Difference between r1.101 and the current
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Ch13
Ch14
Ch15
= CHEM154 일반화학실험2 =
일정:
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분반: CHEM154-29
시험: 2020 12 09 수 10:30~
due: 시험전날
= KECE204 전기회로Ⅱ =
Textbook: Hayt, Kimmerly, Durbin, Engineering Circuit Analysis 8th Edition, Mcgraw-Hill
@@ -49,1946 +53,9 @@
[[복소해석,complex_analysis]]
[[선형대수,linear_algebra]]
// [[거듭제곱근,nth_root]] curr goto [[제곱근,square_root]]
Roots of $z$
[[공학수학2_선형대수]]
$z=r(\cos\theta+i\sin\theta)$
$w=\rho(\cos\phi+i\sin\phi)$
If $z=w^n,$ then $w$ is the n^^th^^ root of $z.$
$r(\cos\theta+i\sin\theta)=\rho^n(\cos(n\phi)+i\sin(n\phi))$
$r=\rho^n$ which means $\rho=r^{\frac{1}{n}}$
$\left.{{\cos\theta=\cos n\phi} \atop {\sin\theta=\sin n\phi}}\right\rbrace \; n\phi=\theta+2k\pi$ (k is integer)
$\phi=\frac{\theta+2k\pi}{n}$
for different $k$ values, we have different arguments([[편각,argument]])
$w_k=r^{\frac{1}{n}}\left[\cos\left(\frac{\theta+2k\pi}{n}\right)+i\sin\left( \frac{\theta+2k\pi}{n} \right) \right]$
where $k=0,1,2,\cdots,n-1$
Find the three cube roots of $i$
$k=0,\; w_0=\frac{\sqrt{3}}{2}+\frac12i$
$k=1,\; w_1=-\frac{\sqrt{3}}{2}+\frac12i$
complex analysis ([[복소해석,complex_analysis]])
- complex variable $z=x+iy$
- complex calculus (integrals)
its conjugate is $\bar{z}=z^*=x-iy$
Polar form ([[극형식,polar_form]])
$z=re^{i\theta}=r(\cos\theta+i\sin\theta)$
$\theta$ : argument ([[편각,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)
de Moivre's Formula ([[드무아브르_공식,de_Moivre_s_formula]])
$z=r(\cos\theta+i\sin\theta)$
$z^n=r^n(\cos n\theta + i\sin n\theta)$
$z=r(\cos\theta+i\sin\theta)$
$w=\rho(\cos\phi+i\sin\phi)$
$w_k=r^{1/n}\left[ \cos\left( \frac{\theta+2k\pi}{n} \right) + i\sin\left( \frac{\theta+2k\pi}{n} \right) \right]$
where $k=0,1,2,\cdots,n-1$
Our goal is to examine the functions of a single complex variable $z=x+iy$ and the calculus of these functions.
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
// (중심점 z,,0,,에서 거리 ρ인 점 z들의 집합 즉 [[원,circle]])
circle centered at $z_0$ contains all $z$ that satisfy $|z-z_0|=\rho$
$z_0=0,$ equation of unit circle centered at origin
$z_0=1+2i,$ 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 $z_0$
$z_0$ is an '''interior point''' of set $S$ if some neighborhood of $z_0$ lies entirely within $S.$
https://i.imgur.com/b9Chqvw.png
// interior point(내점, 안점), boundary point(경계점)
Set $S$ is an '''open set''' if every point $z$ of set $S$ is an interior point.
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
a domain with all, some or none of its boundary points.
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 $f$
$=\underbrace{(x^2-y^2-4x)}_{u(x,y)} + i\underbrace{(2xy-4y)}_{v(x,y)}$
Find the image of line Re(z)=1 given $f(z)=z^2$
$u=1-\left(\frac{v}{2}\right)^2=1-\frac{v^2}{4}$
https://i.imgur.com/VBFvQhT.png
complex function $w=f(z)$ interpreted as mapping([[사상,mapping]]) or transformation([[변환,transformation]]) from z-plane to w-plane.
https://i.imgur.com/vfU9KJ7.png
Limits of a complex function
whenever $0<|z-z_0|<\delta$
https://i.imgur.com/DgFTcio.png
as $z\to z_0,$ it can approach from any direction in the complex plane.
Limit of sum, product, quotient
$\lim_{z\to z_0}f(z)=L_1$
$\lim_{z\to z_0}g(z)=L_2$
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 $h(z)=0$
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 $z_0$
$\frac{d}{dz}cf(z)=cf'(z)$
$\frac{d}{dz}\left[f(z)+g(z)\right]=f'(z)+g'(z)$
$\frac{d}{dz}\left[f(z)g(z)\right]=f(z)g'(z)+g(z)f'(z)$
$\frac{d}{dz}\left[\frac{f(z)}{g(z)}\right]=\frac{g(z)f'(z)-f(z)g'(z)}{[g(z)]^2}$
chain rule ([[연쇄법칙,chain_rule]])
$\frac{d}{dz}f(g(z))=f'(g(z))g'(z)$
$\frac{d}{dz}z^n=nz^{n-1},\;\;n\in\mathbb{Z}$
Note: for it to be differentiable, it must approach the __same__ complex number from __any direction__
Show that $f(z)=x+4iy$ is nowhere differentiable.
$\Delta z=\Delta x+i\Delta y$
$=(x+\Delta x)+i4(y+\Delta y)-x-4iy$
$f'(z_0)=\lim_{\Delta z \to 0}(.....)$
$\lim_{\Delta z\to 0}\frac{f(z+\Delta z)-f(z)}{\Delta z}$
$=\lim_{\Delta z\to 0}\frac{\Delta x+4i\Delta y}{\Delta x+i\Delta y}$
if $\Delta z\to 0$ along line parallel to x-axis
if $\Delta z\to 0$ along line parallel to y-axis,
● approach from different direction gives different values. It is nowhere differentiable.
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 $z_0$
Is $f(z)=z^2+2$ analytic for all $z?$
$f(z)=x^2-y^2+x+i(2xy+y)$
$\frac{\partial u}{\partial x}=2x+1=\frac{\partial v}{\partial y}$
$\frac{\partial u}{\partial y}=-2y=-\frac{\partial v}{\partial x}$ 이 둘은 모든 $x,y$ 에 대해 코시-리만 방정식을 만족.
