Seokyun Ha

Kubernetes Lover
CS && MATH

응용복소함수론(MATH210)

Part I - Complex Functions

  • Complex Variable - Basic
    • Triangle inequality
    • Euler’s formula
    • de Moivre’s formula
    • $n$-th root of $z$
    • Interior / Exteriror / Boundary
  • Complex Analysis - Basic
    • Complex Functions
    • Complex Limit
    • Complex Continuity
    • Complex Derivative
    • Analytic Functions

Cauchy-Riemann Equation

$f(z) = u(x, y) + i v(x, y)$ is analytic in a domain $D$
iff the first partial derivatives of $u$ and $v$ satisfy

\[\begin{aligned} u_x &= v_y \\ u_y &= -v_x \end{aligned}\]

IF $f(z) = u(x, y) + i u(x, y)$ is analytic in a domain $D$, THEN both $u$ and $v$ are harmonic functions.

Laplace’s Equations and Harmonic functions

A real valued function $H(x, y)$ is harmonic in a domain $D$, IF it satisfies Laplace equations:

\[H_{xx} + H_{yy} = 0\]
  • Cauchy-Riemann Equation
    • Laplace’s Equation & Harmonic functions
    • harmonic conjugate
  • Elementary Complex Functions
    • Exponential Functions; $\exp z$
    • Trigonometric Functions; $\cos z$, $\sin z$, $\tan z$
    • Hyperbolic Functions; $\cosh z$, $\sinh z$
    • Logarithm; $\log z$
    • Power Functions; $z^c$

Part II - Contour Integrals

Cauchy-Goursat Theorem

Let $f(z)$ be an analytic function in a domain $D$. THEN,

\[\oint_{C} f(z) dz = 0\]

for any simple closed curve $C$ whose interior is contained in $D$.

Cauchy's Integral Formula $$ f(z_0) = \frac{1}{2\pi i} \oint_{C} \frac{f(z)}{z-z_0} dz $$ // 함숫값 $f(z_0)$를 적분을 통해 구할 수 있다는 의미를 가진다.

#Part III - Power Series

Talyor Series

If $f(z)$ is analytic at $z_0$, and $f(z)$ is analytic through the disk $\lvert z-z_0 \rvert < R_0$.

THEN, $f(z)$ has a power series representation

\[f(z) = \sum^{\infty}_{n=0} a_n (z-z_0)^n \quad (\lvert z-z_0 \rvert < R_0)\]

where

\[a_n = \frac{f^{(n)}(z_0)}{n!} = \frac{1}{2\pi i} \oint_{C} \frac{f(z)}{(z-z_0)^{n+1}} dz\]

Laurent Series

Supp. that a function $f(z)$ is analytic throughout a domain $D$ containing an annular region $R_1 \le \lvert z-z_0 \rvert \le R_2$. THEN, $f(z)$ can be represented as the Laurnet series

\[f(z) = \sum^{\infty}_{n=0} a_n (z-z_0)^n + \sum^{\infty}_{n=1} \frac{b_n}{(z-z_0)^n} \quad (R_1 \le \lvert z-z_0 \rvert < R_2)\]

The coefficient of the Laurent series are given by

\[a_n = \frac{1}{2\pi i} \oint_{C} \frac{f(w)}{(w-z_0)^{n+1}} dw \quad \textrm{and} \quad b_n = \frac{1}{2\pi i} \oint_{C} f(w)(w-z_0)^{n-1} dw\]
  • Convergence Tests
    • ratio test
    • root test
  • Uniformly Convergent
    • $g_n(x) = x^n$ on $[0, 1]$ doesn’t uniformly converge.
  • Talyor Series
  • Laurent Series

Part IV - Residue

Part V - Complex Transformation

Review Problems