공학수학2_복소해석

Up: Class_2020_2

1. 2020-09-03
2. 2020-09-08
3. 2020-09-10
4. 2020-09-15
5. 2020-09-17
6. 2020-09-22
7. 2020-09-24
8. 2020-10-06
9. 2020-10-08
10. 2020-10-13
11. 2020-10-15
12. 2020-10-20
13. 2020-10-22

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1. 2020-09-03 ¶

// 거듭제곱근,nth_root curr goto 제곱근,square_root
Roots of $z$

Given

$z=r(\cos\theta+i\sin\theta)$
$w=\rho(\cos\phi+i\sin\phi)$

If $z=w^n,$ then $w$ is the n^th root of $z.$

$r(\cos\theta+i\sin\theta)=\rho^n(\cos(n\phi)+i\sin(n\phi))$

In this case,

$r=\rho^n$ which means $\rho=r^{\frac{1}{n}}$

and

$\left.{{\cos\theta=\cos n\phi} \atop {\sin\theta=\sin n\phi}}\right\rbrace \; n\phi=\theta+2k\pi$ (k is integer)

$\phi=\frac{\theta+2k\pi}{n}$

for different $k$ values, we have different arguments(편각,argument)
Therefore

$w_k=r^{\frac{1}{n}}\left[\cos\left(\frac{\theta+2k\pi}{n}\right)+i\sin\left( \frac{\theta+2k\pi}{n} \right) \right]$

where $k=0,1,2,\cdots,n-1$

Exercise
Find the three cube roots of $i$

Answer
$k=0,\; w_0=\frac{\sqrt{3}}{2}+\frac12i$
$k=1,\; w_1=-\frac{\sqrt{3}}{2}+\frac12i$
$k=2,\; w_2=-i$

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2. 2020-09-08 ¶

KECE232
complex analysis (복소해석,complex_analysis)

- complex variable $z=x+iy$
- complex functions
- complex calculus (integrals)

// 복소수 가감승제 생략

Conjugate
If $z=x+iy$
its conjugate is $\bar{z}=z^*=x-iy$

$z+\bar{z}=2x$
$z-\bar{z}=2iy$
$z\bar{z}=x^2+y^2$

Cartesian form

$z=x+iy$

Polar form (극형식,polar_form)

$z=re^{i\theta}=r(\cos\theta+i\sin\theta)$
여기서

$r$ : modulus (// =절대값)
$\theta$ : argument (편각,argument)

// 당연하지만 굳이 적으면 위에서

$x=r\cos\theta$
$y=r\sin\theta$

Principal Argument

$z=r(\cos\theta+i\sin\theta)$
여기서 $\theta$ : argument or $\operatorname{arg}z$

If $\theta$ is $-\pi<\theta\le\pi,$ it is the principal argument $\operatorname{Arg}z$

(주치,principal_value, 편각,argument 참조)

Power of $z$
de Moivre's Formula (드무아브르_공식,de_Moivre_s_formula)

$z=r(\cos\theta+i\sin\theta)$

양변을 n제곱하면

$z^n=r^n(\cos n\theta + i\sin n\theta)$
$z^n=r^ne^{in\theta}$

Roots of $z$

$z=r(\cos\theta+i\sin\theta)$
$w=\rho(\cos\phi+i\sin\phi)$

if $z=w^n$

$w_k=r^{1/n}\left[ \cos\left( \frac{\theta+2k\pi}{n} \right) + i\sin\left( \frac{\theta+2k\pi}{n} \right) \right]$
where $k=0,1,2,\cdots,n-1$

Sets in complex planes

Our goal is to examine the functions of a single complex variable $z=x+iy$ and the calculus of these functions.

Terminology

Given a complex value (point) $z_0=x_0+iy_0$ then

$|z-z_0|=\sqrt{(x-x_0)^2+(y-y_0)^2}$

is the 거리,distance between $z$ and $z_0$ that satisfy

$|z-z_0|=\rho$

// (중심점 z₀에서 거리 ρ인 점 z들의 집합 즉 원,circle)

circle centered at $z_0$ contains all $z$ that satisfy $|z-z_0|=\rho$

Example

(a) $|z|=1$

$z_0=0,$ equation of unit circle centered at origin

(b) $|z-(1+2i)|=5$

$z_0=1+2i,$ equation of circle, radius=5, centered at 1+2i

Neighborhood of $z$ (근방,neighborhood)

Points $z$ that satisfies $|z-z_0|<\rho$ lies within the circle of radius $\rho$ centered at $z_0$

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3. 2020-09-10 ¶

Interior point

$z_0$ is an interior point of set $S$ if some neighborhood of $z_0$ lies entirely within $S.$

// interior point(내점, 안점), boundary point(경계점)

Open set

Set $S$ is an open set if every point $z$ of set $S$ is an interior point.

// open set (개집합, 열린 집합)

Connected

any 2 points $z_1$ and $z_2$ in an open set can be connected by a polygonal line(꺾은선) that lies entirely within the set

Domain

open connected set

Region

a domain with all, some or none of its boundary points.

// (영역,region)

Closed Region

a region that contains all its boundary points.

Example: $|z-i|\le2$ is a closed region

Functions of a complex variable

Given $z=x+iy$ and complex function $f$
then $w=f(z)$

$=u+iv$
$=u(x,y)+iv(x,y)$

Example
$f(z)=z^2-4z,\;z=x+iy$

$=(x+iy)^2+4(x+iy)$
$=\underbrace{(x^2-y^2-4x)}_{u(x,y)} + i\underbrace{(2xy-4y)}_{v(x,y)}$

Example
Find the image of line Re(z)=1 given $f(z)=z^2$

Solution

$f(z)=z^2$

we get

$u(x,y)=x^2-y^2$
$v(x,y)=2xy$

substitute $x=1$

$u=1-y^2,v=2y$
$u=1-\left(\frac{v}{2}\right)^2=1-\frac{v^2}{4}$

// 그림으로 그리면

complex function $w=f(z)$ interpreted as mapping(사상,map) or transformation(변환,transformation) from z-plane to w-plane.

Limits of a complex function

Formal def:

$\lim_{z\to z_0}f(z)=L$
if $|f(z)-L|<\epsilon$
whenever $0<|z-z_0|<\delta$

as $z\to z_0,$ it can approach from any direction in the complex plane.

Limit of sum, product, quotient

Suppose

$\lim_{z\to z_0}f(z)=L_1$
$\lim_{z\to z_0}g(z)=L_2$

⑴ $\lim_{z\to z_0}[f(z)+g(z)]=L_1+L_2$
⑵ $\lim_{z\to z_0}f(z)g(z)=L_1L_2$
⑶ $\lim_{z\to z_0}\frac{f(z)}{g(z)}=\frac{L_1}{L_2},\;L_2\ne0$

Continuity at a point (연속성,continuity)

A function $f$ is continuous at a point $z_0$ if

$\lim_{z\to z_0}f(z)=f(z_0)$

● If $f(z)$ and $g(z)$ are continuous at point $z_0,$ their sum and product are also continuous at $z_0,$ the quotient is also continuous at $z_0$ if $g(z_0)\ne0$

● A rational function $f(z)=\frac{g(z)}{h(z)}$ is continuous except at points where $h(z)=0$

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4. 2020-09-15 ¶

Derivative

Derivative of $f$ at $z_0$ is

$f'(z_0)=\lim_{\Delta z\to 0}\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z}$

provided the limit exists.

