0.  引言---回忆

(1)  Cauchy 积分公式 (第三章) \beex \bea f\mbox{ 在 }D\mbox{ 内解析}, \mbox{ 在 }\bar D=D+\p D\mbox{ 上连续}&\ra \int_C \cfrac{f(z)}{z-a}\rd z=2\pi if(a),\quad a\in D\\ &\ra \int_C \cfrac{f(z)}{(z-a)^{n+1}}\rd z=\cfrac{2\pi i}{n!}f^{(n)}(a),\quad a\in D \eea \eeex<script type="math/tex; mode=display">\beex \bea f\mbox{ 在 }D\mbox{ 内解析}, \mbox{ 在 }\bar D=D+\p D\mbox{ 上连续}&\ra \int_C \cfrac{f(z)}{z-a}\rd z=2\pi if(a),\quad a\in D\\ &\ra \int_C \cfrac{f(z)}{(z-a)^{n+1}}\rd z=\cfrac{2\pi i}{n!}f^{(n)}(a),\quad a\in D \eea \eeex</script>

(2)  Laurent 定理 \bex f\mbox{ 以 }a\mbox{ 为孤立奇点}\ra f(z)=\sum_{n=-\infty}^{+\infty} c_n(z-a)^n, \eex<script type="math/tex; mode=display">\bex f\mbox{ 以 }a\mbox{ 为孤立奇点}\ra f(z)=\sum_{n=-\infty}^{+\infty} c_n(z-a)^n, \eex</script> 其中 \bex c_n=\cfrac{1}{2\pi i}\int_{|z-a|=\rho}\cfrac{f(z)}{(z-a)^{n+1}}\rd z, \eex<script type="math/tex; mode=display">\bex c_n=\cfrac{1}{2\pi i}\int_{|z-a|=\rho}\cfrac{f(z)}{(z-a)^{n+1}}\rd z, \eex</script> 特别地, 当 n=1<script type="math/tex">n=1</script> 时, \bex c_{-1}=\cfrac{1}{2\pi i}\int_{|z-a|=\rho}f(z)\rd z.  \eex<script type="math/tex; mode=display">\bex c_{-1}=\cfrac{1}{2\pi i}\int_{|z-a|=\rho}f(z)\rd z. \eex</script>

(3)  它们都可用来计算周线积分, 比如 \dps{I=\int_{|z|=1}\cfrac{\sin z}{z^2}\rd z}<script type="math/tex">\dps{I=\int_{|z|=1}\cfrac{\sin z}{z^2}\rd z}</script>:

a.  \bex I=\cfrac{2\pi i}{1!}(\sin z)‘|_{z=0}=2\pi i.  \eex<script type="math/tex; mode=display">\bex I=\cfrac{2\pi i}{1!}(\sin z)‘|_{z=0}=2\pi i. \eex</script>

b.  \beex \bea &\quad\cfrac{\sin z}{z^2}=\cfrac{1}{z^2}\sex{z-\cfrac{z^3}{3!}+\cdots} =\cfrac{1}{z}-\cfrac{z}{3!}+\cdots\\ &\ra I=2\pi i\cdot c_{-1}=2\pi i.  \eea \eeex<script type="math/tex; mode=display">\beex \bea &\quad\cfrac{\sin z}{z^2}=\cfrac{1}{z^2}\sex{z-\cfrac{z^3}{3!}+\cdots} =\cfrac{1}{z}-\cfrac{z}{3!}+\cdots\\ &\ra I=2\pi i\cdot c_{-1}=2\pi i. \eea \eeex</script> 但 Cauchy 积分定理只能计算复函数在周线内仅有一个极点的情形.

1.  留数

(1)  定义: 设 a<script type="math/tex">a</script> 为 f<script type="math/tex">f</script> 的孤立奇点, 则称积分 \bex \cfrac{1}{2\pi i}\int_{|z-a|=\rho}f(z)\rd z \eex<script type="math/tex; mode=display">\bex \cfrac{1}{2\pi i}\int_{|z-a|=\rho}f(z)\rd z \eex</script> 为 f<script type="math/tex">f</script> 在 a<script type="math/tex">a</script> 的留数, 记作 \underset{z=a}{\Res}f(z)<script type="math/tex">\underset{z=a}{\Res}f(z)</script>.

(2)  \underset{z=a}{\Res}f(z)=c_{-1}<script type="math/tex">\underset{z=a}{\Res}f(z)=c_{-1}</script>.

