Author: 孤筝(lvbowen040427@163.com)

1. Residue Theorem

The Residue Theorem is a pivotal result in complex function theory, built upon the concept of residues. Its core idea states that if a function is analytic everywhere on and within a closed contour enclosing isolated singularities, the integral around the contour equals the sum of the residues at those singularities.

2. Laurent Series

A Laurent series is an expansion of a complex function into an infinite series, including both positive and negative powers. Specifically, a complex function in an annular region can be expressed as:

Here, \( c_n \) are the coefficients, and \( z_0 \) is the expansion point.

3. Higher-Order Derivative Formula

The higher-order derivative formula for complex functions resembles that of real functions but requires careful consideration in the complex plane. If a function is analytic at a point, its higher-order derivatives at that point can be obtained by term-wise differentiation of its power series.

4. Cauchy’s Integral Formula

Cauchy’s Integral Formula is a fundamental result in complex analysis, establishing a relationship between an analytic function and its integral over a contour. Specifically, if \( f(z) \) is analytic inside and on a simple closed contour \( C \), then for any point \( z_0 \) inside \( C \):

Relationships and Connections

1. Residue Theorem and Laurent Series:
   The Residue Theorem computes integrals over closed contours, while the Laurent series helps analyze the behavior of functions near singularities, facilitating residue calculation.

2. Residue Theorem and Higher-Order Derivatives:
   The Residue Theorem can derive higher-order derivatives by term-wise differentiation of the Laurent series expansion around singularities.

3. Residue Theorem and Cauchy’s Integral Formula:
   Cauchy’s Integral Formula evaluates contour integrals, while the Residue Theorem is a special case where the contour encloses finitely many isolated singularities.

1. Proof of the Residue Theorem via Laurent Series

Residue Theorem:

If \( f(z) \) is analytic everywhere inside and on a closed contour \( C \) except for isolated singularities, then:

Laurent Series Expansion:

Proof Steps:

1. Laurent Series Expansion:
   Expand \( f(z) \) around \( z_0 \):
   

   $$f(z) = \sum_{n=-\infty}^{\infty} c_n (z - z_0)^n$$

2. Integral Computation:
   Integrate term-wise over \( C \):
   

   $$\oint_C f(z) \, dz = \oint_C \left( \sum_{n=-\infty}^{\infty} c_n (z - z_0)^n \right) dz$$

3. Interchange Sum and Integral:
   By uniform convergence:
   

   $$\oint_C f(z) \, dz = \sum_{n=-\infty}^{\infty} c_n \oint_C (z - z_0)^n \, dz$$

4. Residue Extraction:
   For \( n \neq -1 \), \( \oint_C (z - z_0)^n \, dz = 0 \). Only \( n = -1 \) contributes:
   

   $$\oint_C f(z) \, dz = 2\pi i \cdot c_{-1}$$

5. Conclusion:
   The residue \( \text{Res}(f, z_0) = c_{-1} \), yielding:
   

   $$\oint_C f(z) \, dz = 2\pi i \cdot \text{Res}(f, z_0)$$

2. Deriving the Higher-Order Derivative Formula from the Residue Theorem

Residue Theorem:

Higher-Order Derivative Formula:

For \( f(z) \) analytic at \( z_0 \), the \( n \)-th derivative is:

Proof Steps:

1. Residue Theorem Setup:
   Consider \( g(z) = \frac{f(z)}{(z - z_0)^{n+1}} \). Its residue at \( z_0 \) is:
   

   $$\text{Res}(g, z_0) = \frac{f^{(n)}(z_0)}{n!}$$

2. Apply Residue Theorem to \( g(z) \):
   

   $$\oint_C \frac{f(z)}{(z - z_0)^{n+1}} \, dz = 2\pi i \cdot \text{Res}(g, z_0) = 2\pi i \cdot \frac{f^{(n)}(z_0)}{n!}$$

3. Solve for \( f^{(n)}(z_0) \):
   

   $$f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z - z_0)^{n+1}} \, dz$$

3. Proving Cauchy’s Integral Formula via the Residue Theorem

Residue Theorem:

Cauchy’s Integral Formula:

Proof Steps:

1. Identify the Residue:
   For \( g(z) = \frac{f(z)}{z - z_0} \), the residue at \( z_0 \) is \( \text{Res}(g, z_0) = f(z_0) \).

2. Apply the Residue Theorem:
   

   $$\oint_C \frac{f(z)}{z - z_0} \, dz = 2\pi i \cdot \text{Res}(g, z_0) = 2\pi i \cdot f(z_0)$$

3. Solve for \( f(z_0) \):
   

   $$f(z_0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z - z_0} \, dz$$

Published on 2023-12-20 at 孤筝の温暖小家, last modified on 2023-12-20

All articles on this blog are licensed under the BY-NC-SA license agreement unless otherwise stated. Please indicate the source when reprinting!