● derivative of $w=f(z)$ is $\frac{dw}{dz}$
● if it is differentiable at $z_0,$ it is also continous at $z_0$

constant rules

$\frac{d}{dz}c=0$
$\frac{d}{dz}cf(z)=cf'(z)$

sum rule

$\frac{d}{dz}\left[f(z)+g(z)\right]=f'(z)+g'(z)$

product rule

$\frac{d}{dz}\left[f(z)g(z)\right]=f(z)g'(z)+g(z)f'(z)$

quotient rule

$\frac{d}{dz}\left[\frac{f(z)}{g(z)}\right]=\frac{g(z)f'(z)-f(z)g'(z)}{[g(z)]^2}$

chain rule (연쇄법칙,chain_rule)

$\frac{d}{dz}f(g(z))=f'(g(z))g'(z)$

power rule

$\frac{d}{dz}z^n=nz^{n-1},\;\;n\in\mathbb{Z}$

// Example 생략

Note: for it to be differentiable, it must approach the same complex number from any direction

Funky example

Show that $f(z)=x+4iy$ is nowhere differentiable.

Soln.
$\Delta z=\Delta x+i\Delta y$
$f(z+\Delta z)-f(z)$

$=(x+\Delta x)+i4(y+\Delta y)-x-4iy$
$=\Delta x+4i\Delta y$

$f'(z_0)=\lim_{\Delta z \to 0}(.....)$
so
$\lim_{\Delta z\to 0}\frac{f(z+\Delta z)-f(z)}{\Delta z}$
$=\lim_{\Delta z\to 0}\frac{\Delta x+4i\Delta y}{\Delta x+i\Delta y}$
if $\Delta z\to 0$ along line parallel to x-axis

this means $\Delta y=0,$
the limit becomes 1

if $\Delta z\to 0$ along line parallel to y-axis,

$\Delta x=0,$
then limit becomes 4

● approach from different direction gives different values. It is nowhere differentiable.

Cauchy-Riemann Equations

Given $f(z)=u(x,y)+iv(x,y),$
(u와 v는 real valued and continuous functions)
It is differentiable at point $z=x+iy$ if

$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$

continuous 1st order partial derivative exist.

// 코시-리만_방정식,Cauchy-Riemann_equation

Analytic Function (해석함수,analytic_function, curr see 함수,function#s-40)

A complex function $w=f(z)$ is said to be analytic at a point $z_0$ if $f$ is differentiable at $z_0$ and at every point in some neighborhood of $z_0$

Example

Is $f(z)=z^2+2$ analytic for all $z?$

Soln.
$f(z)=x^2-y^2+x+i(2xy+y)$
$u(x,y)=x^2-y+x$
$v(x,y)=2xy+y$
$\frac{\partial u}{\partial x}=2x+1=\frac{\partial v}{\partial y}$
$\frac{\partial u}{\partial y}=-2y=-\frac{\partial v}{\partial x}$ 이 둘은 모든 $x,y$ 에 대해 코시-리만 방정식을 만족.
∴ $f(z)$ is analytic for all $z.$

Example

Show that $f(z)=(2x^2+y)+i(y^2-x)$ is not analytic at any point.

Soln.
Identify

$u(x,y)=2x^2+y$
$v(x,y)=y^2-x$

Let's investigate:

$\frac{\partial u}{\partial x}=4x\text{ and }\frac{\partial v}{\partial y}=2y \;\Rightarrow\; \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ only at $y=2x$ - a line!
$\frac{\partial u}{\partial y}=1\text{ and }\frac{\partial v}{\partial x}=-1 \;\Rightarrow\; \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$

It satisfy CR equations only on the line $y=2x$
but for any point $z$ on the line $y=2x,$
there is no neighborhood about $z$ in which CR equations are satisfied

↳ not differentiable in neighborhood
↳ $f$ is nowhere analytic

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5. 2020-09-17 ¶

Criterion for differentiability
⑴ real valued functions $u(x,y)$ and $v(x,y)$ are continuous.
⑵ have continuous 1st order partial derivative in the neighborhood of point $z$
⑶ $u(x,y)$ and $v(x,y)$ satisfy C-R equations at point $z$

즉 복소함수 미분가능성은

실함수 u, v가 연속
z 근방에서 연속인 1st order 편미분을 가짐
u, v는 z에서 코시리만방정식을 만족

Criterion for analyticity
⑴ real valued functions $u(x,y)$ and $v(x,y)$ are continuous.
⑵ have continuous 1st order partial derivatives in domain D
⑶ $u(x,y)$ and $v(x,y)$ satisfy C-R equations at all points of domain D

즉 복소함수 해석가능성은 위 미분가능성에서, 점(point) 대신 domain으로만 바꾸면 됨

To check for differentiability, we use C-R equations.
Is there an easy way to check if $f(z)$ is analytic in a given domain?
(Ans: Yes)

Theorem
If $f(z)=u(x,y)+iv(x,y)$ is analytic in domain D
then $u(x,y)$ and $v(x,y)$ are harmonic functions
in the same domain.

Harmonic functions

A real valued function such as $u(x,y)$ and $v(x,y)$

- has continuous 2nd order partial derivatives in domain D
- satisfies Laplace equations

Laplace Equation

$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$

// 라플라스_방정식,Laplace_equation, 조화함수,harmonic_function

Harmonic conjugate function

We know if $f(z)=u(x,y)+iv(x,y)$ is analytic in $D$ then $u(x,y)$ and $v(x,y)$ are harmonic in $D.$
If $u(x,y)$ is harmonic in $D,$ it is possible to find a function $v(x,y)$ that is also harmonic in $D$ so that $u(x,y)+iv(x,y)$ is analytic in $D.$
$v(x,y)$ is the harmonic conjugate function of $u(x,y)$

(D는 domain)

// 현재 코시-리만_방정식,Cauchy-Riemann_equation페이지에서 conjugate harmonic function 언급되는데 같은 것인듯

Example

(a) verify $u(x,y)=x^3-3xy^2-5y$ is harmonic in the entire complex plane.

Soln. $6x+(-6x)=0$ 으로 라플라스 방정식을 만족.

