91 lines
2.7 KiB
Markdown
91 lines
2.7 KiB
Markdown
# Math416 Lecture 19
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## Continue on the Laurent series
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### Laurent series
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If $f$ is holomorphic in $A(z_0;R_1,R_2)$ then $f=\sum_{n=-\infty}^{\infty} a_n (z-z_0)^n$ where the Laurent series converges on the annulus $A(z_0;R_1,R_2)$
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$$
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\int_{C(z_0,r)} f(z)(z-z_0)^{-k-1} dz = \sum_{n=-\infty}^{\infty} a_n \int_{C(z_0,r)} (z-z_0)^{n-k-1} dz=a_k 2\pi i
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$$
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> $C(z_0,r)$ is a circle centered at $z_0$ with radius $r$
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### Isolated singularities
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A punctured disk at $z_0$ is $A(z_0;0,R)=\{z:0<|z-z_0|<R\}$
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Say a function $f$ has an isolated singularity at $z_0$ if it is holomorphic in a punctured disk $A(z_0;0,R)$
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$f$ has a Laurent series in $A(z_0;0,R)$
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$$
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f(z) = \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n
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$$
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that converges in $A(z_0;0,R)$
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#### Principal part of a Laurent series
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The principal part of a Laurent series is the sum of the terms with negative powers of $(z-z_0)$
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$$
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\sum_{n=-\infty}^{-1} a_n (z-z_0)^n
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$$
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Say the isolated singularity is
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- removable if $a_n=0$ for all $n<0$
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- If $f(z)$ has a removable singularity at $z_0$, then extend $f$ to $\mathbb{D}_{z_0,R}$ by defining $f(z_0)=a_0$. This extended $f$ is holomorphic on $\mathbb{D}_{z_0,R}$ and $f(z)=\sum_{n=0}^{\infty} a_n (z-z_0)^n$ for $z\in \mathbb{D}_{z_0,R}$
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- pole if $a_{-k}\neq 0$ and $a_n=0$ for all $n<-k$
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- A pole with order $1$ is a simple pole
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- essential if the cases above are not true
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Example:
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1. $f(z)=\frac{\sin z}{z}$ has a removable singularity at $z=0$.
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the power series is
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$$
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\sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots
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$$
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So the Laurent series is
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$$
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\frac{\sin z}{z} = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots
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$$
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The singularity is removable by defining $f(0)=1$
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2. $f(z)=\frac{z^2-1}{(z-1)(z-3)}=\frac{(z-1)(z+1)}{(z-1)(z-3)}$
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There are two poles at $z=1$ and $z=3$
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the singularity at $z=1$ is removable by defining $f(1)=1$
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the singularity at $z=3$ is a simple pole with order 1 $f(z)=\frac{z+1}{z-3}=\frac{(z-3)+4}{z-3}=4(z-3)^{-1}+1$
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3. $f(z)=\frac{(z+1)^2(z+2)^3}{(z-1)^2(z-5)^6(z-8)}$
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there are three poles at $z=1,5,8$, the order of the poles are 2, 6, 1 respectively.
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If $f$ has a pole of order $m$ at $z_0$,
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$$
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f(z) = \sum_{n=-m}^{\infty} a_n (z-z_0)^n
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$$
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then $(z-z_0)^m f(z)$ has a removable singularity at $z_0$. Value of holomorphic extension of $(z-z_0)^m f(z)$ at $z_0$ is $a_{-m}$.
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- $f$ is given by a power series in $A(z_0;0,R)$
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- $f=(z-z_0)^{-m} g(z)$ where $g$ is holomorphic and $g(z_0)\neq 0$, $\frac{1}{f}=(z-z_0)^m \frac{1}{g(z)}$ has a pole of order $m$ at $z_0$. So $f$ has a pole of order $m$ at $z_0$ if and only if $\frac{1}{f}$ has a zero of order $m$ at $z_0$
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$e^{1/z}=1+\frac{1}{z}+\frac{1}{2!z^2}+\frac{1}{3!z^3}+\cdots$ has an essential singularity at $z=0$ since it has infinitely many terms with negative powers of $z$.
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