}\cdot\underset{\begin{array}{c} \left|z\right|\to\left|z_{0}\right|\\ \arg\left(\Delta z\right)=\varphi \end{array}}{\lim}\left[\frac{1}{sgn\left(z\right)}\right]=e^{-2i\varphi}.$$ This is an example of a non-differentiable and discontinuous function, for whom the detachment is nevertheless able to provide monotony information.
Example 2. Let $f\left(z\right)=z^{2}.$ The derivative is $f’\left(z_{0}\right)=2z_{0}.$ Particularly, at $z_{0}=0$ the derivative vanishes and we ought to calculate the detachment based on its definition: $$f_{\varphi}^{;}\left(0\right)\equiv e^{-i\varphi}\underset{\begin{array}{c} \left|z\right|\to\left|z_{0}\right|\\ \arg\left(\Delta z\right)=\varphi \end{array}}{\lim}sgn\left[f\left(z\right)-f\left(z_{0}\right)\right]=e^{-i\varphi}\underset{\begin{array}{c} \left|z\right|\to\left|z_{0}\right|\\ \arg\left(\Delta z\right)=\varphi \end{array}}{\lim}sgn\left(z^{2}\right)=e^{-i\varphi}\cdot e^{2i\varphi}=e^{i\varphi}.$$ Thus, similarly to the case in Real Analysis, while the directional derivatives don’t provide monotony information at the stationary point, the directional detachment does.
Example 3. Let $f\left(z\right)=\sqrt{z}.$ The derivative is $f’\left(z_{0}\right)=\frac{1}{2\sqrt{z_{0}}}.$ Particularly, at $z_{0}=0$ the derivative doesn’t exist and we ought to calculate the detachment based on its definition: $$f_{\varphi}^{;}\left(0\right)\equiv\underset{\begin{array}{c} \left|z\right|\to\left|z_{0}\right|\\ \arg\left(\Delta z\right)=\varphi_{2} \end{array}}{\lim}sgn\left[f\left(z\right)-f\left(z_{0}\right)\right]=\underset{\begin{array}{c} \left|z\right|\to\left|z_{0}\right|\\ \arg\left(z\right)=\varphi_{2} \end{array}}{\lim}sgn\left(\sqrt{z}\right)=e^{\frac{i\varphi}{2}}.$$ Thus, similarly to the case in Real Analysis, the detachment may provide monotony information at the derivative’s singularities, for continuous functions as well.
Example 4. Let $$f\left(z\right)=\begin{cases} z\sin\left(z\right), & z\neq0,\\ 0, & z=0. \end{cases}$$ Then $f$ is differentiable at $z=0,$ but the definition of its directional detachment there yields: $$f_{\varphi}^{;}\left(0\right)\equiv\underset{\begin{array}{c} \left|z\right|\to\left|z_{0}\right|\\ \arg\left(\Delta z\right)=\varphi_{2} \end{array}}{\lim}sgn\left[f\left(z\right)-f\left(z_{0}\right)\right]=\underset{\begin{array}{c} \left|z\right|\to\left|z_{0}\right|\\ \arg\left(z\right)=\varphi_{2} \end{array}}{\lim}sgn\left(z\sin\left(z\right)\right)=e^{i\varphi}\underset{\begin{array}{c} \left|z\right|\to\left|z_{0}\right|\\ \arg\left(\Delta z\right)=\varphi_{2} \end{array}}{\lim}sgn\left[\sin\left(z\right)\right],$$ which is undefined. Thus, similarly to the case in Real Analysis, the complex detachment may not provide monotony information at points of infinite oscillations.
Example 5. Let $f\left(z\right)=\Re\left(z\right),$ whose image is the real line. The sign of its directional derivative satisfies: $$sgn\left[\frac{\partial f}{\partial\varphi}\left(z_{0}\right)\right]\equiv sgn\left\{ \underset{\begin{array}{c} \left|z\right|\to\left|z_{0}\right|\\ \arg\left(\Delta z\right)=\varphi \end{array}}{\lim}\left[\frac{\Re\left(\Delta z\right)}{\Delta z}\right]\right\} =sgn\left[\cos\left(\varphi\right)e^{-i\varphi}\right]=\begin{cases} e^{-i\varphi}, & \left|\varphi\right|<\frac{\pi}{2},\\ 0, & \left|\varphi\right|=\frac{\pi}{2},\\ -e^{-i\varphi}, & \text{else}, \end{cases}.$$ Since the derivative may vanish, let us calculate the $\varphi$-detachment directly according to the definition: $$f_{\varphi}^{;}\left(z_{0}\right) \equiv e^{-i\varphi}\underset{\begin{array}{c} \left|z\right|\to\left|z_{0}\right|\\ \arg\left(\Delta z\right)=\varphi \end{array}}{\lim}sgn\left[f\left(z\right)-f\left(z_{0}\right)\right]=e^{-i\varphi}\underset{\begin{array}{c} \left|z\right|\to\left|z_{0}\right|\\ \arg\left(\Delta z\right)=\varphi \end{array}}{\lim}sgn\left[\Re\left(\Delta z\right)\right]=\begin{cases} e^{-i\varphi}, & \left|\varphi\right|<\frac{\pi}{2},\\ 0, & \left|\varphi\right|=\frac{\pi}{2},\\ -e^{-i\varphi}, & \text{else}. \end{cases}$$ This is an example where the detachment agrees with the derivative sign even in cases when it vanishes ($\left|\varphi\right|=\frac{\pi}{2}$ in this example).