(b) Find the harmonic conjugate function of $u(x,y)$

Soln. Since both

$u(x,y)$ - we have this

and

$v(x,y)$ - we are trying to find this

must satisfy CR eqns

$\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=3x^2-3y^2$ ......① and
$\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=6xy+5$ ......②

we use partial integration of eqn ① with respect to $y$

$\frac{\partial v}{\partial y}=3x^2-3y^2$
$v=3x^2y+y^3+h(x)$ - we need to find $h(x)$

// y^3앞에 부호 -인듯

To find $h(x),$

$v(x,y)=3x^2y-y^3+h(x)$
$\frac{\partial v}{\partial x}=6xy+h'(x)$ and substitute into eqn ②

gives

$h'(x)=5$
$h(x)=5x+C$

Therefore

$v(x,y)=3x^2y-y^3+5x+C$

and the analytic fn is

$f(z)=\underbrace{x^3-3xy^2-5y}_{u(x,y)}+i\underbrace{(3x^2y-y^3+5x+C)}_{v(x,y)}$

tmp; 내 생각은

$u_x=3x^2-3y^2=v_y$
$u_y=-6xy-5=-v_x$ i.e. $v_x=6xy+5$

각각 $x,y$ 로 편적분하면

$v=3x^2y-y^3+h_1(x)$
$v=3x^2y+5x+h_2(y)$

이렇게 하는게 아니라 저렇게 원래 식에 넣어야..
TBW

Exponential, logarithmic, trigonometric and hyperbolic functions

Exponential functions

$e^z=e^{x+iy}=e^x(\cos y+i\sin y)=e^x\cos y+ie^x\sin y$

Example
Evaluate $e^{1.7+4.2i}$
Soln.

$x=1.7,y=4.2$
$e^{1.7}\cos4.2=-2.6837$
$e^{1.7}\sin4.2=-4.7710$
$e^{1.7+4.2i}=-2.6837-4.7710i$

Analyticity

$f(z)=e^z$ is analytic for all $z$ because
1.

$u(x,y)=e^{x}\cos y$
$v(x,y)=e^{x}\sin y$

(이 두 함수가) continuous and have continuous 1st order partial derivatives at every point of complex plane
2. also satisfy C-R eqns at all points of the complex plane

Properties of $e^z$

$\bullet\; \frac{d}{dz}e^z=e^z$
$\bullet\; e^{z_1}e^{z_2}=e^{z_1+z_2}$
$\bullet\; \frac{e^{z_1}}{e^{z_2}}=e^{z_1-z_2}$

Periodicity

Unlike real function $e^x,$ complex function $f(z)=e^z$ is periodic with complex period $2\pi i$

$f(z+2\pi i)=f(z)$
$-\pi < y \le \pi$ is the fundamental interval/region for $f(z)=e^z$

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6. 2020-09-22 ¶

Logarithmic function

● Inverse of exponential function

$w=\ln z$ if $z=e^w$

● For $z\ne 0,$ and $\theta=\text{arg}z$

$\ln z=\ln|z| + i(\theta+2n\pi),\;\;\; n=0,\pm1,\pm2,\cdots$

// (절대값 빼먹지 말 것!)

Infinitely many values of $\ln z$
In real calculus, $\ln$ of negative numbers not defined. But in complex calculus, it is defined.

Example

Find the values of

(i) $\ln(-2)$
(ii) $\ln i$
(iii) $\ln(-1-i)$

(i)

$\ln(-2)=0.6932+i(\pi+2n\pi)$

(ii)

$\theta=\textrm{arg}(i)=\frac{\pi}{2},$
$|i|=1,$
$\log_e 1=0$
$\ln i=i\left( \frac{\pi}{2} + 2n\pi \right)$

(iii)

$\ln(-1-i)=0.3466 + i \left( \frac{5\pi}{4} + 2n\pi \right)$

Example

Find all values of $z$ such that $e^z=\sqrt{3}+i$
Hint: $z=\ln(\sqrt{3}+i)$

solving

$z=\ln 2 + i\left( \frac{\pi}{6} + 2n\pi \right)$

// 계산기 없으면 ln2의 값을 쓰지 않아도 된다는 언급.

Principal value

$\text{Ln} z=\ln |z| + i\text{Arg} z$
Arg에서, principal argument는 $(-\pi,\pi]$

// 주치,principal_value

Example

Recall $\ln(-2)=0.6932+i(\pi+2n\pi)$
Since

$\text{Arg}(-2)=\pi$ when $n=0$
$\text{Ln}(-2)=0.6932+\pi i$

- principal branch of $\ln z$ of principal logarithm function

Properties

$\ln(z_1z_2)=\ln z_1+\ln z_2$
$\ln\left(\frac{z_1}{z_2}\right)=\ln z_1-\ln z_2$

Exercise

given $z_1=1,z_2=-1$

Hint: $\ln z_1=2\pi i$

$\ln(z_1z_2)=3\pi i$
$\ln(z_1/z_2)=\pi i$

Complex Powers

recall real calculus:

$x^a=e^{a\ln x}=e^{\ln x^{a}}=x^a$

similarly,

$z^{\alpha}=e^{\alpha \ln z},\;\; z\ne 0$

Use of $\text{Ln}z$ gives principal value of $z^{\alpha}$

Example

Find the value of $i^{2i}$

$z=i,\; \text{arg}z=\frac{\pi}{2},\; \alpha=2i$
$i^{2i}=e^{2i[\ln 1+i(\pi/2 + 2n\pi)]}$

$=e^{-(1+4n)\pi}$

principal value $i^{2i}=e^{-\pi}=0.0432$

Trigonometric functions

For any complex number $z=x+iy$
$\sin z=\frac{e^{iz}-e^{-iz}}{2i}$
$\cos z=\frac{e^{iz}+e^{-iz}}{2}$

Derivatives of complex trig functions

$\frac{d}{dz}\sin z=\cos z$	$\frac{d}{dz}\cos z=-\sin z$
$\frac{d}{dz}\tan z=\sec^2 z$	$\frac{d}{dz}\cot z=-\csc^2 z$
$\frac{d}{dz}\sec z=\sec z \tan z$	$\frac{d}{dz}\csc z=-\csc z\cot z$

Identities

$\sin(-z)=-\sin z$
$\cos(-z)=\cos z$
$\cos^2z+\sin^2z=1$
$\sin(z_1\pm z_2)=\sin z_1 \cos z_2 \pm \cos z_1 \sin z_2$
$\cos(z_1\pm z_2)=\cos z_1\cos z_2\mp \sin z_1 \sin z_2$
$\sin 2z=2\sin z \cos z$
$\cos 2z=\cos^2 z-\sin^2 z$

$\sin z=\sin x\cosh y+i\cos x\sinh y$
$\cos z=\cos x\cosh y-i\sin x\sinh y$
$\cosh^2y-\sinh^2y=1$
$|\sin z|^2=\sin^2 x+\sinh^2 y$
$|\cos z|^2=\cos^2 x+\sinh^2 y$

