F(z+A2)-F(-) f +△z f(5)-f()4 ()在0内解析∴E>0,30>0.k-2<0 时有()f()<6于是:1<=6 F(z+△x)-F(z) f(2 即F(z)=f(z)[ ] ( ) ( ) ( ) 1 ( ) - ( ) 1 ( ) ( ) F z z F z I f z z z z f f z d z z z z f f z d z z x x x x + D - = - D + D = ò D + D £ - ò D ∴ ∵ f z ( )在 s 内解析 ∴ "e > 0, $d x > 0, 0 - < z 时有 f (x e ) - < f z( ) 于是： z z z I d z e x e +D < = D ò 0 ( ) ( ) lim ( ) z F z z F z f z D ® z +D - = ∴ D 即 F¢(z) = f z( )