∴ $f(z)$ is analytic for all $z.$
Show that $f(z)=(2x^2+y)+i(y^2-x)$ is not analytic at any point.
$\frac{\partial u}{\partial x}=4x\text{ and }\frac{\partial v}{\partial y}=2y \;\Rightarrow\; \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ 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
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 $z$
* z 근방에서 연속인 1st order 편미분을 가짐
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?
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
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
$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$
// [[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 $u(x,y)$
// 현재 [[코시-리만_방정식,Cauchy-Riemann_equation]]페이지에서 conjugate harmonic function 언급되는데 같은 것인듯
(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 $u(x,y)$
$u(x,y)$ - ''we have this''
$v(x,y)$ - '' we are trying to find this''
$\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=3x^2-3y^2$ ......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 $h(x)$
$\frac{\partial v}{\partial x}=6xy+h'(x)$ and substitute into eqn ②
$f(z)=\underbrace{x^3-3xy^2-5y}_{u(x,y)}+i\underbrace{(3x^2y-y^3+5x+C)}_{v(x,y)}$
$u_y=-6xy-5=-v_x$ i.e. $v_x=6xy+5$
이렇게 하는게 아니라 저렇게 원래 식에 넣어야..
Exponential, logarithmic, trigonometric and hyperbolic functions
$e^z=e^{x+iy}=e^x(\cos y+i\sin y)=e^x\cos y+ie^x\sin y$
$e^{1.7+4.2i}=-2.6837-4.7710i$
$f(z)=e^z$ is analytic for all $z$ because
(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
$\bullet\; \frac{d}{dz}e^z=e^z$
$\bullet\; e^{z_1}e^{z_2}=e^{z_1+z_2}$
$\bullet\; \frac{e^{z_1}}{e^{z_2}}=e^{z_1-z_2}$
Unlike real function $e^x,$ [[복소함수,complex_function|complex function]] $f(z)=e^z$ is periodic with complex period $2\pi i$
* $-\pi < y \le \pi$ is the fundamental interval/region for $f(z)=e^z$
● Inverse of [[지수함수,exponential_function|exponential function]]
● For $z\ne 0,$ and $\theta=\text{arg}z$
$\ln z=\ln|z| + i(\theta+2n\pi),\;\;\; n=0,\pm1,\pm2,\cdots$
* Infinitely many values of $\ln z$
* In real [[미적분,calculus|calculus]], $\ln$ of negative numbers not defined. But in complex calculus, it is defined.
$\ln(-2)=0.6932+i(\pi+2n\pi)$
$\theta=\textrm{arg}(i)=\frac{\pi}{2},$
$\ln i=i\left( \frac{\pi}{2} + 2n\pi \right)$
$\ln(-1-i)=0.3466 + i \left( \frac{5\pi}{4} + 2n\pi \right)$
Find all values of $z$ such that $e^z=\sqrt{3}+i$
Hint: $z=\ln(\sqrt{3}+i)$
$z=\ln 2 + i\left( \frac{\pi}{6} + 2n\pi \right)$
// 계산기 없으면 ln2의 값을 쓰지 않아도 된다는 언급.
$\text{Ln} z=\ln |z| + i\text{Arg} z$
Arg에서, principal argument는 $(-\pi,\pi]$
// [[주치,principal_value]]
Recall $\ln(-2)=0.6932+i(\pi+2n\pi)$
$\text{Arg}(-2)=\pi$ when $n=0$
$\text{Ln}(-2)=0.6932+\pi i$
- principal branch of $\ln z$ of principal logarithm function
* $\ln(z_1z_2)=\ln z_1+\ln z_2$
* $\ln\left(\frac{z_1}{z_2}\right)=\ln z_1-\ln z_2$
$x^a=e^{a\ln x}=e^{\ln x^{a}}=x^a$
$z^{\alpha}=e^{\alpha \ln z},\;\; z\ne 0$
Use of $\text{Ln}z$ gives [[주치,principal_value|principal value]] of $z^{\alpha}$
Find the value of $i^{2i}$
$z=i,\; \text{arg}z=\frac{\pi}{2},\; \alpha=2i$
$i^{2i}=e^{2i[\ln 1+i(\pi/2 + 2n\pi)]}$
principal value $i^{2i}=e^{-\pi}=0.0432$
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
||$\frac{d}{dz}\sin z=\cos z$ ||$\frac{d}{dz}\cos z=-\sin z$ ||
||$\frac{d}{dz}\tan z=\sec^2 z$ ||$\frac{d}{dz}\cot z=-\csc^2 z$ ||
||$\frac{d}{dz}\sec z=\sec z \tan z$||$\frac{d}{dz}\csc z=-\csc z\cot z$ ||
$\sin(z_1\pm z_2)=\sin z_1 \cos z_2 \pm \cos z_1 \sin z_2$
$\cos(z_1\pm z_2)=\cos z_1\cos z_2\mp \sin z_1 \sin z_2$
$\cos 2z=\cos^2 z-\sin^2 z$
$\sin z=\sin x\cosh y+i\cos x\sinh y$
$\cos z=\cos x\cosh y-i\sin x\sinh y$
$|\sin z|^2=\sin^2 x+\sinh^2 y$
$|\cos z|^2=\cos^2 x+\sinh^2 y$
Zeros of $\sin z$ are real numbers
$z=n\pi+0i,\;\;\; n=0,\pm1,\pm2,\cdots$
Zeros of $\cos z$ are only when
$z=(2n+1)\frac{\pi}{2}\;\;\; n=0,\pm1,\pm2,\cdots$
$\sin(2+i)=\sin 2\cosh 1+i\cos 2\sinh 1$
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 +∞
$\cos z=(e^{iz}+e^{-iz})/2,$ this gives
$\frac{e^{iz}+e^{-iz}}{2}=10$
multiply by $e^{iz},$ we get
$e^{iz}=10\pm 3\sqrt{11}$
$iz=\ln(10\pm 3\sqrt{11})\pm 2n\pi i, \;\;\; n\in\mathbb{Z}$
$z=\frac1{i}\ln(10\pm 3\sqrt{11})\pm 2n \pi$
// [[쌍곡선함수,hyperbolic_function]]
* $\sinh z=\frac{e^{z}-e^{-z}}{2}$
* $\cosh z=\frac{e^{z}+e^{-z}}{2}$
* $\tanh z=\frac{\sinh z}{\cosh z}$
* $\text{sech}z=\frac{1}{\cosh z}$
* $\coth z=\frac{1}{\tanh z}$
* $\text{csch}z=\frac{1}{\sinh z}$
* $\frac{d}{dz}\sinh z=\cosh z$
* $\frac{d}{dz}\cosh z=\sinh z$ (마이너스부호가 없다!)
