Singular Points: Theory & Analysis
Definition: Singular Point
Let \( f: D \subset \mathbb{C} \longrightarrow \mathbb{C} \).
A point \( z_0 \in \mathbb{C} \) is a singular point of \( f \) if it is a limit point of regular points but is not a regular point itself.
Equivalent Definition: Every neighborhood of \( z_0 \) contains a regular point, but \( z_0 \) is not regular.
A point \( z_0 \in \mathbb{C} \) is a singular point of \( f \) if it is a limit point of regular points but is not a regular point itself.
Equivalent Definition: Every neighborhood of \( z_0 \) contains a regular point, but \( z_0 \) is not regular.
Mathematical Formulation:
Let \( D \) be the domain.
\( D_f \) = Set of points where \( f \) is Differentiable.
\( R \) = Set of Regular Points = \( D_f^\circ \) (Interior of \( D_f \)).
\( R’ \) = Derived Set of \( R \) (Limit points of \( R \)).
The Set of Singular Points \( S \) is: \[ \boxed{ S = R’ – R } \] Note: \( R \cap S = \phi \) (A point cannot be both regular and singular).
Let \( D \) be the domain.
\( D_f \) = Set of points where \( f \) is Differentiable.
\( R \) = Set of Regular Points = \( D_f^\circ \) (Interior of \( D_f \)).
\( R’ \) = Derived Set of \( R \) (Limit points of \( R \)).
The Set of Singular Points \( S \) is: \[ \boxed{ S = R’ – R } \] Note: \( R \cap S = \phi \) (A point cannot be both regular and singular).
Q1: Modulus Squared Function
\( f(z) = |z|^2 = z\bar{z} \).
1. Check Differentiability (\(D_f\)):
1. Check Differentiability (\(D_f\)):
Use the Complex C-R condition: \( \frac{\partial f}{\partial \bar{z}} = z \).
Set derivative to zero: \( z = 0 \).
Thus, \( D_f = \{0\} \).
2. Find Regular Points (\(R\)):
Set derivative to zero: \( z = 0 \).
Thus, \( D_f = \{0\} \).
\( R = D_f^\circ = \text{Interior of } \{0\} \).
Since a single point has no interior, \( R = \phi \).
3. Find Singular Points (\(S\)):
Since a single point has no interior, \( R = \phi \).
\( R’ = \phi \).
\( S = R’ – R = \phi – \phi = \phi \).
\( S = R’ – R = \phi – \phi = \phi \).
Conclusion: This function is nowhere analytic, but has NO singular points.
(Every point is neither singular nor regular).
(Every point is neither singular nor regular).
Q2: Piecewise Analytic Function
\[ f(z) = \begin{cases} z^2 & \text{if } xy \neq 0 \\ 0 & \text{if } xy = 0 \end{cases} \]
.
Step 1: Determine \( D_f \) (Differentiable Points)
Case 1: Open Quadrants (\(xy \neq 0\))
Let \( \alpha \) be in a quadrant. There exists a neighborhood \( |z-\alpha| < \delta \) entirely inside the quadrant.
Here \( f(z) = z^2 \) (polynomial), which is differentiable everywhere.
\( f'(\alpha) = 2\alpha \).
Result: Differentiable at all \( xy \neq 0 \).
Let \( \alpha \) be in a quadrant. There exists a neighborhood \( |z-\alpha| < \delta \) entirely inside the quadrant.
Here \( f(z) = z^2 \) (polynomial), which is differentiable everywhere.
\( f'(\alpha) = 2\alpha \).
Result: Differentiable at all \( xy \neq 0 \).
Case 2: Axes excluding Origin (\(xy=0, z \neq 0\))
Let \( \alpha \) be on an axis (\( \alpha \neq 0 \)). \( f(\alpha) = 0 \).
Path 1 (Along Axis): \( f(z)=0 \implies \text{Limit} = 0 \).
Path 2 (From Quadrant): \( f(z)=z^2 \implies \text{Limit} = \frac{\alpha^2}{0} = \infty \).
Result: Not Differentiable.
Let \( \alpha \) be on an axis (\( \alpha \neq 0 \)). \( f(\alpha) = 0 \).
