f(z)=(1-e^(-z))/z^4,问z=0的留数

f(z) = [1 - e^(- z)]/z^4
设g(z) = 1 - e^(- z)
g'(z) = e^(- z),g'(0) = 1
z = 0 是 g(z) 的一阶零点
z^4 是 f(z) 的三阶极点
∴Res[f(z),0] = 1/2!* lim(z→0) d^2/dz^2 [(z - 0)^3 * (1 - e^(- z))/z^4]
= (1/2)lim(z→0) (- z^2 - 2z + 2e^z - 2) * e^(- z)/z^3
= (1/2)(1/3)
= 1/6
或者直接展开：
e^z = 1 + z + z^2/2 + z^3/6 + z^4/24 + ...
e^(- z) = 1 - z + z^2/2 - z^3/6 + z^4/24 - ...
1 - e^(- z) = z - z^2/2 + z^3/6 - z^4/24 + ...
[1 - e^(- z)]/z^4 = (z - z^2/2 + z^3/6 - z^4/24 + ...)/z^4
= 1/z^3 - 1/(2z^2) + 1/(6z) - 1/24 + ...
其中 1/z 的系数为1/6,∴Res[f(z),0] = 1/6