Auxiliary Terminology

Assumptions and abbreviations. We will apply the following notations below:
    • We focus on directional detachments in a given angle $\varphi.$ Therefore, by a $\delta$-“environment” of the point $z_{0}$ we refer to the $\delta$-lengthed line segment starting at $z_{0}$ and tilted by $\varphi$ with respect to $z_{0}.$
    • We assume without loss of generality that all the complex numbers’ arguments are between $0$ and $\pi.$
    • For the two variables $z_{0}$ and $z$ in the definition domain of a function $f,$ we apply the abbreviation $\Delta f=f\left(z\right)-f\left(z_{0}\right).$
    • The value of an operator applied to a function is calculated at $z_{0}$ unless we specifically mention otherwise. For example, $sgn\left(f\right)=sgn\left(f\left(z_{0}\right)\right).$
We assume detachability, but not differentiability nor continuity, of the functions involved in the results below. As a tradeoff, we will apply the following intuitive and useful concepts.

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

Definition 3. Let $f,g:\mathbb{C}\longrightarrow\mathbb{C}$ be a pair of functions, and let $z_{0}\in\mathbb{C}.$ Then $f,g$ are:

    • Bimodular at $z_{0}$ if for each $z$ in its neighborhood: $\left|\Delta f\right|=\left|\Delta g\right|.$
    • Locally bimodular at $z_{0}$ if for each $z$ in its neighborhood, $\left|f\Delta g\right|=\left|g\Delta f\right|.$
    • Spatially bimodular at $z_{0} $ if for each $z$ in its neighborhood, $\left|f\left(z\right)\Delta g\right|=\left|g\left(z\right)\Delta f\right|.$
Example 6. Any complex function is sign-continuous whenever the function’s angle of change is continuous. It may happen at point of non-differentiability, such as cusps. It may also happen at discontinuities, for example if the function’s graph approaches a line in a discontinuous fashion.
Example 7. Any function $g:\mathbb{C}\longrightarrow\mathbb{C}$ is everywhere bimodular to all its reflections with respect to the axes: $$g_{1}\left(z\right)=-f\left(z\right), g_{2}\left(z\right)=\overline{f\left(z\right)}, g_{3}\left(z\right)=-\overline{f\left(z\right)},$$ and to all its rotations by a unit vector, $g_{4}\left(z\right)=f\left(z\right)e^{i\xi}.$
Example 8. Let $f,g:\mathbb{C}\longrightarrow\mathbb{C},$ and let $z_{0}\in\mathbb{C}.$ If there exists an environment where $g$ is a constant multiplication of $f,$ then they are locally and spatially bimodular. More precisely, if $g\neq0$ and $\frac{f}{g}$ is continuous at $z_{0}$ and constant in its neighborhood, then $f,g$ are both locally and spatially bimodular at $z_{0}.$ Let us show local bimodularity. Given $z$ in an environment of $z_{0} $ where $\frac{f}{g}$ is held constant: $$\left|f\left(z_{0}\right)\Delta g\right| =\left|\frac{f\left(z_{0}\right)}{g\left(z_{0}\right)}g\left(z_{0}\right)g\left(z\right)-f\left(z_{0}\right)g\left(z_{0}\right)\right|=\left|\frac{f\left(z\right)}{g\left(z\right)}g\left(z_{0}\right)g\left(z\right)-f\left(z_{0}\right)g\left(z_{0}\right)\right|=\left|g\left(z_{0}\right)\Delta f\right|,$$ where the fourth equality is due to the continuity and constancy of $\frac{f}{g}.$ A similar argument holds for their spatial bimodularity. Note that this condition is sufficient but not necessary for bimodularity.
Remark 1. Let $f:\mathbb{C}\longrightarrow\mathbb{C}$ be a detachable function and $z_{0}\in C.$ Let $\epsilon>0.$ We previously showed with a chord-angle analysis, the detachability of a function $f$ implies that there exists an environment of $z_{0}$ wherein: $$\arg\left(\Delta f\right)\in\left(\arg\left(f^{;}e^{i\varphi}\right)-2\arcsin\left(\frac{\epsilon}{2}\right),\arg\left(f^{;}e^{i\varphi}\right)+2\arcsin\left(\frac{\epsilon}{2}\right)\right).$$ Let us use the following notation for brevity: $$\theta_{f}\equiv\arg\left(\frac{\Delta f}{f^{;}e^{i\varphi}}\right)<2\arcsin\left(\frac{\epsilon}{2}\right)\tag{3}\label{theta_f_bound},$$ where the equality can be rewritten as follows: $$sgn\left(\Delta f\right)=f^{;}e^{i\left(\varphi+\theta_{f}\right)}\tag{4}\label{sgn_df_formula}.$$