Note:
Zeros of $\sin z$ are real numbers

$z=n\pi+0i,\;\;\; n=0,\pm1,\pm2,\cdots$

Zeros of $\cos z$ are only when

$z=(2n+1)\frac{\pi}{2}\;\;\; n=0,\pm1,\pm2,\cdots$

Example

$\sin(2+i)=\sin 2\cosh 1+i\cos 2\sinh 1$

$=1.4031-0.4891i$

Note:
real trig is accustomed to $|\sin x|\le 1,\;|\cos x|\le 1$
but in complex trig, we can have $\cos z=10$
since $\sinh y$ can range from -∞ to +∞

Example

Solve $\cos z=10$

$\cos z=(e^{iz}+e^{-iz})/2,$ this gives
$\frac{e^{iz}+e^{-iz}}{2}=10$
multiply by $e^{iz},$ we get
$e^{2iz}-20e^{iz}+1=0$
For quadratic formula,
$e^{iz}=10\pm 3\sqrt{11}$
$iz=\ln(10\pm 3\sqrt{11})\pm 2n\pi i, \;\;\; n\in\mathbb{Z}$
$z=\frac1{i}\ln(10\pm 3\sqrt{11})\pm 2n \pi$

Hyperbolic functions

// 쌍곡선함수,hyperbolic_function

For $z=x+iy,$

$\sinh z=\frac{e^{z}-e^{-z}}{2}$
$\cosh z=\frac{e^{z}+e^{-z}}{2}$
$\tanh z=\frac{\sinh z}{\cosh z}$
$\text{sech}z=\frac{1}{\cosh z}$
$\coth z=\frac{1}{\tanh z}$
$\text{csch}z=\frac{1}{\sinh z}$

Derivatives

$\frac{d}{dz}\sinh z=\cosh z$
$\frac{d}{dz}\cosh z=\sinh z$ (마이너스부호가 없다!)

Identities

$\sin (z)=(-i)\sinh(iz)$
$\cos (z)=\cosh (iz)$
$\sinh (z)=(-i)\sin(iz)$
$\cosh (z)=\cos(iz)$
$\sinh z=\sinh x \cos y + i \cosh x \sin y$
$\cosh z=\cosh x \cos y + i \sinh x \sin y$

zeros of $\sinh(z)$ and $\cosh(z)$ are pure imaginary

$z=n\pi i$

and

$z=(2n+1)\frac{\pi i}{2},\;\;\; n=0,\pm1,\pm2,\cdots$

Periodicity

$\sin(z+2\pi)=\sin z$
$\cos(z+2\pi)=\cos z$

[edit]

7. 2020-09-24 ¶

Inverse trig functions

// 역삼각함수,inverse_trigonometric_function

Inverse sine:

w=sin^-1(z) if z=sin(w)
$z=\sin w \;\to\; w=\sin^{-1}z$

$\sin^{-1}z=-i\ln\left[iz+(1-z^2)^{1/2}\right]$
$\cos^{-1}z=-i\ln\left[z+i(1-z^2)^{1/2}\right]$
$\tan^{-1}z=\frac{i}{2}\ln\frac{i+2}{i-2}$

첫번째 arcsin example
$\sin^{-1}\sqrt{5}=-i\ln[i\sqrt{5}+(1-(\sqrt{5})^2)^{1/2}]$

...crazy manipulation...
$=\frac{\pi}{2}+2n\pi \pm i\ln(\sqrt{5}+2)$

Derivatives of inverse trig functions

$\frac{d}{dz}\sin^{-1}z=\frac{1}{(1-z^2)^{1/2}}$
$\frac{d}{dz}\cos^{-1}z=-\frac{1}{(1-z^2)^{1/2}}$
$\frac{d}{dz}\tan^{-1}z=\frac{1}{1+z^2}$

Example
Find the derivative of $w=\sin^{-1}z$ at $z=\sqrt{5}$

$\left.\frac{dw}{dz}\right|_{z=\sqrt{5}}=\frac{1}{(1-(\sqrt{5})^2)^{1/2}}=\frac{1}{(-4)^{1/2}}=\pm\frac12i$

Inverse hyperbolic functions

// 역쌍곡선함수,inverse_hyperbolic_function

$\sinh^{-1}z=\ln\left[z+(z^2+1)^{1/2}\right]$
$\cosh^{-1}z=\ln\left[z+(z^2-1)^{1/2}\right]$
$\tanh^{-1}z=\frac12\ln\frac{1+z}{1-z}$

$\frac{d}{dz}\sinh^{-1}z=\frac{1}{(z^2+1)^{1/2}}$
$\frac{d}{dz}\cosh^{-1}z=\frac{1}{(z^2-1)^{1/2}}$
$\frac{d}{dz}\tanh^{-1}z=\frac{1}{1-z^2}$

Example

Find all values of $\cosh^{-1}(-1)$

With $z=-1$
$\cosh^{-1}(-1)$

$=\ln(-1)$
$=\ln 1+(\pi+2n\pi)i$
$=(2n+1)\pi i$ for $n=0,\pm1,\pm2,\cdots$

Contour integral

// 경로적분,contour_integral

We can use $z(t)=x(t)+iy(t),\; a\le t \le b$ to describe a curve $C$ in the complex plane. (여기서 t는 real parameter)

For example
$x=\cos t$
$y=\sin t$
$0\le t \le 2\pi$
gives a 단위원,unit_circle at the origin

// 관련: 매개변수방정식,parametric_equation

Definition

$\bullet\;$ contour or path → piecewise-smooth curve
$\bullet\; \int_C f(z)dz$ → integral of $f(z)$ on contour $C$
$\bullet\; \oint_C f(z)dz$ → integral of $f(z)$ on closed contour $C$
$\bullet\; \int_C f(z)dz$ → also referred to as contour or complex integral

Theorem

If $f$ is continuous on a smooth curve $C$ given by $z(t)=x(t)+iy(t),\,a\le t\le b,$ then

$\int_C f(z)dz=\int_a^b f(z(t))z'(t)dt$

(매우 중요)

Example

Evaluate $\textstyle\int_C\bar{z}dz$ where $C$ is given by $x=3t,\,y=t^2,\,-1\le t\le 4$

Soln.
$z(t)=3t+it^2$
$z'(t)=3+2it$
$f(z(t))=\bar{3t+it^2}=3t-it^2$
$\int_C\bar{z}dt=\int_{-1}^{4}(3t-it^2)(3+2it)dt$

$=\int_{-1}^{4}(2t^3+9t)dt+i\int_{-1}^{4}3t^2dt$
$=195+65i$

Example

Evaluate $\oint_C\frac{1}{z}dz$ where $C$ is a circle $x=\cos t,\,y=\sin t,\,0\le t \le 2\pi$
Soln.
$z(t)=\cos t+i\sin t$

$=e^{it}$

$z'(t)=ie^{it}$
$f(z)=\frac{1}{z}=e^{it}$
$\oint_C \frac{1}{z}dz=\int_0^{2\pi}(e^{-it})ie^{it}dt=i\int_0^{2\pi}dt=2\pi i$

