singular points & poles
2020-2νκΈ°, λνμμ βμμ©λ³΅μν¨μλ‘ β μμ μ λ£κ³ 곡λΆν λ°λ₯Ό μ 리ν κΈμ λλ€. μ§μ μ μΈμ λ νμμ λλ€ :)
Three Types of (isolated) Singular points
Definition. Principal part of Laurent Series
(i) removable singular point
all $b_n=0$
(ii) poles of order $m$
$b_1 \ne 0$ and $b_n=0$ for $n \ge m+1$ 1
(iii) essential singular point
$b_n \ne 0$ for infinitely many $n$ 2
μ΄μ€μμ μ£Όλͺ©ν λ μμ poleμ΄λ€.
\[\frac{b_1}{(z-z_0)} + \frac{b_2}{(z-z_0)^2} + \cdots \frac{b_m}{(z-z_0)^m}\]μ $z_0$λ₯Ό poleλΌκ³ νλ©°, $m$μ order of poleμ΄λΌκ³ νλ€.
removable singular point
Example. removable singular point
Sol.
$f(z)$μ λν λ‘λ κΈμλ₯Ό μ΄ν΄λ³΄λ©΄, $z=0$μ΄ removable singular pointμμ μ μ μλ€.
$f(z)$μ λ‘λ κΈμμ singular pointμΈ $z=0$μ λμ νλ©΄, $f(0)=\frac{\sin(0)}{(0)}=1$μ΄ λλ€.
μ΄λ, $f(z)$μ κ·ΉνμΈ $\underset{z \rightarrow 0}{\lim}{\frac{\sin z}{z}}$ μμ κ·Έκ°μ΄ $1$μ΄λ€.
μ΄ κ²°κ³Όλ₯Ό μΌλ°νν μ λ¦¬κ° λ°λ‘ μλμ μ 리μ΄λ€.
Theorem.
IF $z_0$ is a removable singular point, THEN $f(z_0)$ is defined & $\underset{z \rightarrow z_0}{\lim} f(z) = f(z_0)$.
κ·Έλ¦¬κ³ μμ μ 리λ₯Ό λ°λ λ°©ν₯μΌλ‘ κΈ°μ νλ μ 리λ μ‘΄μ¬νλ€.
Theorem.
IF $z_0$ is an isolated singular point of $f(z)$ and $f(z)$ is bounded for some neighborhood of $z_0$, THEN $z_0$ is a removable singular point.
$f(z)$μ κ·Όλ°©μμ boundedλμ΄ μλ€λ λ§μ 곧 $\underset{z \rightarrow z_0}{\lim} f(z)$κ° μ‘΄μ¬ν¨μ μλ―Ένλ€. κ·Έλ¦¬κ³ μ΄λ₯Ό ν΅ν΄ $z_0$μ removableμ νλ¨ν μλ μλ€λ λ§μ΄λ€!
poles of order $m$
\[f(z) = \sum^{\infty}_{n=0} {a_n (z-z_0)^n} + \left\{ \frac{b_1}{(z-z_0)} + \frac{b_2}{(z-z_0)^2} + \cdots \right\}\]Definition. simple pole
IF $m=1$, THEN it is called a simple pole.
Example.
Evaluate $\underset{z \rightarrow 0}{\lim} {\lvert f(z) \rvert}$.
Sol.
\(\begin{aligned}
f(z) &= \frac{1}{z^2 (z-1)} \\
&= - \frac{1}{z^2} \frac{1}{1-z} \\
&= - \frac{1}{z^2} (1+z+z^2+\cdots) \\
&= - \frac{1}{z^2} - \frac{1}{z} - 1 - z - \cdots \quad (\lvert z \rvert < 1)
\end{aligned}\)
λ°λΌμ $z=0$λ order $2$μ poleμ΄λ€.