Author: 孤筝(lvbowen040427@163.com)

Complex Numbers

1. Representation of complex numbers: $$z = r\cdot e^{i\theta} = r(\cos\theta +i\cdot \sin\theta)$$
2. Elementary functions
   1. Exponential function: $e^z = e^x(\cos y + i \sin y)$
      1. $e^z$ is merely shorthand for $\exp z$ and does not imply exponentiation.
      2. $|e^z| = e^x$, $\text{Arg}(e^z) = y + 2k\pi$
   2. Logarithmic function: $\text{Ln}\,z = \ln|r| + i\,\text{Arg}\,z$
      1. The function is analytic everywhere except at the origin and the negative real axis, and $(\text{Ln}\,z)' = \frac{1}{z}$.
   3. Trigonometric functions
      1. $\cos z = \frac{e^{iz} + e^{-iz}}{2}$, $\sin z = \frac{e^{iz} - e^{-iz}}{2i}$
      2. $\text{ch}\,z = \frac{e^z + e^{-z}}{2}$, $\text{sh}\,z = \frac{e^z - e^{-z}}{2}$

Analytic Functions

1. Definition of differentiability:
   $$\lim_{\Delta z \to 0} \frac{f(z_0 + \Delta z) - f(z_0)}{\Delta z} \text{ exists, then } f(z) \text{ is differentiable at } z_0.$$
2. Definition of analyticity:
   $$f(z) \text{ is analytic at } z_0 \text{ if it is differentiable at } z_0 \text{ and in some neighborhood of } z_0.$$
   Corollary: The sum, difference, product, and quotient of analytic functions are also analytic. The composition of analytic functions is analytic.
3. Necessary and sufficient conditions for differentiability and analyticity:
   $u(x, y)$ and $v(x, y)$ are differentiable and satisfy the Cauchy-Riemann equations:
   $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.$$
   If either condition fails, the function is neither differentiable nor analytic.
   Corollary:
   $$f'(z) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{1}{i} \frac{\partial u}{\partial y} + \frac{\partial v}{\partial y}.$$

Complex Integration

Key Formula

1. Cauchy-Goursat Theorem
   For a simply connected, analytic region, the integral over any closed contour is zero:
   $$\oint_C f(z) dz = 0.$$
2. Composite Contour Theorem—Extension to multiply connected regions
   Let $C$ be a simple closed curve in an analytic, multiply connected region, and $C_1, C_2, \dots, C_n$ be simple closed curves inside $C$ with the same orientation. Then:
   $$\oint_C f(z) dz = \sum_{k=1}^n \oint_{C_k} f(z) dz.$$
3. Cauchy Integral Formula—Expressing the value of a function inside a contour in terms of its boundary values
   If $f(z)$ is analytic in a region $D$ and $C$ is a positively oriented simple closed curve in $D$:
   $$2\pi i \cdot f(z_0) = \oint_C \frac{f(z)}{z - z_0} dz.$$
4. Generalized Cauchy Integral Formula—Using higher-order derivatives to compute integrals
   $$f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z - z_0)^{n+1}} dz.$$

Series

Power Series

1. Two properties of analytic functions
   1. Analytic functions have derivatives of all orders.
   2. Every analytic function can be represented by a power series.
2. Taylor expansion:
   $$f(z) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} z^n.$$
3. Methods for finding Taylor expansions [[Advanced Mathematics#Expansion of Functions into Power Series]]

Laurent Series

1. Bilateral power series
   1. The region of convergence is an annulus $R_1 \lt |z - z_0| \lt R_2$.
2. Laurent expansion:
   $$f(z) = \sum_{n=-\infty}^\infty c_n (z - z_0)^n, \quad c_n = \frac{1}{2\pi i} \oint_C \frac{f(z)}{(z - z_0)^{n+1}} dz.$$
   Corollary: When $n = -1$, $c_{-1} \cdot 2\pi i = \oint_C f(z) dz$.
3. Methods for finding Laurent expansions
   1. Compute $c_n$ directly using the definition (rarely used).
   2. Use algebraic operations or substitutions to transform the Laurent series into the form and convergence domain of a Taylor series.

Residues

Isolated Singularities

1. Definition: $f(z)$ is not analytic at $z_0$ but is analytic in some punctured neighborhood of $z_0$.
2. Classification of isolated singularities (based on negative power terms in the Laurent series)
   1. Removable singularity: No negative power terms. As $z \to z_0$, $f(z)$ approaches a finite limit.
   2. Pole: Finite number of negative power terms (if there are $m$ such terms, $z_0$ is called an $m$-th order pole). As $z \to z_0$, $f(z) \to \infty$.
   3. Essential singularity: Infinite number of negative power terms. The limit of $f(z)$ does not exist.
3. Relationship between poles and zeros
   1. Definition of zeros: For a non-zero analytic function $f(z)$, if it can be expressed as $f(z) = (z - z_0)^m \varphi(z)$, then $z_0$ is called an $m$-th order zero of $f(z)$.
      Necessary and sufficient condition: $f^{(n)}(z_0) = 0$ for $n \lt m$, and $f^{(m)}(z_0) \neq 0$.
   2. If $z_0$ is an $m$-th order zero of $f(z)$, then $z_0$ is an $m$-th order pole of $\frac{1}{f(z)}$.

Residues

1. Definition:
   $$\text{Res}[f(z), z_0] = c_{-1} = \frac{1}{2\pi i} \oint_C f(z) dz.$$
2. Rules for computing residues
   1. If $z_0$ is a simple pole of $f(z)$:
      $$\text{Res}[f(z), z_0] = \lim_{z \to z_0} (z - z_0) f(z).$$
   2. If $z_0$ is an $m$-th order pole of $f(z)$:
      $$\text{Res}[f(z), z_0] = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} \left( (z - z_0)^m f(z) \right).$$
   3. If $f(z) = \frac{P(z)}{Q(z)}$, where $P(z_0) \neq 0$, $Q(z_0) = 0$, and $Q'(z_0) \neq 0$:
      $$\text{Res}[f(z), z_0] = \frac{P(z_0)}{Q'(z_0)}.$$

Published on 2023-11-17 at 孤筝の温暖小家, last modified on 2023-11-17

All articles on this blog are licensed under the BY-NC-SA license agreement unless otherwise stated. Please indicate the source when reprinting!