的等而解析函数 则有 f(z+△2)-f(2) +(e1+ie3)x+(e2+ie4) 当△2→0的,因例≤1,≤1,定式右端的平后两项都趋于零 △ t2- 证毕 当函数f(x)=u(x,y)+i(x,y)在z=x+iy微导研,有 auau 1 au f(a) ar i dy 上理二函概f(x)=u(x,y)+iv(x,y)在其定义域D内解析的充是解 件是:u(x,y)与v(x,y)在D内可微,并且满足柯西-黎在方程. 上理(可微的充分解件若函概f(x,y)在分P的某个邻域D内存在 两个偏导概,且偏导概在该分连续,则函概∫(x,y)在分P可微 (高等概学(下)pp.27) 在本件后面是证明:一个解析函概的导函概f(x)数是解析的,因的 ∫"(x)=tx+itn=ty-iy数是连续的.于是我们有下面的定理 上理二(B)函概f()=u(x,y)+iv(x,y)在其定义域D内解析的充 是解件是:a(x,y)与v(x,y)在D内有一阶连续偏导概,并且满足柯西 黎在方程
6 $%/ )-'+ nc f(z + 4z) − f(z) 4z = ∂u ∂x + i ∂v ∂x +(ε1 + iε3) 4x 4z + (ε2 + iε4) 4y 4z . 4z → 0 "^> 4x 4z 6 1, 4y 4z 6 1, %d+$JnK*fr ^ f 0 (z) = lim 4z→0 f(z + 4z) − f(z) 4z = ∂u ∂x + i ∂v ∂x. t 1X f(z) = u(x, y) + iv(x, y) x z = x + iy ;Us f 0 (z) = ∂u ∂x + i ∂v ∂x = 1 i ∂u ∂y + ∂v ∂y . =# F+ f(z) = u(x, y) + iv(x, y) m )]i D ^B$W7 V& u(x, y) h v(x, y) m D f; ÆweC - hx4 = (f;$57V) F+ f(x, y) m( P $=qi D m n=!+Æ!+m9(mSnF+ f(x, y) m( P f; (<%+T (E) pp. 27) m(J}Wt~ Y=^BF+$!F+f 0 (z)X&^B$, ^ f 0 (z) = ux + ivx = vy − iuy X&mS$f&@{cE}$)i =# (B) F+ f(z) = u(x, y) + iv(x, y) m )]i D ^B$ W7V& u(x, y) h v(x, y) m D cY[mS!+ ÆweC - hx4
§2.2函概解析的定要条件 例1判定下列函数在何处可导,在何处解析 1)f(x)=zRe(2) 2)f(2)= 3)f(a=e(cosy+ ising) 解1)由f(x)=zRe(z)=x2+ixy,得u=x2,v=xy,所以 au dy ar y, ay 显然,这四个偏导数处处连续(由此知u,可微),但概仅当x=y=0 的,它们才满足柯西-黎曼方程. 因此,函数仅在z=0可导,在复平面内处处不解析 2因为u=十=二+y所以 一0 y (x2+y2)2 一可两 z2-2xy (x2+y2)2 (x2+y2)2 一阶偏导数在z≠0处连续(由此知u,v可微)并且满足柯西-黎曼方 程 因此,函数在复平面内除z=0外处处可导,在复平面内除z=0外 处处解析,并且 au. av x2)(1+)-2xy(1-i) (x2+y2)2 (z≠0) 3)因为u= e cosy,v= sing,所以 e Cosy, e s dv cosy dy 一阶偏导数在复平面内处处连续(由此知u,可微,并且满足柯西-黎曼 方程.因此,函数在复平面内处处解析,并且 A'(a)=Ou,ov +io=e(cosy ising )=f(a
§2.2 '+)-" .,( 7 > 1 )EpF+mHf!mH^B 1) f(z) = zRe(z); 2) f(z) = x + y x2 + y 2 + i x − y x2 + y 2 ; 3) f(z) = e x (cosy + isiny); 7 1) b f(z) = zRe(z) = x 2 + ixy, # u = x 2 , v = xy, 2Z ∂u ∂x = 2x, ∂u ∂y = 0, ∂v ∂x = y, ∂v ∂y = x. Gp/=!+mS (b v u, v f;), &_ x = y = 0 "4{weC - hx4 ^ F+_m z = 0 f!m7}Æ^B 2) ^> u = x + y x2 + y 2 , v = x − y x2 + y 2 , 2Z ∂u ∂x = y 2 − x 2 − 2xy (x2 + y 2) 2 , ∂u ∂y = x 2 − y 2 − 2xy (x2 + y 2) 2 , ∂v ∂x = y 2 − x 2 + 2xy (x 2 + y 2 ) 2 , ∂v ∂y = y 2 − x 2 − 2xy (x 2 + y 2 ) 2 . Y[!+m z 6= 0 mS (b v u, v f;), ÆweC - hx4 ^ F+m7} z = 0 :f!m7} z = 0 : ^B Æ f 0 (z) = ∂u ∂x + i ∂v ∂x = (y 2 − x 2 )(1 + i) − 2xy(1 − i) (x 2 + y 2 ) 2 (z 6= 0). 3) ^> u = e x cosy, v = e x siny, 2Z ∂u ∂x = e x cosy, ∂u ∂y = −e x siny, ∂v ∂x = e x siny, ∂v ∂y = e x cosy. Y[!+m7}mS (b v u, v f;), ÆweC - hx 4^ F+m7}^B Æ f 0 (z) = ∂u ∂x + i ∂v ∂x = e x (cosy + isiny) = f(z).