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1 IV 1 IV ] shib@mth.hiroshim-u.c.jp [] 1. z 0 ε δ := ε z 0 z <δ z f 1 (z 0 ) f 1 (z) = z 0 z = z 0 z <δ= ε f 1 z 0 z 0 0 ε 0 <δ< ε +ε z 0 δ z z 0 z <δ z z 0 z 0 z > z 0 δ > z 0 ε +ε z 0 = +ε z 0 > 0 (3 ) z 0 z f (z 0 ) f (z) = z 0 (z z 0 ) z 0 ( z z 0 ) z 0 z z z 0 z δ +ε z 0 <ε

2 IV f z 0 0 z 0 z = re it (r, t R, r>0, 0 t<π) z z = e it lim z 0 f (z) f z =0. z 0 z 0 f 3 z = re it (r, t R, r>0, 0 t<π) lim f z 3(z) = lim z 0 z 0 z = lim r 0 re 3it =0 f 3 (0) = 0 f 3 z =0 x = ξ, y = η ( ) ε δ x ξ <δ, y η <δ e x e ξ < ε, sin y η < εeξ e ε (ξ, η), (x, y) (x ξ) +(y η) δ x ξ <δ, y η <δ (ex cos y e ξ cos η) +(e x sin y e ξ sin η) = e ξ (1 e x ξ ) +4e x ξ sin y η e ξ (εe ξ ) +4e x ξ ( εe ξ e ε δ ε ε R R )

3 z =0 x = ξ, y = η ( ) ε δ x ξ <δ, y η <δ e x e ξ < ε, sin y η < εeξ e ε (ξ, η), (x, y) (x ξ) +(y η) δ x

3 IV 3 {ex cos y} + {e x sin y} = e x x [ ] ( ) 3. 1 f(z) =1 z 1 āz z < 1 f(z) < 1, = (1 )(1 z ) 1 āz f D f D D 1 1 z 1,z D f(z 1 )=f(z ) (1 )(z 1 z )=0 z 1 = z f D D w D z := w + 1+āw z < 1 f(z) =w ( ) f (z) = (1 āz)+ā(z ) = 1 0 (z D) (1 āz) (1 āz) f D

4 IV 4 4. u u(x, y) = u x + u y = x u x + y u y = x (x)+ ( y) = =0 y u (ξ,η) { x(t) = ξt y(t) = ηt 0 t 1 v(ξ,η) v(0, 0) = = = (ξ,η) (0,0) u u dx + y x dy ( u y (x(t),y(t))x (t)+ u ) x (x(t),y(t))y (t) dt (ηt ξ +ξt η) dt =[ξηt ] 1 0 =ξη u v v(x, y) =xy +const.. 5. C {0} z( 0) ε 0 <ε< z D(z, ε) C z =1 z z C X 1 := C \ (, 0] z C X := C \ [0, ) X 1 X = C z, z C C C C

1 v(ξ,η) v(0, 0) = = = (ξ,η) (0,0) 1 0 1 0 u u dx + y x dy ( u y (x(t),y(t))x (t)+ u ) x

5 IV 5 u(x, y) =x y +3y C u x u y = x + xy (x + y ), y (x + y ) = y +3 x y (x + y ), u = +y y 3x x (x + y ), u = +y 3x y y (x + y ) C u =0 v (ξ,η) C z =1 γ v(x, y) = u u dx + y x dy = γ γ = (y 3+ x y γ (x + y ) ydx + xdy 3 dx γ + 1 γ =xy 3x x y (x + y ) dx + xy (x + y ) dy x (x + y ) +const. ) ( ) xy dx + x + dy (x + y ) v(x, y) =xy 3x 1 x x + y +const.

6 IV 6 C v x = u y, v y = u x [ ] (x y ( ) ) (x + y ) dx + xy A(x, y) (x + y ) dy = d x + y { (x + y )A x xa = x y A(x, y) = x (x + y )A y ya = xy (x, y) (r, θ) (1, 0) (R, Θ) γ γ 1 : { r(t) = (1 t)+tr θ(t) = 0 (0 t 1) { r(t) = R γ : θ(t) = tθ (0 t 1) γ 1 + γ X 1,X z =1 z = 1 X 1,X v 1,v v 1,v (X 1 X )

7 IV 7 u C z = x + iy c f(z) = u(x, y)+iv(x, y) ( ) = x y y +3y (x + y ) ( ) x +i xy 3x (x + y ) = z z 3iz i z = z 3iz ı z. + ic 6. (r, θ) r = x + y,θ = rctn y x ( ) x = r r x + θ θ x = cosθ r sin θ r θ = r y r y + θ = sinθ θ y r + cos θ r θ z z = 1 ( r z r i ) θ = 1 z ( r r + i ). θ [ ] z 8 ( ) r = z z, θ = 1 (log z log z) i

8 IV 8 z = r r z + θ θ z, z = r r z + θ θ z =4 { ( 1 z z =4 r z z r + i )} θ =4 1 z ( r r i ){ ( 1 r θ z r + i )} θ = 1 ( r z r i )( r θ r + i ) θ = 1 { r ( r ) + ir r r r r θ i ( r ) } + θ r θ = 1 { r ( r ) } + r r r θ r + 1 r r + 1 r θ C u u =0 r u r ( r u r ) + u θ =0 u θ ( u r u ) =0 r r