Properties of contour integrals

Suppose $f$ and $g$ are continuous in domain $D$ and $c$ is a smooth curve lying entirely in $D$

$\bullet\; \int_c kf(z)dz=k\int_c f(z)dz$ (k는 상수)
$\bullet\; \int_c[f(z)+g(z)]dz=\int_c f(z)dz + \int_c g(z)dz$
$\bullet\; \int_c f(z)dz=\int_{c_1}f(z)dz+\int_{c_2}f(z)dz$

where $c$ is the union of smooth curves $c_1$ and $c_2$

$\bullet\; \int_{-c}f(z)dz=-\int_c f(z)dz$

where $-c$ denotes curve having opposite orientation of $c$

* $c_1$ is defined by $y=x,$ using $x$ as parameter

$z(x)=x+ix$
$z'(x)=1+i$
$f(z(x))=x^2+ix^2$
$\int_{c_1}(x^2+iy^2)dz$

$=\int\nolimits_{x=0}^{x=1}(x^2+ix^2)(1+i)dx$
$=(1+i)^2\int\nolimits_0^1x^2dx$
$=\frac{(1+i)^2}{3}$
$=\frac23i$

* $c_2$ is defined by $x=1,\,1\le y\le 2,$ using $y$ as parameter

$z(y)=1+iy$
$z'(y)=i$
$f(z(y))=1+iy^2$
$\int_{c_2}(x^2+iy^2)dz$

$=\int\nolimits_1^2(1+iy^2)idy$
$=-\int\nolimits_1^2y^2dy+i\int\nolimits_1^2dy$
$=-\frac73+i$

$\int_c(x^2+iy^2)dz=\frac23i+(-\frac73+i)=-\frac73+\frac53i$

[edit]

8. 2020-10-06 ¶

Bounding Theorem

● Sometimes we just want to figure out the bounding values of a contour integral
● If $f$ is continuous on smooth curve $C$

and if $|f(z)|\le M$ for all $z$ on $C$
then $\left|\int_Cf(z)dz\right|\le ML$ where $L$ is the length of $C$

(also referred to as

ML inequality)

Ex.

Find an upper bound for the absolute value of

$\oint_C\frac{e^z}{z+1}dz,$

where $C$ is the circle $|z|=4$

Soln.

Length of $C$ circle of radius 4 is $8\pi$
Now we need to find $|f(z)|$

$f(z)=\frac{e^z}{z+1}$
$|f(z)|=\left| \frac{e^z}{z+1} \right|$

Invoking(적용) $|z_1+z_2|\ge |z_1|-|z_2|$ (모르는 부등식,inequality이므로 정리 TODO)
// 수업 끝 질문답변에 의하면 실수 뿐만 아니라 복소수에서도 성립하는 부등식
Let $z_1=z,z_2=1$
we get $|z+1|\ge |z|-|1|=4-1=3$
This means $|f(z)|=\left| \frac{e^z}{z+1} \right| \le \frac{|e^z|}{|z|-1} = \frac{|e^z|}{3}$
and we have $|f(z)|\le \frac{|e^z|}{3}$
Let $|e^z|=|e^x(\cos y+i\sin y)|=e^x$
For circle $|z|=4,$ the maximum $x$ is $4$
hence $|f(z)|\le \frac{e^4}{3} = M$

Therefore

$\left| \oint_C \frac{e^z}{z+1} dz \right| \le \frac{8\pi e^4}{3}$

Simply Connected Domains

every simple closed contour $C$ lying entirely in domain $D$ can be shrunk to a point without leaving $D$

(simple: no crossing!)

Multiple Connected Domain

- not simply connected domains

Cauchy-Goursat Theorem

● If $f$ is analytic in a simply connected domain $D,$
for every simple closed contour $C$ in $D$

$\oint_C f(z)dz=0$

or
● If $f$ is analytic at all points within and on a simple closed contour $C,$ then

$\oint_C f(z)dz=0$

// 코시-구르사_정리,Cauchy-Goursat_theorem

Example

Evaluate $\oint_C \frac1{z^2}dz$ where $C$ is an ellipse $(x-2)^2+\frac{(y-5)^2}{4}=1$
(centered at x=2, y=5)

Soln.
$f(z)=\frac{1}{z^2}$ is analytic everywhere except at $z=0$
But $z=0$ is not a point interior to or on contour $C$
Therefore invoking Cauchy-Goursat Theorem

$\oint_C\frac1{z^2}dz=0$

Cauchy-Goursat Theorem for multiple connected domains

$\oint_C f(z)dz = \oint_{C_1} f(z)dz$

Deformation of contours
● evaluate integral over a funky simple closed contour by replacing with a convenient contour.

Example

Evaluate $\oint_C\frac{1}{z-i}dz$ where $C$ is this (초록색 C)

Soln.
● At $z=i,\; f(z)$ is not analytic
● Contour $C$ is too funky, let's deform it.
● We replace $C$ with $C_1,$ a circle centered at $i$ radius $1$ (노란색 C₁)

Equation of contour $C_1$

$x=\cos t,\; y=1+\sin t,\; 0\le t \le 2\pi$

This means $z=\cos t+i(1+\sin t)=i+e^{it}$
then $z-i=e^{it}$ and $dz=ie^{it}dt$
Hence $\oint\nolimits_C\frac{1}{z-i}dz$

$=\oint\nolimits_{C_1}\frac{1}{z-i}dz$
$=\int\nolimits_0^{2\pi}\frac{ie^{it}}{e^{it}}dt$
$=i\int\nolimits_0^{2\pi}dt$
$=2\pi i$

Generalization

$\oint_C\frac{1}{(z-z_0)^n}dz=\begin{cases}2\pi i,&n=1\\0,&n \textrm{ an integer }\ne 1\end{cases}$

1:18
Q: Elaborate $|e^z|=|e^x(\cos y+i\sin y)|=e^x$
A: Elaboration. $|e^z|$

$=|e^xe^{iy}|$
$=|e^x(\cos y+i\sin y)|$
$=\underbrace{|e^x|}_{e^x} \underbrace{|\cos y+i\sin y|}_{=1}$

[edit]

9. 2020-10-08 ¶

Cauchy-Goursat Theorem for multiple connected domains

$\oint_C f(z)dz=\sum_{k=1}^n \oint_{C_k} f(z)dz$

(C_k는 different contours)

Example

Evaluate $\oint_C \frac{1}{z^2+1}dz$ where $C$ is circle $|z|=3$

Soln.
- not analytic at $z=i,\;z=-i$
- these two points lie within contour $C$
- we will replace contour $C$ with contours $C_1$ and $C_2$

We can write

$\oint_C\frac{1}{z^2+1}dz=\oint_{C_1}\frac{1}{z^2+1}dz+\oint_{C_2}\frac{1}{z^2+1}dz$

