f(z)=1⼀(z+1)在f(z)的孤立奇点的去心邻域展开为洛朗级数 求大神,怎么做?

2024-11-14 15:07:00
推荐回答(1个)
回答1:

在z=1处化:
令t=z-1, 则z=t+1
f(z)=1/t(t+1-3)
=1/t(t-2)
=0.5/(t-2)-0.5/t
=-0.25/(1-t/2)-0.5/t
=-0.25[1+t/2+t^2/4+t^3/8+...]-0.5/t
此即为在z=1处展开。

在z=3处化,也同理:
令t=z-3, 则z=t+3
f(z)=1/t(t+3-1)
=1/t(t+2)
=0.5/t-0.5/(t+2)
=0.5/t-0.25/(1+t/2)
=0.5/t-0.25[1-t/2+t^2/4-t^3/8+..]