(3)  Cauchy 留数定理: \bex (\mbox{大范围积分}) \int_Cf(z)\rd z=2\pi i\sum_{k=1}^n \underset{z=a_k}{\Res}f(z). \eex<script type="math/tex; mode=display">\bex (\mbox{大范围积分}) \int_Cf(z)\rd z=2\pi i\sum_{k=1}^n \underset{z=a_k}{\Res}f(z). \eex</script>

2.  计算

(1)  设 a<script type="math/tex">a</script> 为 f<script type="math/tex">f</script> 的 n<script type="math/tex">n</script> 阶极点, 即 \bex f(z)=\cfrac{\phi(z)}{(z-a)^n},\quad \phi(a)\neq 0, \eex<script type="math/tex; mode=display">\bex f(z)=\cfrac{\phi(z)}{(z-a)^n},\quad \phi(a)\neq 0, \eex</script> 则 \bex \underset{z=a}{\Res}f(z) =\cfrac{\phi^{(n-1)}(a)}{(n-1)!}. \eex<script type="math/tex; mode=display">\bex \underset{z=a}{\Res}f(z) =\cfrac{\phi^{(n-1)}(a)}{(n-1)!}. \eex</script>

(2)  设 a<script type="math/tex">a</script> 为 f<script type="math/tex">f</script> 的一阶极点, \phi(z)=(z-a)f(z)<script type="math/tex">\phi(z)=(z-a)f(z)</script>, 则 \bex \underset{z=a}{\Res}f(z)=\phi(a). \eex<script type="math/tex; mode=display">\bex \underset{z=a}{\Res}f(z)=\phi(a). \eex</script>

(3)  设 a<script type="math/tex">a</script> 为 f<script type="math/tex">f</script> 的二阶极点, \phi(z)=(z-a)^2f(z)<script type="math/tex">\phi(z)=(z-a)^2f(z)</script>, 则 \bex \underset{z=a}{\Res}f(z)=\phi‘(a). \eex<script type="math/tex; mode=display">\bex \underset{z=a}{\Res}f(z)=\phi‘(a). \eex</script>

(4)  设 a<script type="math/tex">a</script> 为 f=\cfrac{\phi}{\psi}<script type="math/tex">f=\cfrac{\phi}{\psi}</script> 的一阶极点 (\phi(a)\neq 0,\ \psi(a)=0,\ \psi‘(a)\neq 0<script type="math/tex">\phi(a)\neq 0,\ \psi(a)=0,\ \psi‘(a)\neq 0</script>), 则 \bex \underset{z=a}{\Res}f(z)=\cfrac{\phi(a)}{\psi‘(a)}. \eex<script type="math/tex; mode=display">\bex \underset{z=a}{\Res}f(z)=\cfrac{\phi(a)}{\psi‘(a)}. \eex</script>

(5)  例

a.  \dps{\int_{|z|=2}\cfrac{5z-2}{z(z-1)^2}\rd z}<script type="math/tex">\dps{\int_{|z|=2}\cfrac{5z-2}{z(z-1)^2}\rd z}</script>.

b.  \dps{\int_{|z|=n}\tan \pi z\rd z\ (n\in\bbZ^+)}<script type="math/tex">\dps{\int_{|z|=n}\tan \pi z\rd z\ (n\in\bbZ^+)}</script>.

c.  \dps{\int_{|z|=1}\cfrac{\cos z}{z^3}\rd z}<script type="math/tex">\dps{\int_{|z|=1}\cfrac{\cos z}{z^3}\rd z}</script>.

d.  \dps{\int_{|z|=1} e^\frac{1}{z^2}\rd z}<script type="math/tex">\dps{\int_{|z|=1} e^\frac{1}{z^2}\rd z}</script>.

e.  \dps{\underset{z=1}{\Res} e^{\frac{1}{z-1}},\quad \underset{z=1}{\Res}\cfrac{z^{2n}}{(z-1)^n},\quad \underset{z=1}{\Res}\cfrac{e^z}{z^2-1},\quad \underset{z=-1}{\Res}\cfrac{e^z}{z^2-1}}<script type="math/tex">\dps{\underset{z=1}{\Res} e^{\frac{1}{z-1}},\quad \underset{z=1}{\Res}\cfrac{z^{2n}}{(z-1)^n},\quad \underset{z=1}{\Res}\cfrac{e^z}{z^2-1},\quad \underset{z=-1}{\Res}\cfrac{e^z}{z^2-1}}</script>.

作业: P 262 T 1 (1)  (2)  (3) .