Path 1 (Along Axis): \( f(z)=0 \implies \text{Limit} = 0 \).
Path 2 (From Quadrant): \( f(z)=z^2 \implies \text{Limit} = \frac{\alpha^2}{0} = \infty \).
Result: Not Differentiable.
Case 3: The Origin (\(z=0\))
Using definition: \( \lim_{z \to 0} \frac{f(z)}{z} \).
Bound the function: \( \left| \frac{f(z)}{z} \right| \le \left| \frac{z^2}{z} \right| = |z| \).
As \( z \to 0 \), limit is 0.
Result: Differentiable at 0.
Using definition: \( \lim_{z \to 0} \frac{f(z)}{z} \).
Bound the function: \( \left| \frac{f(z)}{z} \right| \le \left| \frac{z^2}{z} \right| = |z| \).
As \( z \to 0 \), limit is 0.
Result: Differentiable at 0.
\( D_f = \{ z : xy \neq 0 \} \cup \{0\} \)
Step 2: Calculate Sets R and S
Regular Points (\(R\)):
\( R = \text{Interior of } D_f \).
0 is not an interior point (neighborhoods contain axes).
\( R = \{ z : xy \neq 0 \} \).
\( R = \text{Interior of } D_f \).
0 is not an interior point (neighborhoods contain axes).
\( R = \{ z : xy \neq 0 \} \).
Limit Points (\(R’\)):
\( R’ = \mathbb{C} \) (Quadrants are dense in \(\mathbb{C}\)).
\( R’ = \mathbb{C} \) (Quadrants are dense in \(\mathbb{C}\)).
Singular Points (\(S\)):
\( S = R’ – R = \mathbb{C} – \{ \text{Quadrants} \} \).
\( \boxed{ S = \{ z : xy = 0 \} } \) (The Axes).
\( S = R’ – R = \mathbb{C} – \{ \text{Quadrants} \} \).
\( \boxed{ S = \{ z : xy = 0 \} } \) (The Axes).
Rational Functions & Singularities
Let \( H(z) = \frac{F(z)}{G(z)} \). Singularities are generally where \( G(z) = 0 \).
Check: If \( \lim_{z \to \alpha} H(z) \) exists (finitely), then \( \alpha \) is a Removable Singularity (not a singular point if defined properly).
Check: If \( \lim_{z \to \alpha} H(z) \) exists (finitely), then \( \alpha \) is a Removable Singularity (not a singular point if defined properly).
| Function \( f(z) \) | Conditions | Singular Points (\(S\)) |
|---|---|---|
| \( \frac{\sin z}{z} \) | Undefined at \( z=0 \) | \( \{0\} \) |
| \( \begin{cases} \frac{\sin z}{z} & z \neq 0 \\ 1 & z = 0 \end{cases} \) | Limit exists and equals function value | \( \phi \) (None) |
| \( \frac{3z+6}{z^2-3z+2} \) | Roots of \( z^2-3z+2 \) | \( \{1, 2\} \) |
| \( \frac{z}{\sin z} \) | Roots of \( \sin z \) | \( \{ n\pi : n \in \mathbb{Z} \} \) |
| \( \begin{cases} \frac{z}{\sin z} & z \neq 0 \\ 1 & z = 0 \end{cases} \) | Removable at 0, infinite elsewhere | \( \{ n\pi : n \in \mathbb{Z} – \{0\} \} \) |
Singular Points of Trigonometric Functions
1. \( f(z) = \sin z \)
Analysis:
The function \( \sin z \) is a standard entire function. It is defined and differentiable for all complex numbers \( z \in \mathbb{C} \).
Sets:
- \( D_f = \mathbb{C} \) (Differentiable everywhere)
- \( R = \mathbb{C} \) (Regular everywhere)
- \( R’ = \mathbb{C} \)
Calculation:
\( S = R’ – R = \mathbb{C} – \mathbb{C} = \emptyset \)
\( S = \emptyset \) (No Singular Points)
2. \( f(z) = \sin\left(\frac{1}{z}\right) \)
Analysis:
This is a composition of functions.
\( \sin(w) \) is analytic everywhere, but \( w = 1/z \) is undefined at \( z=0 \).