* $\sin (z)=(-i)\sinh(iz)$
* $\sinh (z)=(-i)\sin(iz)$
* $\sinh z=\sinh x \cos y + i \cosh x \sin y$
* $\cosh z=\cosh x \cos y + i \sinh x \sin y$
zeros of $\sinh(z)$ and $\cosh(z)$ are pure imaginary
$z=(2n+1)\frac{\pi i}{2},\;\;\; n=0,\pm1,\pm2,\cdots$
// [[역삼각함수,inverse_trigonometric_function]]
w=sin^^-1^^(z) if z=sin(w)
$z=\sin w \;\to\; w=\sin^{-1}z$
$\sin^{-1}z=-i\ln\left[iz+(1-z^2)^{1/2}\right]$
$\cos^{-1}z=-i\ln\left[z+i(1-z^2)^{1/2}\right]$
$\tan^{-1}z=\frac{i}{2}\ln\frac{i+2}{i-2}$
$\sin^{-1}\sqrt{5}=-i\ln[i\sqrt{5}+(1-(\sqrt{5})^2)^{1/2}]$
$=\frac{\pi}{2}+2n\pi \pm i\ln(\sqrt{5}+2)$
Derivatives of inverse trig functions
$\frac{d}{dz}\sin^{-1}z=\frac{1}{(1-z^2)^{1/2}}$
$\frac{d}{dz}\cos^{-1}z=-\frac{1}{(1-z^2)^{1/2}}$
$\frac{d}{dz}\tan^{-1}z=\frac{1}{1+z^2}$
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]]
$\sinh^{-1}z=\ln\left[z+(z^2+1)^{1/2}\right]$
$\cosh^{-1}z=\ln\left[z+(z^2-1)^{1/2}\right]$
$\tanh^{-1}z=\frac12\ln\frac{1+z}{1-z}$
$\frac{d}{dz}\sinh^{-1}z=\frac{1}{(z^2+1)^{1/2}}$
$\frac{d}{dz}\cosh^{-1}z=\frac{1}{(z^2-1)^{1/2}}$
$\frac{d}{dz}\tanh^{-1}z=\frac{1}{1-z^2}$
Find all values of $\cosh^{-1}(-1)$
$=(2n+1)\pi i$ for $n=0,\pm1,\pm2,\cdots$
// [[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)''
gives a [[단위원,unit_circle]] at the origin
// 관련: [[매개변수방정식,parametric_equation]]
$\bullet\;$ contour or path → piecewise-smooth [[곡선,curve|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
If $f$ is continuous on a smooth curve $C$ given by $z(t)=x(t)+iy(t),\,a\le t\le b,$ then
$\int_C f(z)dz=\int_a^b f(z(t))z'(t)dt$
Evaluate $\textstyle\int_C\bar{z}dz$ where $C$ is given by $x=3t,\,y=t^2,\,-1\le t\le 4$
$f(z(t))=\bar{3t+it^2}=3t-it^2$
$\int_C\bar{z}dt=\int_{-1}^{4}(3t-it^2)(3+2it)dt$
$=\int_{-1}^{4}(2t^3+9t)dt+i\int_{-1}^{4}3t^2dt$
Evaluate $\oint_C\frac{1}{z}dz$ where $C$ is a circle $x=\cos t,\,y=\sin t,\,0\le t \le 2\pi$
$f(z)=\frac{1}{z}=e^{it}$
$\oint_C \frac{1}{z}dz=\int_0^{2\pi}(e^{-it})ie^{it}dt=i\int_0^{2\pi}dt=2\pi i$
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 $c$
https://i.imgur.com/HwRmb0h.png
* $c_1$ is defined by $y=x,$ using $x$ as parameter
$=\int\nolimits_{x=0}^{x=1}(x^2+ix^2)(1+i)dx$
$=(1+i)^2\int\nolimits_0^1x^2dx$
* $c_2$ is defined by $x=1,\,1\le y\le 2,$ using $y$ as parameter
$=\int\nolimits_1^2(1+iy^2)idy$
$=-\int\nolimits_1^2y^2dy+i\int\nolimits_1^2dy$
$\int_c(x^2+iy^2)dz=\frac23i+(-\frac73+i)=-\frac73+\frac53i$
● 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 Google:ML+inequality)
Find an upper bound for the absolute value of
$\oint_C\frac{e^z}{z+1}dz,$
where $C$ is the circle $|z|=4$
* Length of $C$ circle of radius 4 is $8\pi$
* Now we need to find $|f(z)|$
$|f(z)|=\left| \frac{e^z}{z+1} \right| $
Invoking(적용) $|z_1+z_2|\ge |z_1|-|z_2|$ (모르는 [[부등식,inequality]]이므로 정리 TODO)
// 수업 끝 질문답변에 의하면 실수 뿐만 아니라 복소수에서도 성립하는 부등식
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 $|f(z)|\le \frac{e^4}{3} = M$
$\left| \oint_C \frac{e^z}{z+1} dz \right| \le \frac{8\pi e^4}{3}$
every simple closed contour $C$ lying entirely in domain $D$ can be shrunk to a point without leaving $D$
https://i.imgur.com/5djgUzG.png
Multiple Connected Domain
- not simply connected domains
https://i.imgur.com/PBsdWrm.png
● If $f$ is analytic in a simply connected domain $D,$
for every simple closed contour $C$ in $D$
● If $f$ is analytic at all points within and on a simple closed contour $C,$ then
// [[코시-구르사_정리,Cauchy-Goursat_theorem]]
Evaluate $\oint_C \frac1{z^2}dz$ where $C$ is an ellipse $(x-2)^2+\frac{(y-5)^2}{4}=1$
$f(z)=\frac{1}{z^2}$ 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
Cauchy-Goursat Theorem for multiple connected domains
$\oint_C f(z)dz = \oint_{C_1} f(z)dz$
● evaluate integral over a funky simple closed contour by replacing with a convenient contour.
