4)设曲线C的长度为L,函数∫(z)在C上满 足‖f(z)≤M,那么 f(2)d2|≤/f(2)lds≤ 证明设|△k=|2k-xk-1,且△Sk为这两点之间的 弧段的长度.显然|△2≤△Sk.所以 f()△≤∑(A≤∑|)△s k=1 令6=max{△s}0.得4) ­‚ C ÝǑ L, ¼ê f(z) 3 C þ÷ v |f(z)| 6 M, o     Z C f(z)dz     6 Z C |f(z)|ds 6 ML. y²  |△zk| = |zk − zk−1|, … △sk Ǒùü:ƒm lãÝ. w, |△zk| 6 △sk. ¤±      Xn k=1 f(ζk)△zk      6 Xn k=1 |f(ζk)△zk| 6 Xn k=1 |f(ζk)|△sk - δ = max 16k6n {△sk} → 0,      Z C f(z)dz     6 Z C |f(z)|ds. ÏǑ3 C þ |f(z)| 6 M,  Z C |f(z)|ds 6 M Z C ds = ML. 13/127