4)设曲线C的长度为L,函数∫(2)在C上满 足|f(2)≤M,那么 f(2)d2|≤/f(2)lds≤ML 证明设|△xk=|2k-2k-1|,且△Sk为这两点之间的 弧段的长度.显然|△k≤△Sk.所以 f()△≤∑(A|≤∑)△s k=1 令6=max{△sk}→0,得 f()d2|≤/|f()ds 因为在C上|f(2)≤M,得到 f(2)ds≤M/d=ML3
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:m 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
例1计算积分 (2+mk/ (az+ig)d 其中C1,C2分别是沿y=x与y=x2从原点到点1+i 的曲线
~ 1 OÈ© Z C1 (x 2 + iy)dz, Z C2 (x 2 + iy)dz, Ù¥ C1, C2 ©O´÷ y = x y = x 2 l:: 1 + i . y O x 1+i y=x y=x2 14/127
解路线C1的参数方程 z=t+it,t从0到1, 于是dz=(1+a)ldt +1)(1+2)d +(1+0)/tdt
y O x 1+i y=x y=x2 ) ´ C1 ëê§: z = t + it, tl01, u´ dz = (1 + i)dt. ¤± Z C1 (x 2 + iy)dz = Z 1 0 (t 2 + it)(1 + i)dt = (1 + i) Z 1 0 t 2 dt + i(1 + i) Z 1 0 tdt = − 1 6 + 5 6 i. 15/127
解路线C1的参数方程: z=t+it,t从0到1, 于是dz=(1+a)ldt 所以 +igd (t2+i)(1+i) 695× (1+i) dt+(1+i)/tdt 1
y O x 1+i y=x y=x2 ) ´ C1 ëê§: z = t + it, tl01, u´ dz = (1 + i)dt. ¤± Z C1 (x 2 + iy)dz = Z 1 0 (t 2 + it)(1 + i)dt = (1 + i) Z 1 0 t 2 dt + i(1 + i) Z 1 0 tdt = − 1 6 + 5 6 i. 15/127
路线C2的参数方程 z=t+it2,t从0到1
y O x 1+i y=x y=x2 ´ C2 ëê§: z = t + it2 , tl01, u´ dz = (1 + 2it)dt. ¤± Z C2 (x 2 + iy)dz = Z 1 0 (t 2 + it2 )(1 + 2it)dt = (1 + i) Z 1 0 t 2 dt + 2i(1 + i) Z 1 0 t 2 dt = − 1 6 + 5 6 i. 16/127