Remark 2. Similarly, if $f$ is sign-continuous, then given $\epsilon>0,$ the limit definition ensures the existence of an environment such that for each $z$ there, $\left|sgn\left(f\left(z\right)\right)-sgn\left(f\right)\right|<\frac{\epsilon}{2},$ therefore $sgn\left(f\left(z\right)\right)\in\left(sgn\left(f\right)-\epsilon,sgn\left(f\right)+\epsilon\right).$ Let us use the following notation:

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

which can be rewritten as $$sgn\left(f\left(z\right)\right)=sgn\left(f\left(z_{0}\right)\right)e^{i\theta}.\label{sgn_f_formula}\tag{5}$$

Let us cite the following known result without proof.

Lemma 2. Let $z_{1},z_{2}\in\mathbb{C}.$ If $\left|z_{1}\right|=\left|z_{2}\right|$ such that $0<\arg\left(z_{1}\right),\arg\left(z_{2}\right)<\pi$ then: $$\arg\left(z_{1}+z_{2}\right)=\frac{\arg\left(z_{1}\right)+\arg\left(z_{2}\right)}{2}.\tag{6}\label{arg_sum}$$

Lemma 3. Let $f,g:\mathbb{C}\longrightarrow\mathbb{C}$ be detachable functions 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. By the following transitions:

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

where the first equality is due to a chord-angle analysis of the angle $\frac{\theta_{f}+\theta_{g}}{2}$, the inequality is due to the upper bound provided in formula ($\ref{theta_f_bound}$),
applied to both $\theta_{f},\theta_{g}$, combined with the monotony of the $\sin$ function near zero.$\,\,\,\,\blacksquare$