https://i.imgur.com/3Lir8yY.png
Evaluate $\oint_C\frac{1}{z-i}dz$ where $C$ is this (초록색 C)
● 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 C,,1,,)
https://i.imgur.com/iGSR8IC.png
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 $\oint\nolimits_C\frac{1}{z-i}dz$
$=\oint\nolimits_{C_1}\frac{1}{z-i}dz$
$=\int\nolimits_0^{2\pi}\frac{ie^{it}}{e^{it}}dt$
$=i\int\nolimits_0^{2\pi}dt$
$\oint_C\frac{1}{(z-z_0)^n}dz=\begin{cases}2\pi i,&n=1\\0,&n \textrm{ an integer }\ne 1\end{cases}$
Q: Elaborate $|e^z|=|e^x(\cos y+i\sin y)|=e^x$
$=\underbrace{|e^x|}_{e^x} \underbrace{|\cos y+i\sin y|}_{=1}$
Cauchy-Goursat Theorem for multiple connected domains
$\oint_C f(z)dz=\sum_{k=1}^n \oint_{C_k} f(z)dz$
(C,,k,,는 different contours)
Evaluate $\oint_C \frac{1}{z^2+1}dz$ where $C$ is circle $|z|=3$
- not analytic at $z=i,\;z=-i$
- these two points lie within contour $C$
- we will replace contour $C$ with contours $C_1$ and $C_2$
https://i.imgur.com/XP8fUfF.png
$\oint_C\frac{1}{z^2+1}dz=\oint_{C_1}\frac{1}{z^2+1}dz+\oint_{C_2}\frac{1}{z^2+1}dz$
$\oint_C\frac{1}{z^2+1}dz=\frac{1}{2i}\oint_C\left[\frac{1}{z-i}-\frac{1}{z+i}\right]dz$ ([[부분분수,partial_fraction]])
$\oint_C\frac{1}{z^2+1}dz=\frac{1}{2i}\oint_{C_1}\left[\frac{1}{z-i}-\frac{1}{z+i}\right]dz+\frac{1}{2i}\oint_{C_2}\left[\frac{1}{z-i}-\frac{1}{z+i}\right]dz$
$=\frac{1}{2i}\oint_{C_1}\frac{1}{z-i}dz-\frac{1}{2i}\oint_{C_1}\frac{1}{z+i}dz+\frac{1}{2i}\oint_{C_2}\frac{1}{z-i}dz-\frac{1}{2i}\oint_{C_2}\frac{1}{z+i}dz$
$=\frac{1}{2i}\underbrace{\oint_{C_1}\frac{1}{z-i}dz}_{2\pi i}-\frac{1}{2i}\underbrace{\oint_{C_1}\frac{1}{z+i}dz}_{=0}+\frac{1}{2i}\underbrace{\oint_{C_2}\frac{1}{z-i}dz}_{=0}-\frac{1}{2i}\underbrace{\oint_{C_2}\frac{1}{z+i}dz}_{2\pi i}$
$=0:$ 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
$\oint_C \frac{1}{z^2+1}dz=\pi-\pi=0$
If $f$ is analytic in $D$ then
$\int_c f(z)dz=\int_{c} f(z)dz$ // 하나는 c,,1,,인듯
the contour integral $\int_c f(z)dz$ is __independent of path__
https://i.imgur.com/ZD3Uh2C.png
Evaluate $\int_c 2zdz$ where $c$ is given as follows
https://i.imgur.com/VuKrwSV.png
● $f(z)=2z$ is analytic throughout the domain. We can replace $c$ by $c_1.$
● terminal point $z=-1+i$
https://i.imgur.com/tpOsAvO.png
Contour $c_1$ is given by $x=-1,\, 0\le y\le 1$
This gives $z=-1+iy,\; 0\le y\le 1$
$\int_c 2zdz=\int_{c_1}2zdz=-2\int_0^1ydy-2i\int_0^1dy=-1-2i$
Note: path independent contour can be written as
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
$\int_C f(z)dz=F(z_1)-F(z_0)$
Evaluate $\int_C \cos z dz,$ where $C$ is any contour with initial point $z=0$ and terminal point $z=2+i$
$F(z)=\sin z$ is the antiderivative of $f(z)=\cos z$
$\int_C \cos z dz=\int_0^{2+i}\cos zdz$
$=\left[\sin z\right]_{0}^{2+i}$
Note: if contour $C$ is closed, $z_0=z_1,$ then
Evaluate $\int_C \frac{1}{z} dz$ where $C$ is given as follows
Domain is simply connected domain defined by
https://i.imgur.com/d6KCs2r.png
Given the domain, $\text{Ln}z$ is the antiderivative of $\frac{1}{z}$
(Ln(z) is not analytic on non-positive real axis)
$\int_3^{2i}\frac{1}{z}dz=\left.\text{Ln}z\right|_3^{2i}=\text{Ln}2i-\text{Ln}3$
$\text{Ln}2i=\ln2+\frac{\pi}{2}i$ and $\text{Ln}3=\ln 3$
$\int_3^{2i}\frac{1}{z}dz=\ln\frac23+\frac{\pi}{2}i$
원점을 제외한 1사분면이 도메인이라고....(오른쪽 그림)
https://i.imgur.com/U0oir8X.png
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 $D$
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
$\oint_C \frac{f(z)}{z-z_0}dz=2\pi i f(z_0)$
$z_0$ is a given/known point in the complex plane
$f(z)$ evaluated at $z=z_0$
Google:cauchy+integral+formula
https://mathworld.wolfram.com/CauchyIntegralFormula.html
Evaluate $\oint\frac{z^2-4z+4}{z+i}dz$ where $c$ is $|z|=2$
It is analytic at all points in the domain
* $z_0=-i$ is an interior point of $C$
* By Cauchy's integral formula,
$\oint_C\frac{z^2-4z+4}{z+i}dz=2\pi i f(-i)$
Evaluate $\oint_C\frac{z}{z^2+a}dz$ where $C$ is the circle $|z-2i|=4$
https://i.imgur.com/ZuY4bZx.png
https://i.imgur.com/YBnEqsd.png
Cauchy's Integral Formula (AKA for derivatives or general form)
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
$\oint_C\frac{f(z)}{(z-z_0)^{n+1}}dz=\frac{2\pi i}{n!}f^{(n)}(z_0)$
Evaluate $\oint_C\frac{z+1}{z^4+4z^3}dz$ where $C$ is $|z|=1$
https://i.imgur.com/rnuzVVO.png
https://i.imgur.com/5c3yeZj.png
https://i.imgur.com/UKDBoda.png
https://i.imgur.com/OTz3aIB.png
'''''// 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 $z_0=i,\;n=1,\;f(z)=\frac{z^3+3}{z}$
$I_2=\oint_{c_2}\frac{\frac{z^3+3}{2}}{(z-i)^2}dz=\frac{2\pi i}{1!}f'(i)$
https://i.imgur.com/jj1a7P5.png
$\oint_c\frac{z^3+3}{z(z-i)^2}dz=-I_1+I_2$
https://i.imgur.com/xIu8Pt9.png
// [[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.