Since

$\oint_C\frac{1}{z^2+1}dz=\frac{1}{2i}\oint_C\left[\frac{1}{z-i}-\frac{1}{z+i}\right]dz$ (부분분수,partial_fraction)

we have

$\oint_C\frac{1}{z^2+1}dz=\frac{1}{2i}\oint_{C_1}\left[\frac{1}{z-i}-\frac{1}{z+i}\right]dz+\frac{1}{2i}\oint_{C_2}\left[\frac{1}{z-i}-\frac{1}{z+i}\right]dz$
$=\frac{1}{2i}\oint_{C_1}\frac{1}{z-i}dz-\frac{1}{2i}\oint_{C_1}\frac{1}{z+i}dz+\frac{1}{2i}\oint_{C_2}\frac{1}{z-i}dz-\frac{1}{2i}\oint_{C_2}\frac{1}{z+i}dz$

다시 쓰면

$=\frac{1}{2i}\underbrace{\oint_{C_1}\frac{1}{z-i}dz}_{2\pi i}-\frac{1}{2i}\underbrace{\oint_{C_1}\frac{1}{z+i}dz}_{=0}+\frac{1}{2i}\underbrace{\oint_{C_2}\frac{1}{z-i}dz}_{=0}-\frac{1}{2i}\underbrace{\oint_{C_2}\frac{1}{z+i}dz}_{2\pi i}$

$=0:$ because $C_1,C_2$ does not contain the point $z=-i,$ hence the function $\frac{1}{z+i}$ is analytic on and within the contour

Therefore

$\oint_C \frac{1}{z^2+1}dz=\pi-\pi=0$

Independence of the path

If $f$ is analytic in $D$ then

$\int_c f(z)dz=\int_{c} f(z)dz$ // 하나는 c₁인듯

the contour integral $\int_c f(z)dz$ is independent of path

Example

Evaluate $\int_c 2zdz$ where $c$ is given as follows

Soln.
● $f(z)=2z$ is analytic throughout the domain. We can replace $c$ by $c_1.$
● initial point $z=-1$
● terminal point $z=-1+i$

Contour $c_1$ is given by $x=-1,\, 0\le y\le 1$
This gives $z=-1+iy,\; 0\le y\le 1$

$dz=idy$

Therefore

$\int_c 2zdz=\int_{c_1}2zdz=-2\int_0^1ydy-2i\int_0^1dy=-1-2i$

Note: path independent contour can be written as

$\int_{z_0}^{z_1}f(z)dz$

Fundamental Theorem for Contour Integral

If $f$ is continuous in domain $D$ and $F$ is the antiderivative of $f$ in $D$ for any contour $C$ in $D$ with initial point $z_0,$ terminal point $z,$ we get

$\int_C f(z)dz=F(z_1)-F(z_0)$

Example

Evaluate $\int_C \cos z dz,$ where $C$ is any contour with initial point $z=0$ and terminal point $z=2+i$

Soln.
$F(z)=\sin z$ is the antiderivative of $f(z)=\cos z$
This means

$\int_C \cos z dz=\int_0^{2+i}\cos zdz$
$=\left[\sin z\right]_{0}^{2+i}$
$=\sin(2+i)$

Note: if contour $C$ is closed, $z_0=z_1,$ then

$\oint_C f(z)dz=0$

Example

Evaluate $\int_C \frac{1}{z} dz$ where $C$ is given as follows
Domain is simply connected domain defined by

$x=\textrm{Re}(z)>0$
$y=\textrm{Im}(z)>0$

Soln.
Given the domain, $\text{Ln}z$ is the antiderivative of $\frac{1}{z}$
(Ln(z) is not analytic on non-positive real axis)
Therefore

$\int_3^{2i}\frac{1}{z}dz=\left.\text{Ln}z\right|_3^{2i}=\text{Ln}2i-\text{Ln}3$

and

$\text{Ln}2i=\ln2+\frac{\pi}{2}i$ and $\text{Ln}3=\ln 3$

hence

$\int_3^{2i}\frac{1}{z}dz=\ln\frac23+\frac{\pi}{2}i$

원점을 제외한 1사분면이 도메인이라고....(오른쪽 그림)

Existence of an antiderivative

If $f$ is analytic in a simply connected domain $D,$ then $f$ has an antiderivative in $D$
There exist a function $F$ such that $F'(z)=f(z)$ for all $z$ in $D$

[edit]

10. 2020-10-13 ¶

Cauchy's Integral Formula

Let $f$ be analytic in a simply connected domain $D$ and
let $C$ be a simple closed contour lying entirely with $D$
If $z_0$ is any point within (interior to) $C,$ then

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

where

$z_0$ is a given/known point in the complex plane
$f(z)$ evaluated at $z=z_0$

cauchy integral formula
https://mathworld.wolfram.com/CauchyIntegralFormula.html

Example

Evaluate $\oint\frac{z^2-4z+4}{z+i}dz$ where $c$ is $|z|=2$

Soln.

$f(z)=z^2-4z+4$
It is analytic at all points in the domain
$z_0=-i$ is an interior point of $C$
By Cauchy's integral formula,
$\oint_C\frac{z^2-4z+4}{z+i}dz=2\pi i f(-i)$
$=2\pi i(3+4i)$
$=2\pi(-4+3i)$

Example

Evaluate $\oint_C\frac{z}{z^2+a}dz$ where $C$ is the circle $|z-2i|=4$

Soln.

Cauchy's Integral Formula (AKA for derivatives or general form)
// 더 일반적으로

Let $f$ be analytic in a simply connected domain $D$
Let $C$ be a simple closed contour lying entirely within $D$
If $z_0$ is any point interior to $C,$ then

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

Example

Evaluate $\oint_C\frac{z+1}{z^4+4z^3}dz$ where $C$ is $|z|=1$

Soln.

// 이하 tex 대충 작성, chk

For $I_1,$ we identify $z_0=0$ and $f(z)=\frac{z^3+3}{(z-i)^2$
Therefore $I_1=\oint_{c_1}\frac{z^3+3}{\frac{(z-i)^2}{z}}dz=2\pi if(0)=-6\pi i$

For $I_2,$ we identify $z_0=i,\;n=1,\;f(z)=\frac{z^3+3}{z}$

$f'(z)=(2z^3-3)/z^2$

$I_2=\oint_{c_2}\frac{\frac{z^3+3}{2}}{(z-i)^2}dz=\frac{2\pi i}{1!}f'(i)$

$=2\pi i(3+2i)$
$=2\pi(-2+3i)$

Combining

$\oint_c\frac{z^3+3}{z(z-i)^2}dz=-I_1+I_2$

$=6\pi i+2\pi(-2+3i)$
$=4\pi(-1+3i)$

[edit]

11. 2020-10-15 ¶

Series and Residues
// 급수,series and 유수,residue

We want to express complex functions as series.
This will allow us to use Cauchy's Residue Theorem to compute complex integral of certain functions.