Therefore, the function is undefined and not differentiable at \( z=0 \).
Sets:
- \( D_f = \mathbb{C} \setminus \{0\} \)
- \( R = \mathbb{C} \setminus \{0\} \)
- \( R’ = \mathbb{C} \) (Since 0 is a limit point of \( \mathbb{C} \setminus \{0\} \))
Calculation:
\( S = R’ – R = \mathbb{C} – (\mathbb{C} \setminus \{0\}) = \{0\} \)
\( S = \{0\} \)
(Note: \( z=0 \) is an Essential Singularity).
3. \( f(z) = \frac{1}{\sin(1/z)} \)
Analysis:
Singularities occur where the function is undefined. This happens in two cases:
- Argument Undefined: \( 1/z \) is undefined at \( z=0 \).
- Denominator Zero: \( \sin(1/z) = 0 \).
Solving Denominator Zeroes:
\[ \sin\left(\frac{1}{z}\right) = 0 \implies \frac{1}{z} = n\pi \quad (\text{for } n \in \mathbb{Z}) \]
\[ z = \frac{1}{n\pi} \quad (\text{where } n = \pm 1, \pm 2, \dots) \]
(Note: \( n \neq 0 \) because \( 1/z \) cannot be 0).
The Accumulation Point:
As \( n \to \infty \), the points \( z_n = \frac{1}{n\pi} \) get closer and closer to 0.
Thus, \( z=0 \) is a limit point of singularities (a non-isolated singularity).
Sets:
- \( R = \mathbb{C} \setminus \left( \{0\} \cup \{ \frac{1}{n\pi} \} \right) \)
- \( R’ = \mathbb{C} \)
\( S = \{0\} \cup \left\{ \frac{1}{n\pi} : n \in \mathbb{Z} \setminus \{0\} \right\} \)
(Note: The set \( S \) contains infinitely many isolated points accumulating at 0).
Finding Singularities of Nested Functions
Function:
\[ f(z) = \sin\left( \frac{1}{\cos\left( \frac{1}{e^{1/z}-1} \right) – 1} \right) \]
Layer 1: The Innermost Term (1/z)
The function contains the term \( 1/z \).
This is undefined when the denominator is zero.
This is undefined when the denominator is zero.
\[ z = 0 \]
\( S_1 = \{0\} \)
Layer 2: The Middle Fraction
The term \( \frac{1}{e^{1/z}-1} \) is undefined when its denominator is zero.
\[ e^{1/z} – 1 = 0 \implies e^{1/z} = 1 \]
Using the periodicity of the exponential function (\( e^{2n\pi i} = 1 \)):
\[ \frac{1}{z} = 2n\pi i \quad (\text{where } n \in \mathbb{Z}) \]
Note: \( n \neq 0 \) because \( 1/z \) cannot be 0. \[ z = \frac{1}{2n\pi i} = -\frac{i}{2n\pi} \]
Note: \( n \neq 0 \) because \( 1/z \) cannot be 0. \[ z = \frac{1}{2n\pi i} = -\frac{i}{2n\pi} \]
\( S_2 = \left\{ \frac{1}{2n\pi i} : n \in \mathbb{Z} \setminus \{0\} \right\} \)
Layer 3: The Outer Fraction
The term \( \frac{1}{\cos(\dots) – 1} \) is undefined when its denominator is zero.
\[ \cos\left( \frac{1}{e^{1/z}-1} \right) – 1 = 0 \implies \cos\left( \frac{1}{e^{1/z}-1} \right) = 1 \]
The general solution for \( \cos(W) = 1 \) is \( W = 2k\pi \).
\[ \frac{1}{e^{1/z}-1} = 2k\pi \quad (\text{where } k \in \mathbb{Z}) \]
Note: \( k \neq 0 \) because the LHS is a reciprocal (cannot be 0).