// [[유수,residue]] [[유수정리,residue_theorem]]
https://m.blog.naver.com/mindo1103/221977412976
// [[기하급수,geometric_series]]
$\sum_{k=1}^{\infty}az^{k-1}=a+az+az^2+\cdots+az^{n-1}+\cdots$
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
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 $\sum_{k=1}^{\infty} \frac{(1+2i)^k}{5^k}=............$
https://i.imgur.com/1AH9Xyh.png
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.
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]]
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
$\sum_{k=0}^{\infty} a_k (z-z_0)^k = a_0 + a_1(z-z_0) + a_2(z-z_0)^2 + \cdots$
$a_k$ : complex constants
$(z-z_0)$ : power series in $z-z_0$ meaning it is centered at $z_0$
https://i.imgur.com/0DJCr6T.png
Circle of Convergence // 이건 캡쳐할 수 밖에
https://i.imgur.com/3n224JP.png
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
Similar evaluation for root test $\lim_{n\to\infty}\sqrt[n]{|a_n|}$
https://i.imgur.com/ZLbliMU.png
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
https://i.imgur.com/oe16ylX.png
* 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
$f(z)=\sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}z^k$
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 $D$
https://i.imgur.com/vh3KhnW.png
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 $a_k=b_k$
https://i.imgur.com/wF2C2RW.png
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]]
Expand $f(z)=\frac1{1-z}$ in a Taylor series with center $z_0=2i$
Using $f(z)=\sum_{k=0}^{\infty}\frac{f^{(k)}(z_0)}{k!}(z-z_0)^k$
$f''(z)=\frac{2!}{(1-z)^3}$
$f'''(z)=\frac{3!}{(1-z)^4}$ (패턴을 파악)
$f^{(n)}(z)=\frac{n!}{(1-z)^{n+1}}$ and
$f^{(n)}(2i)=\frac{n!}{(1-2i)^{n+1}}$
$\frac{1}{1-z}=\sum_{k=0}^{\infty}\frac{1}{(1-2i)^{k+1}}(z-2i)^k$
Since the [[거리,distance|distance]] from center $z_0=2i$ to nearest singularity $z=1$ is $\sqrt{5},$
we conclude that circle of convergence is $|z-2i|=\sqrt{5}$
https://i.imgur.com/1TjToKJ.png
''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
https://i.imgur.com/GLeWwRa.png
https://i.imgur.com/460agRM.png
If $f$ which is analytic in annular(고리 모양의) domain defined by
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
https://i.imgur.com/dzdyYBw.png
Expand $f(z)=\frac{8z+1}{z(1-z)}$ in a Laurent series valid for $0<|z|<1$
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)
https://i.imgur.com/kYw5QB1.png
Expand $f(z)=e^{3/z}$ in a Laurent series valid for $0<|z|$
$e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$
https://i.imgur.com/z9iFnCZ.png
Expand $f(z)=\frac{1}{(z-1)^2(z-3)}$ in a Laurent series valid for $0<|z-1|<2$
We only want powers of $z-1$ ( given $0<|z-1|<2$ )
We need to express $z-3$ in terms of $z-1$
$f(z)=\frac1{(z-1)^2(z-3)}$
$=\frac1{(z-1)^2}\cdot\frac1{-2+(z-1)}$
$=\frac{-1}{2(z-1)^2}\cdot\frac1{1-\frac{z-1}{2}}$
https://i.imgur.com/KbvRbYe.png
Recall $\frac1{1-z}=1+z+z^2+z^3+\cdots$
We replace $z$ with $\frac{(z-1)}{2}$
$f(z)=\frac{-1}{2(z-1)^2}\left[ 1+\frac{z-1}{2} +\frac{(z-1)^2}{2^2} +\frac{(z-1)^3}{2^3} +\cdots\right]$
$=\underbrace{-\frac1{2(z-1)^2}-\frac1{4(z-1)}}_{\text{principal}}\underbrace{-\frac18-\frac1{16}(z-1) \cdots}_{\textrm{analytic}}$
Can we do it for $0<|z-3|<2$ ?
https://i.imgur.com/fnYZd5i.png
// 로랑 급수에는 principal 과 analytic 파트가 있었다.