μ΄λ, $\underset{z \rightarrow 0}{\lim} {\lvert f(z) \rvert}$λ μμ μ°¨μ ν μ μν΄
\[\underset{z \rightarrow 0}{\lim} {\lvert f(z) \rvert} = \infty\]μ¦, $\underset{z \rightarrow 0}{\lim} {\lvert f(z) \rvert}$κ° bounded λμ΄ μμ§ μμΌλ―λ‘ $z=0$λ removableμ΄ μλλ€!
Theorem.
IF $z_0$ is a pole of $f(z)$, THEN
proof.
μ¦λͺ
μ μμ£Ό κ°λ¨νλ€.
μ΄λ, $\left\{ \sum^{\infty}_{n=0} {a_n (z-z_0)^{n+m}} + \left( b_1 (z-z_0)^{m-1} + b_2 (z-z_0)^{m-2} + \cdots + b_m \right) \right\}$λ power seriesμ΄λ―λ‘ analytic νλ€.
$f(z)$μ λν΄ $z \rightarrow z_0$λ‘ κ·Ήνμ μ·¨νλ©΄,
\[\underset{z \rightarrow z_0}{\lim} {\lvert f(z) \rvert} = \infty \cdot b_m = \infty\]κ° λλ€.
essential singular point
Example.
What is $\underset{z \rightarrow 0}{\lim} e^{1/z}$ ?
Sol.
$e^z$μμ $1/z$λ₯Ό λμ
ν΄μ£Όλ©΄ μ½κ² λ‘λ κΈμλ₯Ό μ»μ μ μλ€.
μ΄μ $\underset{z \rightarrow 0}{\lim} e^{1/z}$λ₯Ό ꡬν΄λ³΄μ. κ³Όμ μ΄ μ‘°κΈ tricky νλ€.
Let $z=x+iy$, THEN $\exp z = e^x \cdot e^{iy}$
μ΄λ, $e^{iy}$λ $2{\pi}$-periodicμ΄λ€.
$e^{1/z}$μμ $0 < \lvert z \rvert < r$ ($r$ is small)μ Imageλ₯Ό μκ°ν΄λ³΄μ.
κ·Έλ¬λ©΄, $r$μ μ무리 μκ² μ‘μλ $2{\pi}$-stripμ ν¬ν¨νκ² λκ³ , λ°λΌμ $e^{1/z}$λ 0μ μ μΈν 볡μ νλ©΄ μ μ²΄κ° λλ€!!
λ°λΌμ $\underset{z \rightarrow 0}{\lim} e^{1/z}$μ $0$μ μ μΈν 볡μ νλ©΄μ μ΄λ κ°μ΄λ λ μ μλ€!
Theorem. (Great) Picardβs Theorem
IF $f(z)$ is analytic except at $z_0$, and $z_0$ is an isolated essential singular point, THEN in any $\epsilon$-neighborhood of $z_0$, $f(z)$ takes every value, with at most one exceptional value.
// μ¦λͺ μμ΄ μκ°λ§ νκ³ λμ΄κ°λ€!
zeros and poles
Consider
\[f(z) = \frac{g(z)}{h(z)}\]where $g(z)$ and $h(z)$ are analytic.
THEN, the singular points of $f(z)$ are zeros of $h(z)$.
(μμ£Ό λΉμ°ν μ§μ μ΄λ€!)
Definition.
Let $f$ be analytic at $z_0$. IF $f(z_0)=0$ and there is a positive integer $m$ such that
THEN $f$ is said to have a zero of order $m$ at $z_0$.
IF $m=1$, THEN it is called a simple zero.
μ¦, $f(z_0)=0$μ΄κ³ , $fβ(z_0) \ne 0$μΈ κ²½μ°λ₯Ό λ§νλ€.
Example. Talyor series
Taylor series where $f(z)$ is a zero of order $m$ at $z_0$.
ν μΌλ¬ κΈμμμ κ³μ $a_k$μ λν΄ μ΄ν΄λ³΄μ.
\[a_k = \frac{f^{(k)}(z_0)}{k!}\]μ΄λ, $f(z)$κ° $z_0$μμ order $m$μ΄λ―λ‘ $k = 0, 1, \cdots, m-1$μμ $a_k=0$μ΄λ€.