Lemma 4. Let $f,g:\mathbb{C}\longrightarrow\mathbb{C}$ be detachable functions at $z_{0}\in\mathbb{C},$ where $f$ is sign-continuous there. 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. From the sign continuity of $f$ we know that for each $z$ close enough to $z_{0}$: $$\left|sgn\left(f\left(z\right)\right)-sgn\left(f\right)\right|<\frac{\epsilon}{2},$$ therefore $sgn\left(f\left(z\right)\right)\in\left(sgn\left(f\right)-\epsilon,sgn\left(f\right)+\epsilon\right),$ and a chord-angle analysis yields that we can bound the angle $\theta=\arg\left(f\left(z\right)\right)-\arg\left(f\left(z_{0}\right)\right),$ from above, by $2\arcsin\left(\frac{\epsilon}{4}\right).$ Thus, it holds that: $$\begin{align} \left|e^{\frac{i}{2}\left(\theta+\theta_{f}+\theta_{g}\right)}-1\right| &=2\sin\left(\frac{\theta+\theta_{f}+\theta_{g}}{4}\right)<2\sin\left(\frac{2\arcsin\left(\frac{\epsilon}{4}\right)+2\arcsin\left(\frac{\epsilon}{4}\right)+2\arcsin\left(\frac{\epsilon}{4}\right)}{4}\right)=2\sin\left(\frac{3}{2}\arcsin\left(\frac{\epsilon}{4}\right)\right) \\ &<2\sin\left(2\arcsin\left(\frac{\epsilon}{4}\right)\right)=4\sin\left(\arcsin\left(\frac{\epsilon}{4}\right)\right)\cos\left(\arcsin\left(\frac{\epsilon}{4}\right)\right)\leq4\sin\left(\arcsin\left(\frac{\epsilon}{4}\right)\right)=\epsilon.\,\,\,\,\blacksquare \end{align}$$

Algebraic Rules

Claim 5. Constant multiple rule. Let $f:\mathbb{C}\longrightarrow\mathbb{C}$ be detachable at the point $z_{0}\in\mathbb{C}$, and let $c\in\mathbb{C}$. Then $cf$ is also detachable there and:

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

Proof. Directly from the definition of the detachment: $$\begin{align*} \left(cf\right)^{;}\left(z_{0}\right) \equiv\underset{z\to z_{0}}{\lim}sgn\left[\left(cf\right)\left(z\right)-\left(cf\right)\left(z_{0}\right)\right] &=\underset{z\to z_{0}}{\lim}sgn\left[c\Delta f\right] \\ &=\underset{z\to z_{0}}{\lim}sgn\left(c\right)sgn\left(\Delta f\right) \\ &=sgn\left(c\right)\underset{z\to z_{0}}{\lim}sgn\left(\Delta f\right) \\ &=sgn\left(c\right)f^{;}=sgn\left(cf^{;}\right).\,\,\,\,\blacksquare \end{align*}$$
Claim 6. Sum and difference rules. Let $f,g:\mathbb{C}\longrightarrow\mathbb{C}$ be detachable at $z_{0}\in\mathbb{C}$ and $f,g$ be bimodular at $z_{0}.$ Then $f\pm g$ is also detachable there and: $$\left(f\pm g\right)^{;}=sgn\left(f^{;}\pm g^{;}\right).$$