Recall the principal part of Laurent series
$\sum_{k=1}^{\infty}a_{-k}(z-z_0)^{-k}=\sum_{k=1}^{\infty}\frac{a_{-k}}{(z-z_0)^k}$
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''
$\frac{\sin z}{z}=1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots,\;z=0$ 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.
for $z=z_0,\,\sum_{k=1}^{\infty}\frac{a_k}{(z-z_0)^k}$ contain infinitely many nonzero terms
$\underbrace{\cdots+\frac{a_{-2}}{(z-z_0)^2}+\frac{a_{-1}}{z-z_0}}_{\textrm{nonzero terms\\ in principal part}}+a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots$
Example: $f(z)=e^{3/z}=1+\frac{3}{z}+\frac{3^2}{z!z^2}+\frac{3^3}{3!z^3}+\cdots$
https://i.imgur.com/JeMjncE.png
$\frac{a_{-n}}{(z-z_0)^n}+\frac{a_{-(n-1)}}{(z-z_0)^{n-1}}+\cdots+\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+\cdots$
$a+z=z_0,$ it has pole of order $n$
If $n=1,$ it is a simple pole
analytic function $f(z)$ has a zero of order $n$ if
$f(z_0)=0,\;f'(z_0)=0,\; f''(z_0)=0,\; \cdots f^{(n-1)}(z_0)=0$
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 $z=z_0$
https://i.imgur.com/eYiRBkA.png
Given rational function $\frac{2z+5}{(z-1)(z+5)(z-2)^4}$
https://i.imgur.com/LnJugsr.png
// Google:rational+function+pole+zero
// Q&A: Google:isolated+singularity 설명만으론 이해가 잘 안 감
// tmp https://m.blog.naver.com/heejoo_kang/220805648869
$f(z)=\sum_{k=-\infty}^{\infty}a_k(z-z_0)^k$ // ''z,,0,, is isolated singularity''
$=\cdots+\frac{a_{-2}}{(z-z_0)^2}+\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+\cdots$
coefficient $a_{-1}$ of $\frac{1}{z-z_0}$ in the Laurent series is called the '''residue''' of $f$ at isolated singularity $z_0$
https://i.imgur.com/3dNCRtE.png
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
$\textrm{Res}(f(z),z_0)=\frac1{(n-1)!}\lim_{z\to z_0}\left[ \frac{d^{n-1}}{dz^{n-1}}\left[ (z-z_0)^n f(z) \right]\right]$
https://i.imgur.com/g4PGtO8.png
given $f(z)=\frac1{(z-1)^2(z-3)}$
$\operatorname{Res}(f(z),3)=\lim_{z\to 3}(z-3)f(z)$
$=\lim_{z\to 3}\frac1{(z-1)^2}=\frac14$
$\operatorname{Res}(f(z),1)=\frac1{1!}\lim_{z\to 1}\frac{d}{dz}(z-1)^2f(z)$
$=\lim_{z\to 1}\frac{d}{dz}\frac{1}{z-3}$
$=\lim_{z\to 1}\frac{-1}{(z-3)^2}=-\frac14$
https://i.imgur.com/vBV7reA.png
$D:$ simply connected domain
$C:$ simple closed contour lying entirely within $D$
function $f:$ analytic on and within $C$
except for finite number of singular points $z_1,z_2,z_3,\cdots,z_n$ within $C$
$\oint_C f(z)dz=2\pi i \sum_{k=1}^{n}\operatorname{Res}(f(z),z_k)$
// [[유수정리,residue_theorem]]
https://i.imgur.com/hYs5boo.png
Evaluate $\oint_C \frac1{(z-1)^2(z-3)}dz$ and $C$ is circle $|z|=2$
Soln. Since pole $z=1$ lies within circle $|z|=2$
$\oint_C\frac1{(z-1)^2(z-3)}dz=2\pi i \textrm{Res}(f(z),1)$
https://i.imgur.com/DsnLRz3.png
Evaluate $\oint_C\frac{2z+6}{z^2+4}dz$ where contour $C$ is $|z-i|=2$
Soln. $\oint_C\frac{2z+6}{z^2+4}dz=\oint_C\frac{2z+6}{(z-2i)(z+2i)}dz$
''simple poles at $z=-2i$ and $2i$''
''but only $z=2i$ lies within the contour''
Therefore $\oint_C\frac{2z+6}{z^2+4}dz=2\pi i\textrm{Res}(f(z),2i)$
https://i.imgur.com/iO04B63.png
$\textrm{Res}(f(z),2i)=\lim_{z\to 2i}(z-2i)\frac{2z+6}{(z-2i)(z+2i)}$
$=\frac{6+4i}{4i}=\frac{3+2i}{2i}$
Therefore $\oint_C \frac{2z+6}{z^2+4}dz=2\pi i\left(\frac{3+2i}{2i}\right)=\pi(3+2i)$
What if $f$ is not a rational function?
given $f(z)=\frac{g(z)}{h(z)}, g\textrm{ and }h$ analytic at $z=z_0$
$h(z)$ has zero of order 1 at $z_0$
$\Rightarrow f(z)$ has simple pole at $z=z_0$
then $\textrm{Res}(f(z),z_0)=\frac{g(z_0)}{h'(z_0)}$
// 참고로 시험에는 rational fn만 나옴
https://i.imgur.com/I1RIMbB.png
== 2020-10-27 (Linear Algebra) ==
It is abstract. Geometrical representations have limitations.
원점 (0, 0, 0)을 지나는 [[평면,plane]]은 벡터공간 $\mathbb{R}^3$ 의 '''부분공간'''이다.
* non empty [[부분집합,subset|subset]] that satisfies requirement for [[벡터공간,vector_space|vector space]]
* [[선형결합,linear_combination|linear combination]] stay in '''subspace'''
"whatever happens in the '''subspace''' stays in the '''subspace'''."
Given matrix A is part of a system of 3 equations and 2 unknowns
$\begin{bmatrix}1&0\\5&4\\2&4\end{bmatrix}\begin{bmatrix}u\\v\end{bmatrix}=\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}$
only a very thin subset of possible b's with(''will인가?'') satisfy the equation
$u\begin{bmatrix}1\\5\\2\end{bmatrix}+v\begin{bmatrix}0\\4\\4\end{bmatrix}=\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}$
* all possible combination of columns
* referred to as '''column space'''(a subspace) C(A) of $\mathbb{R}^n$
https://i.imgur.com/3732MTN.png
Nullspace contains all vectors x that gives Ax=0
$\begin{bmatrix}1&0&1\\5&4&9\\2&4&6\end{bmatrix} \begin{bmatrix}c\\c\\-c\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}$ ....''여기서 두번째 행렬''
'''nullspace''' is a line
where $c$ is any scalar or number.
(''Suppose this describes a physical system that we are trying to find values for some components.'')