(참고)
// 유수,residue 유수정리,residue_theorem

유수_(복소해석학)
https://m.blog.naver.com/mindo1103/221977412976

유수정리

Geometric Series
// 기하급수,geometric_series

$\sum_{k=1}^{\infty}az^{k-1}=a+az+az^2+\cdots+az^{n-1}+\cdots$
When is a series useful?

If $|z|<1,$ series converges to $\frac{a}{1-z}$
If $|z|\ge 1,$ series diverges

Examples (valid for |z|<1)

$\frac{1}{1-z}=1+z+z^2+z^3+\cdots$
$\frac{1}{1+z}=1-z+z^2-z^3+\cdots$
$\frac{1-z^n}{1-z}=1+z+z^2+z^3+\cdots+z^{n-1}$
$\frac{1}{1-z}=1+z+z^2+z^3+\cdots+z^{n-1}+\frac{z^n}{1-z}$ - Q: 첫번째와 네번째 식의 차이? A: let me double check

Example

Given $\sum_{k=1}^{\infty}\frac{(1+2i)^k}{5^k}=\frac{1+2i}{5}+\frac{(1+2i)^2}{5^2}+\frac{(1+2i)^3}{5^3}+\cdots$
We see that $a=\frac{1+2i}{5},\,z=\frac{1+2i}{5}$
Since $|z|=\frac{\sqrt{5}}{5}<1,$ the series converges.
Therefore $\sum_{k=1}^{\infty} \frac{(1+2i)^k}{5^k}=............$

Convergence and Divergence

If $\sum_{k=1}^{\infty} z_k$ converges, then $\lim_{n\to\infty}z_n=0$
If $\sum_{k=1}^{\infty} z_k$ diverges, then $\lim_{n\to\infty}z_n\ne 0$
If $\sum_{k=1}^{\infty}|z_k|$ converges, then $\sum_{k=1}^{\infty}z_k$ is absolutely convergent.

// 수렴,convergence

Ratio Test

Given $\sum_{k=1}^{\infty}z_k$ is a series of non-zero complex terms such as

$\lim_{n\to\infty}\left| \frac{z_{n+1}}{z_n} \right| = L$

(i) $L<1,$ series converges absolutely
(ii) $L>1,$ series diverges
(iii) $L=1,$ inconclusive

Q: (iii)에서 converge하는 case가 있는가? A: 나중에. TBW

// 수렴판정법,convergence_test

Root Test

Given $\sum_{k=1}^{\infty}z_k$ is a series of complex terms such that

$\lim_{n\to\infty}\sqrt[n]{|z_n|}=L$

(i) $L<1,$ series converges absolutely
(ii) $L>1,$ series diverges
(iii) $L=1,$ inconclusive

// oscillation(진동,oscillation,vibration) 등 언급됨

Power Series

$\sum_{k=0}^{\infty} a_k (z-z_0)^k = a_0 + a_1(z-z_0) + a_2(z-z_0)^2 + \cdots$
여기서 시그마 옆의
$a_k$ : complex constants
$(z-z_0)$ : power series in $z-z_0$ meaning it is centered at $z_0$
$k$ : power
필기에는 없지만 $z_0$ : center

// 멱급수,power_series

Circle of Convergence // 이건 캡쳐할 수 밖에

Ratio Test

Given power series $\sum_{k=0}^{\infty}a_k(z-z_0)^k,$
(i) $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L\ne 0$ , radius of convergence $R=1/L$
(ii) $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=0$ , radius of convergence is $\infty$
(iii) $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\infty$ , radius of convergence is zero

여기서 노랑색으로 강조 (아래 참조)

Root Test

Similar evaluation for root test $\lim_{n\to\infty}\sqrt[n]{|a_n|}$

Example

Given power series $\sum_{k=1}^{\infty}\left(\frac{6k+1}{2k+5}\right)^k(z-2i)^k$
identifying $a_n=\left( \frac{6n+1}{2n+5} \right)^n$ and
root test $\lim_{n\to\infty}\sqrt[n]{|a_n|}=\lim_{n\to\infty}\frac{6n+1}{2n+5}=\lim_{n\to\infty}\frac{6+\frac{1}{n}}{2+\frac{5}{n}}=3$
This gives $R=\frac13$ and the circle of convergence is $|z-2i|=\frac13$
and the series converges for $|z-2i|<\frac13$

// 이하 series 두 개를 살펴본다. Taylor & Laurent

Taylor Series

We can use power series to represent analytic function within its circle of convergence.
A power series $\sum_{k=0}^{\infty}a_k(z-z_0)^k$ represents a continuous function $f$ within its circle of convergence $|z-z_0|=R,\;R\ne 0$
A power series $\sum_{k=0}^{\infty}a_k(z-z_0)^k$ can be differentiated or integrated term by term within its circle of convergence $|z-z_0|=R,\,R\ne 0$ for every contour $C$ lying entirey within the circle of convergence.

// 테일러_급수,Taylor_series

Introducing Taylor Series

$f(z)=\sum_{k=0}^{\infty}\frac{f^{(k)}(z_0)}{k!}(z-z_0)^k$

Introducing Maclaurin Series

$f(z)=\sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}z^k$

Taylor's Theorem

If $f$ is analytic in domain $D,\,z_0$ is a point in domain $D$
then $f$ can be represented by

$f(z)=\sum_{k=0}^{\infty}\frac{f^{(k)}(z_0)}{k!}(z-z_0)^k$

valid for the largest circle $C$ that lies entirely in $D$

Power series expansion of a function with center $z_0$ is unique.
If $\sum_{k=0}^{\infty}a_k(z-z_0)^k$ and $\sum_{k=0}^{\infty}b_k(z-z_0)^k$ represent the some function, then $a_k=b_k$

[edit]

12. 2020-10-20 ¶

Example of Maclaurin Series

$e^z=1+\frac{z}{1!}+\frac{z^2}{2!}+\cdots=\sum_{k=0}^{\infty}\frac{z^k}{k!}$
$\sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots=\sum_{k=0}^{\infty}(-1)^k \frac{z^{2k+1}}{(2k+1)!}$
$\cos z=1-\frac{z^2}{2!}+\frac{z^4}{4!}-\cdots=\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k}}{(2k)!}$

// 매클로린_급수,Maclaurin_series

Example

Expand $f(z)=\frac1{1-z}$ in a Taylor series with center $z_0=2i$

Soln.
Using $f(z)=\sum_{k=0}^{\infty}\frac{f^{(k)}(z_0)}{k!}(z-z_0)^k$

$f'(z)=\frac1{(1-z)^2}$
$f''(z)=\frac{2!}{(1-z)^3}$
$f'''(z)=\frac{3!}{(1-z)^4}$ (패턴을 파악)

We conclude

$f^{(n)}(z)=\frac{n!}{(1-z)^{n+1}}$ and
$f^{(n)}(2i)=\frac{n!}{(1-2i)^{n+1}}$

Hence

$\frac{1}{1-z}=\sum_{k=0}^{\infty}\frac{1}{(1-2i)^{k+1}}(z-2i)^k$

Since the distance from center $z_0=2i$ to nearest singularity $z=1$ is $\sqrt{5},$
we conclude that circle of convergence is $|z-2i|=\sqrt{5}$

