如何求复数函数洛朗展？
用部分分式:
f(z) = 1/((z^2+1)(z+3))
= 1/10·(1/(z+3)-(z-3)/(z^2+1))
= 1/10·(1/(z+3)-1/2·(1+3i)/(z-i)-1/2·(1-3i)/(z+i))
= 1/10·1/(z+3)-1/20·((1+3i)/(z-i)+(1-3i)/(z+i))
= 1/30·1/(1+z/3)-1/(20z)·((1+3i)/(1-i/z)-(1-3i)/(1+i/z))
= 1/30·∑{0 ≤ n} (-z/3)^n - 1/(20z)·∑{0 ≤ n} ((1+3i)·(i/z)^n+(1-3i)·(-i/z)^n) ...全部
用部分分式:
f(z) = 1/((z^2+1)(z+3))
= 1/10·(1/(z+3)-(z-3)/(z^2+1))
= 1/10·(1/(z+3)-1/2·(1+3i)/(z-i)-1/2·(1-3i)/(z+i))
= 1/10·1/(z+3)-1/20·((1+3i)/(z-i)+(1-3i)/(z+i))
= 1/30·1/(1+z/3)-1/(20z)·((1+3i)/(1-i/z)-(1-3i)/(1+i/z))
= 1/30·∑{0 ≤ n} (-z/3)^n - 1/(20z)·∑{0 ≤ n} ((1+3i)·(i/z)^n+(1-3i)·(-i/z)^n) (1 < |z| < 3)
= 1/30·∑{0 ≤ n} (-1)^n/3^n·z^n - 1/(20z)·∑{n ≤ 0} ((1+3i)·(i/z)^(-n)+(1-3i)·(-i/z)^(-n))
= 1/30·∑{0 ≤ n} (-1)^n/3^n·z^n - 1/(20z)·∑{n ≤ 0} ((1+3i)·i^(-n)+(1-3i)·(-i)^(-n))·z^n
= 1/30·∑{0 ≤ n} (-1)^n/3^n·z^n - 1/20·∑{n ≤ 0} ((1+3i)·(-i)^n+(1-3i)·i^n)·z^(n-1)
= 1/30·∑{0 ≤ n} (-1)^n/3^n·z^n - 1/20·∑{n ≤ -1} ((1+3i)·(-i)^(n+1)+(1-3i)·i^(n+1))·z^n
= 1/30·∑{0 ≤ n} (-1)^n/3^n·z^n - 1/20·∑{n ≤ -1} ((3-i)·(-i)^n+(3+i)·i^n)·z^n。
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