If $AX_p=b$ and $AX_n=0,$
$X_p$ : particular solution
$(X_p+X_n)$ : our complete solution
https://i.imgur.com/Jzg0HS5.png
$\begin{bmatrix}1&3&3&2\\2&6&9&7\\-1&-3&3&4\end{bmatrix}\begin{bmatrix}u\\v\\w\\y\end{bmatrix}=\begin{bmatrix}1\\5\\5\end{bmatrix}$
// ''여기서 u,v,w,y를 푸는 것이 목적''
We will first re-arrange the above equation into what we called an echelon matrix U.
https://i.imgur.com/HrtHOSt.png
https://i.imgur.com/I5zgSBq.png
https://i.imgur.com/rLHeB9d.png
https://i.imgur.com/hpAjHxQ.png
Linear Independence, Basis and Dimensions
The numbers m(rows) and n(columns) do not give the true size of the linear system.
- it can have zero rows and columns
- combinations of rows of columns
- gives the true size of the [[선형계,linear_system|linear system]]
- is the number of [[추축,pivot|pivot]]s in the elimination process
- genuinely independent rows in matrix A
Given $c_1v_1+c_2v_2+\cdots+c_kv_k=0$
① If the equation can only be satisfied by having $c_1=0,c_2=0,\cdots,c_k=0$ then we say $v_1,v_2,\cdots,v_k$ are '''linearly independent'''.
② If any of the coefficients is a non-zero, then we say $v_1,v_2,\cdots,v_k$ are '''linearly dependent'''.
// [[선형독립,linear_independence]]
$A=\begin{bmatrix}1&3&3&2\\2&6&9&5\\-1&-3&3&0\end{bmatrix}$
If we perform $c_2=c_2-3c_1$
$A=\begin{bmatrix}1&0&3&2\\2&0&9&5\\-1&0&3&0\end{bmatrix}$ // columns are linearly independent
여기서 x: nullspace of A, N(A) must be {zero vectors} if the columns of A are independent.
$A=\begin{bmatrix}1&3&2\\2&9&5\\-1&3&0\end{bmatrix},$
$\begin{bmatrix}1&3&2\\2&9&5\\-1&3&0\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
only $x_1=x_2=x_3=0$ satisfy equation
https://i.imgur.com/q0yZY9a.png
* When we say vectors $w_1,w_2,\cdots,w_l$ '''span''' the space $V,$ it means vector space $V$ consists of all linear combinations of $w_1,w_2,\cdots,w_l.$
$\operatorname{Space}V=c_1w_1+c_2w_2+\cdots+c_lw_l$
for some coefficient $c_i$
* Column space of A is space spanned by its column.
* Row space of A is space spanned by its rows.
// [[생성,span]] [[열공간,column_space]] [[행공간,row_space]]
* A '''basis''' of space $V$ is a set of vectors where
- they are linearly independent ''(not too many vectors)''
- they span the space $V$ ''(not too few vectors)''
* Every vector in the space $V$ is a unique combination of basis vectors
* If columns of matrix are independent, they are a basis for the column space (and they span it as well)
Dimension of a vector space
* A space has infinitely many different bases(←plural for basis)
* The number of basis vectors is a property of the space
(fixed for a given space V)
* number of vectors in the bases = '''dimension''' of space
- a maximal independent set
- cannot be made larger without losing independence
- cannot be made smaller and still span the space
Four Fundamental Subspaces
Given matrix A is m×n matrix
① Column space of A denoted by C(A) ~ dimension is the rank r
② Null space of A denoted by N(A) ~ dimension is n-r
③ Row space of A is the column space of A^^T^^ ~ dimension is r
④ Left nullspace of A is the nullspace of A^^T^^ ~ dimension is m-r
It contains all vectors y such that A^^T^^y=0 denoted by N(A^^T^^)
2번에서 Ax=0에서 x nullspace가 맞는지 CHK
4번에서 y가 left nullspace이다.
* Pivot columns of A are a basis for its column space
* If sets of columns in A are independent, it corresponding columns in echelon matrix V are also independent.
$V=\begin{bmatrix}d_1&*&*&*&*&0\\0&0&0&d_2&*&0\\0&0&0&0&0&d_3\\0&0&0&0&0&0\end{bmatrix}$
* Assumed columns 1,4,6 are independent columns
* Columns 1,4,6 are basis for C(A)
* Row rank = column rank (important theorem in linear algebra)
* If rows of square matrix are independent, the columns are also independent
https://i.imgur.com/sfkIdsA.png
Is dimension of subspace made by 2 vectors (1 2 1)^^T^^ and (1 0 0)^^T^^ two? Even if the number of variables is three and the plane is on vector space dimension 3?
