Question

罗朗级数展开求详解

Answer 1

f(z) = -2/((z-4)(z-2)) = 1/(z-2)-1/(z-4).

当|z-1| > 1,

1/(z-2) = 1/((z-1)-1) = 1/(z-1)·1/(1-1/(z-1)) = 1/(z-1)·∑{0 ≤ n} 1/(z-1)^n = ∑{1 ≤ n} 1/(z-1)^n.

当|z-1| < 3,

1/(z-4) = -(1/3)/(1-(z-1)/3) = -(1/3)·∑{0 ≤ n} ((z-1)/3)^n = -(1/3)·∑{0 ≤ n} (z-1)^n/3^n.

于是对1 < |z-1| < 3, f(z) = 1/(z-2)-1/(z-4) = ∑{1 ≤ n} 1/(z-1)^n+∑{0 ≤ n} (z-1)^n/3^(n+1).

即为f(z)在1 < |z-1| < 3内的Laurent级数展开.