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1、第二章解析函数1 .用导数定义,求下列函数的导数: (x) = z Re Z.解:因limzO(zz)-(z) = lim zOz(z + z) Re(z + z) -zRezzlimzOz Re z + z Re z + z Re zz nn a ReAz、=hm(Rez + Rez + z)zOZlim(Rez + z20Rezz)=lim(Rez + zOy()x + zy当z w时,上述极限不存在,故导数不存在;当z = 0时,上述极限为0,故导数为0.2 .下列函数在何处可导?何处不可导?何处解析?何处不解析?1 1) f(z) = zz2.解:f(z) = z z2 = z z z
2、 = z 2 z= (x2 + y2)(x+iy)= x(x2 + y2) + iy(x2 + y2),这里 w(, y) = (2 + y2), v(x, y) = y(x2 +)2).ux = x2 + y2 + 2x2, vv = x2 + y2 + 2y2,y = 2xy,匕=2xy.ux = vy,uy =-vx,当且当x=y = 0,而“,%,匕,4 均连续,故 (z) = zz2.仅在z = 0处可导,处处不解析.2 2) f(z) = x3 -3xy2 +z(3x2y-y3).解:这里 w(x,y) = x3-3xy2,v(x,y) = 3x2y-yux = 3x2 -3y2,
3、4 = -6 孙匕=6孙 vy = 3x2 - 3j2,四个偏导数均连续且 = vv(v = -匕处处成立,故(z)在整个复平面上处处可导,也处处解析.3 .确定下列函数的解析区域和奇点,并求出导数.(1)空*(gd至少有一不为零).cz + d解:当cw时,(z) 二丝妙除z = -4外在复平面上处处解析,z = -4为奇点,cz + d cc(=cz + d_ (az + b)cz + d)- (cz + d)(az + b)(cz + )2_ acz + d) -c(az + /?) _ ad-ch(cz + d (cz + d)?当c=o时,显然有a。,故(z)=生也在复平面上处处解析
4、,且r(z)=q.dd4 .若函数(z)在区域。内解析,并满足下列条件之一,试证(z)必为常数.(1) (z)在区域。内解析; V = w2;(3) arg(z)在。内为常数;(vu)(y! u) 设(z) = + ,由条件知arctan上=C,从而U法 二0 办 =0l + (vw)2 ,l + (vw)2计算得2zv 须、/ 2L(UV)! r汝、毡=0,4 +2zv u x/ 9一 UV)! I匕内一=。,-+ u-化简,利用C-R条件得u u 八uv = 0,y xu u cuv = 0.x y所以包=磔=0,同理型=0,即在。中为常数,故/在。中为常x yx y数.(4)法一:设。0
5、,则 = (c-bu)o,求导得u _ b v u _ b v1 . - 1 , 1 - ,x a x y a y由C-R条件u _b uv _b vx a yx a y故,v必为常数,即(z)在D中为常数.设q = 0S0,0则加=c,知-为常数,又由C-R条件知也必为常数,所以(z)在。中为常数.法二:等式两边对x,y求偏导得:au +/?v = 0v n,由C R条件,我们有auy + bvy = 0(a -bi b aaur -bux. =0,八,即burauv =06Z2+72O,故=人=0,从而为常数,即有(Z)在。中为常数.5 .设,在区域。内解析,试证:弓+副/ir(z).证:
6、设 (z) =u + iv, | (z)2= w2 + v2,r(z)=一畔,z)2=(2+(2.ox yox oy而2 27 a2 9 9 a2 9 9(+ v)U)二0( +v) + -(wv-)x yxy_ (、)2u z6v7 62v 加、,2u zv9 2v=2 (厂 +w + ()-+ v+ ()- +w + (厂 + v LOx Sc x x y y y y J乂 (Z)解析,则实部及虚部U均为调和函数,故w = + = 0,v = = O,x1 y1x2 y2( + l(z)2= 4(2 + (2) = 41 ft(z) |2.ox yox oy6 .由下列条件求解解析函数(z
7、) = + iv.(l)w = (-y)(2+ 4肛 + J2 );解:因T+6所3/,所以v = j (3x2 + 6xy-3y2 )dy=3x2y + 3y2 -y3 + e(x),乂 = 6孙 + 3y2 + “(x),而包=3x2 - 6盯一3)*,所以,(x) = -3x2,则xxe() = 3 + c.故(z) = w + zv= (x-y)(x2 + 4xy + y2) z(3x2y + 3xy2 -y3 -x3 C)= (l-z)x2 (x + z)-(l-z)(x + iy) - 2x2y(l z) - 2xy2(-i) + Ci=z(l -z)(x2 - y2) - 2ay
8、z zz(l - /) + Ci= (l-z)z(x2 -y2 - 2xy) Ci= (l-z)z3 + Cz(2) v = 2x + 3x;解:因 = 2y + 3, = 2乂由(z)解析,有xoyu v -=2x,及 y= J2xdx = x2 (y).又= -2y -3,而=(y),所以“(y) = -2y-3,则 (y) = -y1 -3y + C.y xy故 (z) = %2 -y2 - 3y + C + i(2xy + 3x). = 2(fy(2) = T解:因学=2y学=2(x-l),由(z)的解析性,有二=-纵=-2(1),x yx yv = -2(x-l)d = -(x-l)
9、2 0(y),又?= 2y,而? = (y),所以“(y) = 2 0(y) = / +C,则y oxyv = -(x-l)2 + y2 +C,故(z) = 2(x-l)y + z(-(x -l)2 + y2 + C),由 /(2) = -i 得 /(2) = z(- l + C) = T 推出 C =().即(z) = 2(x-l)y + z(,2 -x2 +2x-l)= z(-z2+2z-1) = -z(z-1)2.7 .设u = eA siny,求p的值使口为调和函数,并求出解析函数(z) = w + zv.解:要使u(xy)为调和函数,则有% =0.即p2epx sin y - epx
10、 sin y = 0,所以p = l时,u为调和函数,要使/解析,则有 = %、= -vv.u(x, y) = uxdx = epx cos ydx = -epx cos y + My),u = -epx sin y + “(y) = -pepx sin y.P所以(y) = (p)e,x sin y, (y) = (p)epx cos y + C.即(x, y) = pepx cos y + C,故(z)ex (cos j + zsiny) + C = 2 + C,-ex(cosy-sin y) + C = -ez +C, p = -l.8 .试解方程:(1) 血;“ 解:ez = 1 +
11、V3z = 2(cos-+ sin-) =333k = 0,l,2z = In2+ /(2 + -), Z = 0,l,2.iz,解:z = e冗 乃= cos-+ sn-= z9.求下列各式的值。(1) cos z;(2)。(3 + 旬;解:Ln(-3 + 4z) = In 5 + zArtg(-3 + 4z)4=ln 5 i(2k + - arctan ).(一尸;解:(1 - o1+,(1+z)Lh(1-)(l+2+-2,+ In J+2A不一工Inf2+-2,e 444cos(ln 夜-工)+ i sin(ln V-)44(4) 33t;解:q3-j _ (3-)In3 _ (3-Xln32)=e(3T)E3 既=n3+2E e句n3=27e” (cos ln3-zsin In 3).