https://i.imgur.com/gzFJKG8.png
Google:circuit+network+matrix
https://i.imgur.com/pTwtFRf.png
https://i.imgur.com/fdxjKeB.png
$Ax=0,\;x=\begin{bmatrix}c\\c\\c\\c\end{bmatrix}$
Is b in the column space of A (what values of b will satisfy Ax=b)
https://i.imgur.com/I2aGJrQ.png
https://i.imgur.com/dewmp9e.png
https://i.imgur.com/z3t5RPQ.png
https://i.imgur.com/aMdwzyZ.png
https://i.imgur.com/UQhTlND.png
https://i.imgur.com/IxHTfwb.png
// [[선형변환,linear_transformation]]
https://i.imgur.com/pLjEokj.png
Transformations represented by matrices
https://i.imgur.com/QD74p2V.png
The same transformation can be described by another set of basis vectors
https://i.imgur.com/O5QPzv5.png
// basis가 [[다항식,polynomial]]... 행렬로 하는 미적분? $A_{\textrm{diff}}$
https://i.imgur.com/xB5KWXK.png
And we can do the same for integration - integration matrix $A_{\textrm{int}}$
https://i.imgur.com/Bx84XIO.png
Transformation of the plane
https://i.imgur.com/dqlAxfs.png
Projection (onto x axis) // [[사영,projection]]
https://i.imgur.com/RmTTVQQ.png
https://i.imgur.com/bL6NqSd.png
(θ-line은 x축에서 반시계방향으로 회전한 선)
https://i.imgur.com/UF8RNRH.png
Reflection about mirror θ-line
https://i.imgur.com/QaclZdd.png
https://i.imgur.com/9J5xzRD.png
Orthogonal vectors and subspaces
Given vector $\vec{x}=( x_1,x_2,\cdots,x_n )$
Length squared is $||\vec{x}||^2=x_1^2+x_2^2+\cdots+x_n^2$
Given $\vec{x}=\begin{bmatrix}1\\2\\-3\end{bmatrix}$
length square of $\vec{x}$ is $\vec{x}{}^T\vec{x}=[1\;2\;3]\begin{bmatrix}1\\2\\-3\end{bmatrix}=14$
// x^^T^^는 x의 전치(see [[전치행렬,transpose_matrix]])
https://i.imgur.com/Rhe5H45.png
// $|x|$ for scalars, $||x||$ for vectors 언급함. [[절대값,absolute_value]] vs [[노름,norm]] 관계가 저거?? CHK
https://i.imgur.com/bcQxB3T.png
$\vec{v_1}=(\cos\theta,\sin\theta)$
$\vec{v_2}=(-\sin\theta,\cos\theta)$
https://i.imgur.com/9bQnGf9.png
orthogonal unit vectors or orthonormal vectors in ℝ^^2^^
https://i.imgur.com/2BRuEtl.png
https://i.imgur.com/iOTG1R1.png
https://i.imgur.com/5lQiQm3.png
The space of all vectors orthogonal to subspace V of ℝ^^n^^
Notation: V^^⊥^^ or "V perp" (perp는 perpendicular)
[[영공간,null_space|Nullspace]] is orthogonal complement of [[행공간,row_space|row space]]
// Google:orthogonal.complement
https://i.imgur.com/NKdo2JA.png
https://i.imgur.com/dTlZ5jU.png
Inner products and cosines
// [[내적,inner_product]] and [[코사인,cosine]]
// [[코사인법칙,law_of_cosines]]
https://i.imgur.com/YXPWXRe.png
// [[사영,projection]] [[직선,line]]
Projection of vector $b$ onto line in the direction of vector $a$
$\vec{p}=\hat{x}\vec{a}=\frac{a^{\top}b}{a^{\top}a}a$
and all vectors a and b satisfy Schwarz(sic) inequality which is
$|a^{\top}b| \le ||a|| \, ||b||$
https://i.imgur.com/WEdTy5M.png
// [[코시-슈바르츠_부등식,Cauchy-Schwartz_inequality]] [[방향,direction]]
https://i.imgur.com/ZQwqPUD.png
https://i.imgur.com/q72ZJyR.png
https://i.imgur.com/W1sHaKy.png
https://i.imgur.com/A6NTTnv.png
Projections and Least Squares
https://i.imgur.com/LO93ClW.png
Least Squares Problems with Several Variables
https://i.imgur.com/IqHKGxW.png
https://i.imgur.com/zUfBukb.png
Cross Product Matrix A^^T^^A
https://i.imgur.com/L6KnioQ.png
Least Square Fitting of Data
https://i.imgur.com/8c65HYX.png
https://i.imgur.com/tgzmcA2.png
https://i.imgur.com/AcNnhBf.png
https://i.imgur.com/acIOHEv.png
Orthogonal Bases and Gram-Schmidt
Recall orthonormal vectors are orthogonal unit vectors
* If Q(square or rectangular) has orthonormal columns then Q^^T^^Q=I
* If Q is a [[정사각행렬,square_matrix|square matrix]], it is called "orthogonal matrix"
* We will see that orthonormal vectors are very convenient to work with.
https://i.imgur.com/Wm2uK5s.png
https://i.imgur.com/8LD1s22.png
Rectangular Matrices with Orthonormal Columns
If Q has orthonormal columns, the least squares problem becomes easy
https://i.imgur.com/qCfSv05.png
// [[그람-슈미트_과정,Gram-Schmidt_process]]
https://i.imgur.com/0Y6eQU3.png
https://i.imgur.com/yRpUrrJ.png
https://i.imgur.com/xHZSk7s.png
Eigenvalues and Eigenvectors
// [[고유값,eigenvalue]] [[고유벡터,eigenvector]]
https://i.imgur.com/Yo3UfMU.png
https://i.imgur.com/FufFyMZ.png
https://i.imgur.com/2ryE1Bn.png
https://i.imgur.com/GLIh9HY.png
https://i.imgur.com/Eo0YZTa.png
https://i.imgur.com/I5pyfep.png
// [[대각행렬,diagonal_matrix]]
https://i.imgur.com/PyHEsDl.png
https://i.imgur.com/dDM6nYX.png
https://i.imgur.com/Jm6fsXk.png
Diagonalization of a Matrix
// [[대각화,diagonalization]] curr goto [[대각행렬,diagonal_matrix]]
https://i.imgur.com/AUHDb9n.png
https://i.imgur.com/RsrHziZ.png
https://i.imgur.com/mIaTuR0.png
https://i.imgur.com/G7dQwhj.png
https://i.imgur.com/HwJwAhd.png
https://i.imgur.com/3sQwMUC.png
// [[푸리에_급수,Fourier_series]]
https://i.imgur.com/WitHE6B.png
https://i.imgur.com/F8L8xvX.png
Orthogonality between two functions
https://i.imgur.com/BkQz0gB.png
https://i.imgur.com/2Btx2Kw.png
Fast Fourier Transform (FFT)
// [[푸리에_변환,Fourier_transform]]
https://i.imgur.com/sD29I1b.png
https://i.imgur.com/ytTetQU.png
== 2020-12-01 Review 1 ==
https://i.imgur.com/Z1vHrN0.png
https://i.imgur.com/uJOdWjW.png
https://i.imgur.com/RP4oTYi.png
https://i.imgur.com/QTx0Ys1.png
https://i.imgur.com/txFRCVy.png
https://i.imgur.com/lkeCOTT.png
https://i.imgur.com/a4gTXIp.png
https://i.imgur.com/5mhj3Ve.png
https://i.imgur.com/8yIPQDk.png
Taught by: Prof. Beelee Chua
= KECE206 전자기학 =
@@ -2027,5 +94,4 @@
[[전기장,electric_field]]
[[자기장,magnetic_field]]
[[벡터장,vector_field]]
![[ISBN-0133356817]](http://images.amazon.com/images/P/0133356817.01.MZZZZZZZ.gif "Amazon:ISBN-0133356817")
Fundamentals of Applied Electromagnetics 7e