Solving for \( z \): \[ e^{1/z} – 1 = \frac{1}{2k\pi} \] \[ e^{1/z} = 1 + \frac{1}{2k\pi} \] Take the natural log (Complex Logarithm \( \ln + 2m\pi i \)): \[ \frac{1}{z} = \ln\left( 1 + \frac{1}{2k\pi} \right) + 2m\pi i \quad (m \in \mathbb{Z}) \] \[ z = \frac{1}{\ln\left( 1 + \frac{1}{2k\pi} \right) + 2m\pi i} \]
Solving for \( z \): \[ e^{1/z} – 1 = \frac{1}{2k\pi} \] \[ e^{1/z} = 1 + \frac{1}{2k\pi} \] Take the natural log (Complex Logarithm \( \ln + 2m\pi i \)): \[ \frac{1}{z} = \ln\left( 1 + \frac{1}{2k\pi} \right) + 2m\pi i \quad (m \in \mathbb{Z}) \] \[ z = \frac{1}{\ln\left( 1 + \frac{1}{2k\pi} \right) + 2m\pi i} \]
\( S_3 = \left\{ \frac{1}{\ln(1 + \frac{1}{2k\pi}) + 2m\pi i} : k,m \in \mathbb{Z}, k \neq 0 \right\} \)
Total Set of Singular Points \( S \):
\[ S = S_1 \cup S_2 \cup S_3 \]
This set includes the origin \( \{0\} \), the simple poles from the exponential, and the complex nested singularities from the cosine term.
\[ S = S_1 \cup S_2 \cup S_3 \]
This set includes the origin \( \{0\} \), the simple poles from the exponential, and the complex nested singularities from the cosine term.
Classification of Singularities
Definitions:
- \( S \): The set of all singular points of \( f(z) \).
- \( S’ \) (or \( S_{\text{dash}} \)): The derived set of \( S \) (the set of all limit/accumulation points of \( S \)).
Type 1: Isolated Singularity
A singular point \( z_0 \) is Isolated if it belongs to \( S \) but is NOT a limit point of other singularities.
Condition:
\[ z_0 \in S \setminus S’ \]
(Belongs to \( S \), but not \( S’ \))
Meaning: There exists a neighborhood around \( z_0 \) containing no other singularity.
Type 2: Non-Isolated Singularity
A singular point \( z_0 \) is Non-Isolated if it is a limit point of other singularities.
Condition:
\[ z_0 \in S \cap S’ \]
(Belongs to both \( S \) and \( S’ \))
Meaning: Every neighborhood of \( z_0 \) contains other singularities.
Examples
Ex 1: Simple Poles
\[ f(z) = \frac{1}{z(z-1)} \]
Set \( S \): The function is undefined at \( 0 \) and \( 1 \).
\[ S = \{0, 1\} \]
Set \( S’ \): Since \( S \) is a finite set, it has no limit points.
\[ S’ = \emptyset \]
Classification:
For \( z=0 \): \( 0 \in S \) and \( 0 \notin S’ \). \(\implies\) Isolated.
For \( z=1 \): \( 1 \in S \) and \( 1 \notin S’ \). \(\implies\) Isolated.
For \( z=1 \): \( 1 \in S \) and \( 1 \notin S’ \). \(\implies\) Isolated.
Ex 2: The Classic Non-Isolated Case
\[ f(z) = \frac{1}{\sin(\frac{\pi}{z})} \]
Find \( S \):
Singularities occur when denominator is zero:
\[ \sin\left(\frac{\pi}{z}\right) = 0 \implies \frac{\pi}{z} = n\pi \implies z = \frac{1}{n} \quad (n \in \mathbb{Z} \setminus \{0\}) \]
Also, \( z=0 \) is a singularity (undefined argument).
\[ S = \{0\} \cup \left\{ 1, \frac{1}{2}, \frac{1}{3}, \dots \right\} \]
Find \( S’ \):
The sequence \( \frac{1}{n} \) approaches 0 as \( n \to \infty \).
Thus, 0 is a limit point. The points \( 1/n \) themselves are isolated.
\[ S’ = \{0\} \]
Classification:
- For \( z_n = 1/n \): These are in \( S \), but NOT in \( S’ \). \[ \frac{1}{n} \in S \setminus S’ \implies \textbf{Isolated} \]
- For \( z = 0 \): This is in \( S \), AND it is in \( S’ \). \[ 0 \in S \cap S’ \implies \textbf{Non-Isolated} \]