数学物理方法解析函数习题课

1、幂函数w幂函数wz三角函: sin z,: shz,f(z)u(x,y)iv(x,y)wn 角1.1 cos isin角1.1 cos isin ,0 argz ,| z | z x iy 1cos isinx 1cos, y sin则(1cos)2 sin24sin2(1)|z| z 2sin222cos2220 -1cos1,sin0 -1cos1,sinx1cos 0y0,一象sinyarg z 1cos| z arctg(ctg ) arctgtg22222已知z xiy,已知z xiy,求cosz的模关键是求u, v.(下面请同学做cosz 1eiz 1eixy eixy22 1ey

2、(cosxisinx) ey(cosxisinx) 1ey (cosxisin x)ey(cosxisin 21(ey ey)cosxi(ey ey)sinx u1(ey 2vu1(ey 2vye2|coszu222(eyey)2co2 x(eyey)2si2e2ye2yarctgv arctg(eeyarctgv arctg(eeyeuey若x0y 0,则当0 x u 0v2 arctg v u x,u 0,v当2 arctg u三.三.:2(1) | z-a |:2(1) | z-a | z-b a.设z xiya x1iy1 b x2 iy2则za x x1iy y1zb (x x2)i

3、(y y2由| z-a | z-b (x x1)2 (y y1)2222b.根据向量表示也易于得|b.根据向量表示也易于得|zaba0(2)a2a2|(2)a2a2|z|C,a 0,c|z|z|Rez答：不可以Rez不是向| z | z |2 2(| | |2121212x | z | z |2 2(| | |2121212x (y 2211(x (y 2211 2x2 2 222 y 2)(y211222(|12(4).0arg(z1 且2(4).0arg(z1 且2Rez 4i)令w z1 x10 arg(z 1) 40argw4y yx1x4x1) 0,由右y x0 y x(5).w1将

4、y24w:zx(5).w1将y24w:zx1解：令wuivxxux2yyvy2x,v 由x,v 由x2 y2 4(a)代入：u x4u,y 44 4代入(a4u24v2 1即u2v2(1). 11i)(1i)(1). 11i)(1i) x则xiy)21x 21x2y2 2xy 2 21 y或ii)设1i ei则或ii)设1i ei则|1i2 2karc(1i) arctg1 4i(2k) 1i ( 2k i 24eii) 1 eii) 1 (e1(2).sini 1 i(e2 (3).sin z eiz eiz z 12(3).sin z eiz eiz z 12 4i16 (23)i2i(2

5、kiz Ln(2 Ln(2 Ln(23)i Ln(232i( 2ki( 2kiArg(233)e223)i( 2k23)( 2kz iLn(22五性f z)在区域f(z五性f z)在区域f(z)在区域处可导则称w 1.如何判i.定义于可导的联ii.充分必要条R:u,v 连xyxy如(sin z) z存1已知u x2 y2 xy, f(i) 1已知u x2 y2 xy, f(i) 1f 求 2x y, 2y x, yuxx u 由CR:2yx,2x dv vxdxvydx (2yx)dx(2x d(2xy)2v 2xyC2y f (z) yxyi(2xy 222又f (i) 1i(即y f (z

6、) yxyi(2xy 222又f (i) 1i(即x 0y 1 i iC 1i即C 222yx12 2xyi,xyi x2222y f (z) yxyi(2xyC222 z2 1iz2 1i (1 i)z2 i2222z3,3 5i3(2k) 解：w5353z3,3 5i3(2k) 解：w5353(2k 5531 353 (2k 5即1 物线族y2 C2 2Cx(C 0),求等物线族y2 C2 2Cx(C 0),求等 C0),C 0)y解C x,xyxv yx2v 1 x yyx故令v(xy F(tt y2 则v Fxy2x故令v(xy F(tt y2 则v Fxy2x2v F 2F 3 y2)y2(x2：2v yF 2F 3(x2 y2)y2带方程： y2 xF(t)F带方程： y2 xF(t)F(t) xF1 积分：F (t) 亦即F t在积分：F(tv(x,y) F(t) C1t xyx C极坐标形v(,)cos 2sin极坐标形v(,)cos 2sin 122由CR条件u cos11 2sin 112du u d u 111ddu u d u 111d22Cd2 21