λ°λΌμ $f(z)$λ
\[\begin{aligned} f(z) &= \left( a_0 + a_1 (z-z_0) + a_2 (z-z_0)^2 + \cdots + a_{m-1} (z-z_0)^{m-1} \right) + a_m (z-z_0)^m + \cdots \\ &= 0 + a_m (z-z_0)^m + \cdots \end{aligned}\]μ΄λ, λ¨μ ν λ€μμ $(z-z_0)^m$λ₯Ό λ¬Άμ μ μλ€.
\[f(z) = (z-z_0)^m \left( a_m + a_{m+1} (z-z_0)^{m+1} + \cdots \right)\]$g(z) = a_m + a_{m+1} (z-z_0)^{m+1} + \cdots$λΌκ³ νμ. κ·Έλ¬λ©΄, $g(z_0) \ne 0$μ΄ λλ€.
μ¦, $f(z)$κ° μ νν $m$κ° μ€κ·Όμ κ°μ§λ€.
μ 리νλ©΄, ν¨μ $f(z)$κ° a zero of order $m$ at $z_0$μ΄λΌλ©΄, ν¨μ $f(z)$λ $m$κ° μ€κ·Όμ κ°μ§λ€.
Theorem.
IF $z_0$ is a zero of order $m$, THEN there is an analytic function $g(z)$ such that
Theorem.
Let $f(z)$ be analytic at $z=z_0$ and have a zero of $m$-th order at $z=z_0$. THEN $1/f(z)$ has a pole of $m$-th order at $z=z_0$. And so does $g(z)/f(z)$ provided $g(z)$ is analytic at $z=z_0$ and $g(z_0) \ne 0$.
proof.
By previous theorem, $f(z)=(z-z_0)^{m}g(z)$ where $g(z)$ is analytic, and $g(z_0) \ne 0$.
Therefore,
\[\frac{1}{f(z)} = \frac{1}{(z-z_0)^m} \cdot \frac{1}{g(z)}\]μ΄λ, $\frac{1}{g(z)}$κ° $z_0$μμ analyticμ΄λ―λ‘ Talyor seriesλ‘ νν κ°λ₯νλ€.
\[\begin{aligned} \frac{1}{f(z)} &= \frac{1}{(z-z_0)^m} \cdot \frac{1}{g(z)} \\ &= \frac{1}{(z-z_0)^m} \cdot \left(a_0 + a_1 (z-z_0) + \cdots \right) \\ &= \frac{a_0}{(z-z_0)^m} + \frac{a_1}{(z-z_0)^{m-1}} + \cdots \end{aligned}\]λ°λΌμ $1/f(z)$λ order $m$μ poleμ κ°μ§λ€. $\blacksquare$
Example. $\cot z$
μ΄λ, $\sin z$κ° $z=0, \pm \pi, \pm 2\pi, \cdots$ μμ 0μ΄ λλ©°, λͺ¨λ simple zero μ΄λ€.
Therefore, $\cot z$ has pole at $0, \pm \pi, \pm 2\pi, \cdots$ of order $1$.
Example.
Find poles of $f(z)$
Sol.
ν¨μμ zerosμ poleμ λν μ 리μ μν΄ $f(z)$μ poleμ $z^4+4 = 0$μ΄ λλ μ§μ μ΄ λλ€.
\[\begin{aligned} z^4+4 &= 0 \\ z^4 &= -4 \\ (re^{i\theta})^4 &= 4 e^{i(\pi + 2n\pi)} \\ r^4 \cdot e^{i(4\theta)} &= 4 \cdot e^{i(\pi + 2n\pi)} \end{aligned}\] \[r = \sqrt{2}, \quad \theta = \frac{\pi}{4} + \frac{1}{2}k\pi, \quad (k=0, 1, 2, 3) \\\] \[z = 1 + i, \; 1-i, \; -1+i, \; -1-i\]