Proof. Let $\epsilon>0.$ Without loss of generality assume that $\arg\left(f^{;}\right)\leq\arg\left(g^{;}\right).$ Let us prove the formula for the sum, and leave the one for the difference as an exercise. There exist $\delta-environments of z_{0}$ such that for each $z$ there, the following conditions hold:
Environment size ($\delta$) Condition Reason
$\delta_1$
$\left|e^{\frac{i}{2}\left(\theta_{f}+\theta_{g}\right)}-1\right|<\epsilon$
Lemma 3, since f and g are detachable
$\delta_2$
$\left|\Delta f\right|=\left|\Delta g\right|$
$f,g$ are bimodular
Then, for each $z$ in the $\min\left\{ \delta_{1},\delta_{2}\right\}$ -environment, it holds that: $$\begin{align*} \left|sgn\left(\left(f+g\right)\left(z\right)-\left(f+g\right)\left(z_{0}\right)\right)-sgn\left(f^{;}+g^{;}\right)e^{i\varphi}\right| &=\left|sgn\left(\Delta f+\Delta g\right)-sgn\left(f^{;}+g^{;}\right)e^{i\varphi}\right| \\ &=\left|sgn\left(\left|\Delta f\right|f^{;}e^{i\theta_{f}}e^{i\varphi}+\left|\Delta \\ g\right|g^{;}e^{i\theta_{g}}e^{i\varphi}\right)-sgn\left(f^{;}+g^{;}\right)e^{i\varphi}\right| \\ &=\left|sgn\left(\left|\Delta f\right|f^{;}e^{i\theta_{f}}+\left|\Delta g\right|g^{;}e^{i\theta_{g}}\right)-sgn\left(f^{;}+g^{;}\right)\right| \\ &=\left|e^{i\arg\left[\left|\Delta f\right|f^{;}e^{i\theta_{f}}+\left|\Delta g\right|g^{;}e^{i\theta_{g}}\right]}-e^{\frac{i}{2}\left[\arg\left(f^{;}\right)+\arg\left(g^{;}\right)\right]}\right| \\ &=\left|e^{\frac{i}{2}\left[\arg\left(f^{;}e^{i\theta_{f}}\right)+\arg\left(g^{;}e^{i\theta_{g}}\right)\right]}-e^{\frac{i}{2}\left[\arg\left(f^{;}\right)+\arg\left(g^{;}\right)\right]}\right| \\ &=\left|e^{\frac{i}{2}\left[\arg\left(f^{;}\right)+\theta_{f}+\arg\left(g^{;}\right)+\theta_{g}\right]}-e^{\frac{i}{2}\left[\arg\left(f^{;}\right)+\arg\left(g^{;}\right)\right]}\right| \\ &=\left|e^{\frac{i}{2}\left[\arg\left(f^{;}\right)+\arg\left(g^{;}\right)\right]}\left\{ e^{\frac{i}{2}\left(\theta_{f}+\theta_{g}\right)}-1\right\} \right| \\ &=\left|e^{\frac{i}{2}\left(\theta_{f}+\theta_{g}\right)}-1\right|<\epsilon, \end{align*}$$ where the second transition is due to the notation in formula $\ref{sgn_df_formula}$, the fifth transition is due to the third condition above and formula ($\ref{arg_sum}$). The inequality is due to the first condition.$\,\,\,\,\blacksquare$