그림 있어서 캡쳐

Laurent Series

singularity or singular point $z=z_0$

(이게 뜻이 뭐냐면:) Complex function $f$ is not analytic at this point
$f$ cannot be expanded as in power series with $z_0$ as center.
need a new kind of series.
$f(z)=$ (principal part) + (analytic part)

$=\sum_{k=1}^{\infty} a_{-k} (z-z_0)^{-k} + \sum_{k=0}^{\infty}a_k(z-z_0)^k$

$f(z)=\sum_{k=-\infty}^{\infty}a_k(z-z_0)^k$

// 로랑_급수,Laurent_series - 테일러급수의 일반화

Laurent's Theorem

If $f$ which is analytic in annular(고리 모양의) domain defined by

$r < |z-z_0| < R,$

then $f$ can be represented by

$f(z)=\sum_{k=-\infty}^{\infty}a_k(z-z_0)^k$

and the coefficients $a_k$ are given by

$a_k=\frac{1}{2\pi i}\oint_C\frac{f(s)}{(s-z_0)^{k+1}}ds,\;\;\;k=0,\pm1,\pm2,\cdots$

where $C$ is a simple closed contour
lies entirely within $D$ and has $z_0$ in its interior

Example

Expand $f(z)=\frac{8z+1}{z(1-z)}$ in a Laurent series valid for $0<|z|<1$

Soln.
Recall $\frac{1}{1-z}=1+z+z^2+z^3+\cdots$
then $f(z)=\frac{8z+1}{z(1-z)}=\frac{8z+1}{z}\frac{1}{(1-z)}=\left(8+\frac{1}{z}\right)(1+z+z^2+z^3+\cdots)$
multiply out and collect like terms

$f(z)=8+8z+8z^2+8z^3+\cdots+\frac{1}{z}+1+z+z^2+\cdots$

$=\frac{1}{z}+9+9z+9z^2+\cdots$ (.... converges for |z|<1)

and valid for $0<|z|<1$

Example

Expand $f(z)=e^{3/z}$ in a Laurent series valid for $0<|z|$

Soln. Recall

$e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$

TBW

Example

Expand $f(z)=\frac{1}{(z-1)^2(z-3)}$ in a Laurent series valid for $0<|z-1|<2$

Soln.
We only want powers of $z-1$ ( given $0<|z-1|<2$ )
We need to express $z-3$ in terms of $z-1$
$f(z)=\frac1{(z-1)^2(z-3)}$

$=\frac1{(z-1)^2}\cdot\frac1{-2+(z-1)}$
$=\frac{-1}{2(z-1)^2}\cdot\frac1{1-\frac{z-1}{2}}$

Recall $\frac1{1-z}=1+z+z^2+z^3+\cdots$
We replace $z$ with $\frac{(z-1)}{2}$
and we have
$f(z)=\frac{-1}{2(z-1)^2}\left[ 1+\frac{z-1}{2} +\frac{(z-1)^2}{2^2} +\frac{(z-1)^3}{2^3} +\cdots\right]$

$=\underbrace{-\frac1{2(z-1)^2}-\frac1{4(z-1)}}_{\text{principal}}\underbrace{-\frac18-\frac1{16}(z-1) \cdots}_{\textrm{analytic}}$

Can we do it for $0<|z-3|<2$ ?

[edit]

13. 2020-10-22 ¶

Zeros and Poles

// 로랑 급수에는 principal 과 analytic 파트가 있었다.

Recall the principal part of Laurent series

$\sum_{k=1}^{\infty}a_{-k}(z-z_0)^{-k}=\sum_{k=1}^{\infty}\frac{a_{-k}}{(z-z_0)^k}$

Removable Singularity

For $z=z_0,$ all coefficients $a_{-k}$ are zero
Laurent series: $a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots$

// only the analytic part of Laurent series is left

Example:

$\frac{\sin z}{z}=1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots,\;z=0$ is a removable singularity

Q: in sinz/z, z=0 is removable singularity. But is it clear before we convert it?
A: not really. not obvious.

Essential Singularity

for $z=z_0,\,\sum_{k=1}^{\infty}\frac{a_k}{(z-z_0)^k}$ contain infinitely many nonzero terms
Laurent series:

$\underbrace{\cdots+\frac{a_{-2}}{(z-z_0)^2}+\frac{a_{-1}}{z-z_0}}_{\textrm{nonzero terms\\ in principal part}}+a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots$

Example: $f(z)=e^{3/z}=1+\frac{3}{z}+\frac{3^2}{z!z^2}+\frac{3^3}{3!z^3}+\cdots$

Pole of order $n$

for $z=z_0$
$\frac{a_{-n}}{(z-z_0)^n}+\frac{a_{-(n-1)}}{(z-z_0)^{n-1}}+\cdots+\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+\cdots$
$a+z=z_0,$ it has pole of order $n$
If $n=1,$ it is a simple pole

Zero of order $n$

analytic function $f(z)$ has a zero of order $n$ if
$f(z_0)=0,\;f'(z_0)=0,\; f''(z_0)=0,\; \cdots f^{(n-1)}(z_0)=0$
but $f^n(z_0)\ne 0$

Example: $f(z)=z\sin z^2=z^3\left[1-\frac{z^4}{3!}+\frac{z^8}{5!}-\cdots\right]$
If $z_0$ is a zero of non-trivial function $f,$
then $\frac1{f(z)}$ has an isolated singularity at $z=z_0$

// 추가로 노란 annotation 참고

Example

Given rational function $\frac{2z+5}{(z-1)(z+5)(z-2)^4}$
....그림참조

rational function pole zero
// Q&A:

isolated singularity설명만으론 이해가 잘 안 감
// kms 단어는 고립된 특이점
// tmp https://m.blog.naver.com/heejoo_kang/220805648869

Residue

Recall
$f(z)=\sum_{k=-\infty}^{\infty}a_k(z-z_0)^k$ // z₀ is isolated singularity

$=\cdots+\frac{a_{-2}}{(z-z_0)^2}+\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+\cdots$

(위 $a_{-1}$ 에서)
coefficient $a_{-1}$ of $\frac{1}{z-z_0}$ in the Laurent series is called the residue of $f$ at isolated singularity $z_0$

// 유수,residue

Residue at a simple pole

If $f$ has a simple pole at $z=z_0,$ then
$\textrm{Res}(f(z),z_0)=\lim_{z\to z_0}(z-z_0)f(z)$

Residue at pole of order $n$

If $f$ has a pole of order $n$ at $z=z_0,$ then
$\textrm{Res}(f(z),z_0)=\frac1{(n-1)!}\lim_{z\to z_0}\left[ \frac{d^{n-1}}{dz^{n-1}}\left[ (z-z_0)^n f(z) \right]\right]$