Claim 4. Product rule. Let $f,g:\mathbb{C}\longrightarrow\mathbb{C}$ be spatially bimodular and detachable at the point $z_{0}\in\mathbb{C}.$ Assume that $f$ is sign-continuous, and $g,g(z_0)$ are bimodular, or vice versa, at $z_{0}.$ Then $fg$ is also detachable at $z_{0},$ and:

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

Proof. Let $\epsilon>0,$ and without loss of generality assume that $f$ is sign continuous and $g,g\left(z_{0}\right)$ are bimodular at $z_{0}.$ There exist $\delta$-environments such that for each $z$ there, the following conditions hold:
Environment size ($\delta$) Condition Reason
$\delta_1$
$\left|e^{\frac{i}{2}\left(\theta+\theta_{f}+\theta_{g}\right)}-1\right|<\epsilon$
Lemma 4, since $f,g$ are detachable and $f$ is sign-continuous
$\delta_{2}$
$\left|g\left(z\right)\right|=\left|g\right|$
$g,g(z_0)$ are bimodular
$\delta_3$
$\left|f(z)\Delta g\right|=\left|g(z)\Delta f\right|$
$f,g$ are spatially bimodular
Let $z$ in the $\min\left\{ \delta_{1},\delta_{2},\delta_{3}\right\}$-environment. Then: $$\begin{align*} &\left|sgn\left[\left(fg\right)\left(z\right)-\left(fg\right)\left(z_{0}\right)\right]-sgn\left[f^{;}sgn\left(g\right)+g^{;}sgn\left(f\right)\right]e^{i\varphi}\right|\\ &=\left|sgn\left[f\left(z\right)g\left(z\right)-f\left(z\right)g\left(z_{0}\right)+f\left(z\right)g\left(z_{0}\right)-f\left(z\right)g\left(z_{0}\right)\right]-sgn\left[f^{;}sgn\left(g\right)+g^{;}sgn\left(f\right)\right]e^{i\varphi}\right|\\ &=\left|sgn\left[f\left(z\right)\Delta g+g\left(z_{0}\right)\Delta f\right]-sgn\left[f^{;}sgn\left(g\right)+g^{;}sgn\left(f\right)\right]e^{i\varphi}\right|\\ &=\left|sgn\left[\left|f\left(z\right)\right|sgn\left(f\left(z\right)\right)\left|\Delta g\right|sgn\left(\Delta g\right)+\left|g\left(z_{0}\right)\right|sgn\left(g\right)\left|\Delta f\right|sgn\left(\Delta f\right)\right]-sgn\left[f^{;}sgn\left(g\right)+g^{;}sgn\left(f\right)\right]e^{i\varphi}\right|\\ &=\left|sgn\left[\left|f\left(z\right)\right|sgn\left(f\right)e^{i\theta}\left|\Delta g\right|g^{;}e^{i\theta_{g}}e^{i\varphi}+\left|g\left(z\right)\right|sgn\left(g\right)\left|\Delta f\right|f^{;}e^{i\theta_{f}}e^{i\varphi}\right]-sgn\left[f^{;}sgn\left(g\right)+g^{;}sgn\left(f\right)\right]e^{i\varphi}\right|\\ &=\left|sgn\left[sgn\left(f\right)e^{i\theta}g^{;}e^{i\theta_{g}}+sgn\left(g\right)f^{;}e^{i\theta_{f}}\right]-sgn\left[f^{;}sgn\left(g\right)+g^{;}sgn\left(f\right)\right]\right|\\ &=\left|e^{i\left[sgn\left(f\right)e^{i\theta}g^{;}e^{i\theta_{g}}+sgn\left(g\right)f^{;}e^{i\theta_{f}}\right]}-e^{i\left[f^{;}sgn\left(g\right)+g^{;}sgn\left(f\right)\right]}\right|\\ &=\left|e^{\frac{i}{2}\left[\arg\left(f\right)+\theta+\arg\left(g^{;}\right)+\theta_{g}+\arg\left(g\right)+\arg\left(f^{;}\right)+\theta_{f}\right]}-e^{\frac{i}{2}\left[\arg\left(f\right)+\arg\left(g^{;}\right)+\arg\left(g\right)+\arg\left(f^{;}\right)\right]}\right|\\ &=\left|e^{\frac{i}{2}\left[\arg\left(f\right)+\arg\left(g^{;}\right)+\arg\left(g\right)+\arg\left(f^{;}\right)\right]}\left[e^{\frac{i}{2}\left(\theta+\theta_{f}+\theta_{g}\right)}-1\right]\right|=\left|e^{\frac{i}{2}\left(\theta+\theta_{f}+\theta_{g}\right)}-1\right|<\epsilon, \end{align*}$$ where the fourth transition is due to the notation in formula ($\ref{sgn_df_formula}$) and due to the second and third conditions above, the seventh transition is dye to lemma 2, and the inequality is due to lemma 4.$\,\,\,\,\blacksquare$

Claim 5. Quotient rule. Let $f,g\neq0:\mathbb{C}\longrightarrow\mathbb{C}$ be locally bimodular and detachable at the point $z_{0}\in\mathbb{C},$ where $g$ is sign-continuous at $z_{0}.$ Then $\frac{f}{g}$ is also detachable there, and:

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

Proof. First, let us develop the expression by the definition of the limit: $$\begin{align*} \left(\frac{f}{g}\right)^{;} &\equiv e^{-i\varphi}\underset{z\to z_{0}}{\lim}sgn\left[\left(\frac{f}{g}\right)\left(z\right)-\left(\frac{f}{g}\right)\left(z_{0}\right)\right] \\ &=\underset{z\to z_{0}}{\lim}sgn\left[\frac{f\left(z\right)g\left(z_{0}\right)e^{-i\varphi}-f\left(z_{0}\right)g\left(z\right)e^{-i\varphi}}{g\left(z\right)g\left(z_{0}\right)}\right] \\ &=\underset{z\to z_{0}}{\lim}\frac{sgn\left[f\left(z\right)g\left(z_{0}\right)e^{-i\varphi}-f\left(z_{0}\right)g\left(z\right)e^{-i\varphi}\right]}{sgn\left(g\left(z\right)\right)sgn\left(g\right)} \\ &=\frac{1}{sgn\left(g\right)^{2}}\underset{z\to z_{0}}{\lim}sgn\left[g\left(z_{0}\right)\Delta fe^{-i\varphi}-f\left(z_{0}\right)\Delta ge^{-i\varphi}\right] \\ &=\frac{1}{sgn\left(g\right)^{2}}\underset{z\to z_{0}}{\lim}sgn\left[g\left(z_{0}\right)\left|\Delta f\right|f^{;}e^{i\theta_{f}}-f\left(z_{0}\right)\left|\Delta g\right|g^{;}e^{i\theta_{g}}\right] \\ &=\frac{1}{sgn\left(g\right)^{2}}\underset{z\to z_{0}}{\lim}sgn\left[sgn\left(g\right)\left|g\left(z_{0}\right)\right|\left|\Delta f\right|f^{;}e^{i\theta_{f}}-sgn\left(f\right)\left|f\left(z_{0}\right)\right|\left|\Delta g\right|g^{;}e^{i\theta_{g}}\right], \end{align*}$$ where the fourth transition is because $g$ is sign-continuous at $z_{0},$ and the fifth is due to the notation in formula ($\ref{sgn_df_formula}$).
Next, let us show that: $$\underset{z\to z_{0}}{\lim}sgn\left[sgn\left(g\right)\left|g\left(z_{0}\right)\right|\left|\Delta f\right|f^{;}e^{i\theta_{f}}-sgn\left(f\right)\left|f\left(z_{0}\right)\right|\left|\Delta g\right|g^{;}e^{i\theta_{g}}\right]=sgn\left[sgn\left(g\right)f^{;}-sgn\left(f\right)g^{;}\right],$$ with an $\epsilon-\delta$ analysis. Let $\epsilon>0.$ There exist $\delta$-environments of $z_{0}$ such that for each $z$ there, the following conditions hold:
Environment size ($\delta$) Condition Reason
$\delta_1$
$\left|e^{\frac{i}{2}\left(\theta_{f}+\theta_{g}\right)}-1\right|<\epsilon$
Lemma 3, since $f$ and $g$ are detachable
$\delta_2$
$\left|f\left(z\right)\Delta g\right|=\left|g\left(z\right)\Delta f\right|$
$f,g$ are locally bimodular

Then, for each $z$ in the $\min\left\{ \delta_{1},\delta_{2}\right\}$-environment, it holds that:

$$\begin{align*}
&\left|sgn\left[sgn\left(g\right)\left|g\left(z_{0}\right)\right|\left|\Delta f\right|f^{;}e^{i\theta_{f}}-sgn\left(f\right)\left|f\left(z_{0}\right)\right|\left|\Delta g\right|g^{;}e^{i\theta_{g}}\right]-sgn\left[sgn\left(g\right)f^{;}-sgn\left(f\right)g^{;}\right]\right| \\
&=\left|sgn\left[sgn\left(g\right)f^{;}e^{i\theta_{f}}-sgn\left(f\right)g^{;}e^{i\theta_{g}}\right]-sgn\left[sgn\left(g\right)f^{;}-sgn\left(f\right)g^{;}\right]\right| \\
&=\left|e^{i\arg\left[sgn\left(g\right)f^{;}e^{i\theta_{f}}-sgn\left(f\right)g^{;}e^{i\theta_{g}}\right]}-e^{i\arg\left[sgn\left(g\right)f^{;}-sgn\left(f\right)g^{;}\right]}\right| \\
&=\left|e^{i\frac{\arg\left(gf^{;}e^{i\theta_{f}}\right)+\arg\left(-fg^{;}e^{i\theta_{g}}\right)}{2}}-e^{i\frac{\arg\left(gf^{;}\right)+\arg\left(-fg^{;}\right)}{2}}\right| \\
&=\left|e^{i\frac{\arg\left(g\right)+\arg\left(f^{;}\right)+\arg\left(e^{i\theta_{f}}\right)+\arg\left(f\right)+\arg\left(g^{;}\right)+\arg\left(e^{i\theta_{g}}\right)-\pi}{2}}-e^{i\frac{\arg\left(g\right)+\arg\left(f^{;}\right)+\arg\left(f\right)+\arg\left(g^{;}\right)-\pi}{2}}\right| \\
&=\left|e^{i\frac{\arg\left(g\right)+\arg\left(f^{;}\right)+\arg\left(f\right)+\arg\left(g^{;}\right)-\pi}{2}}\left(e^{i\frac{\left(\theta_{f}+\theta_{g}\right)}{2}}-1\right)\right|=\left|e^{i\frac{\left(\theta_{f}+\theta_{g}\right)}{2}}-1\right|<\epsilon,
\end{align*}$$

where the first transition is due to the second condition above, and the inequality is due to lemma 3, which can be applied because of the first and second conditions.$\,\,\,\,\blacksquare$

References

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

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