9 IV 9, b u u(r) = log r + b u (x, y) u(x, y) = log(x + y )+b v(x, y) = y x + y dx + x x + y dy = rctn y + c (c R) x u θ + c (c R) [ ] C 9. z 1 1 r cos θ + r 0 u(r, θ) = 1 r 1 r cos θ + r = r + 1 r r + 1 r θ u r = 4r +(1+r )cosθ (1 r cos θ + r ) 3, u = 4{cos θ r(r +3)cosθ +3r 1}, r (1 r cos θ + r ) 3 u = r(1 r ){(1 + r )cosθ +r cos θ 4r}. θ (1 r cos θ + r ) 3 u D u

10 IV 10 [ ] 1 r cos θ + r =(1 r) +r(1 cos θ) 0 r =1,θ=0(modπ) u C \{(1, 0)} (r r z + i θ ). (u + iv) =0 v r = 1 r u θ, v θ = r u r. f(z) =R(x, y)(cos Θ(x, y)+i sin Θ(x, y)) R, Θ ( ) { u = R cos Θ R = u + v v = R cos Θ Θ = rctn v u R x = R u u x + R v v y Θ = Θ u y u x + Θ v v y = cosθ u v sin Θ x y = cos Θ u R x sin Θ v R y u, v R x = R Θ y R y R y = R Θ x.

11 IV 11 R f R, Θ r, θ 11. [, b] M := mx ϕ = ϕ(ξ ), m := min ϕ = [,b] [,b] ϕ(ξ ) ξ,ξ [, b] M = m (c (, b) ) m<m m 1 b 1 b b b ϕ(t) dt M ϕ(t) dt = ϕ(c) c ξ ξ <c<b ϕ 0 if 0 x < 1 ϕ(x) = 1 1 if < x < f(z) =u(x, y)+iv(x, y) γ γ : x = x(t),y = y(t) ( t b) x() =x(b), y() =y(b)

12 IV 1 b f(z)f (z) dz = [(u(x(t),y(t)) iv(x(t),y(t))][(u x (x(t),y(t)) γ = = b b + iv x (x(t),y(t))][x (t)+iy (t)]dt [(uu x + vv x )+i(uv x vu x )][x (t)+iy (t)]dt [(uu x + vv x )x (t) (uv x vu x )y (t)]dt + i b [(uu x + vv x )y (t)+(uv x vu x )x (t)]dt b = [(uu x + vv x )x (t)+(uu y + vv y )y (t)]dt b + i b d dt (u + v )dt + i =[(u + v )] t=b t= + i = i b [(uv x vu x )x (t)+(uu x + vv x )y (t)]dt b b [(uv x vu x )x (t)+(uu x + vv x )y (t)]dt [(uv x vu x )x (t)+(uu x + vv x )y (t)]dt [(uv x vu x )x (t)+(uu x + vv x )y (t)]dt

x + vv x )x (t)+(uu y + vv y )y (t)]dt b + i b d dt (u + v )dt + i =[(u + v )] t=b t= + i = i b [(uv x vu x )x (t)+(uu x + vv x )y

13 IV 13 Re f(z)f (z) dz = fdf + γ = γ γ γ fdf fdf + fd f = 13. w = u + iv dudv = udv = f( ) f( ) f( ) γ d( ff)=0 udv f( ) dudv f( D) w = f(e it ), 0 t π, z (t) = sin t + i cos t dv =(v x ( sin t)+v y cos t) dt =( v x sin t + u x cos t) dt =Im[(u x + iv x )( sin t + i cos t)] dt =Im[f (z(t))z (t)] dt

14 IV 14 π udv = (Re f)(z(t))im [f (z(t))z (t)] dt f( ) = = 0 π 0 π 0 Im [(Re f)(z(t))f (z(t))z (t)] dt [ f(z(t)) + f(z(t)) Im ( = 1 π Im f(z(t))f (z(t))z (t) dt 0 ] f (z(t))z (t) dt + 1 π Im [f(z(t))f (z(t))z (t)] dt 0 = 1 4 Im [f(z(t)) ] π π Im [f(z(t))f (z(t))z (t)] dt = 1 π Im [f(z(t))f (z(t))z (t)] dt = 1 Im 0 C f(z)f (z) dz 1 f(z)f (z)dz i C [ ] ( ) u = f(z)+f(z), v = 0 f(z) f(z) i

[f(z(t))f (z(t))z (t)] dt 0 = 1 4 Im [f(z(t)) ] π 0 + 1 π Im [f(z(t))f (z(t))z (t)] dt = 1 π Im

15 IV 15 f(z)+f(z) f (z)dz f (z)d z C i = 1 f(z)f (z)dz f(z)f 4i (z)d z + f(z)f (z)dz f(z)f (z)d z C 0 f(z)f (z)dz = 1 d(f(z)) =0, f(z)f (z)d z = 1 d(f(z)) =0. C C 1 f(z)f(z)d z + f(z)f (z)dz = 1 4i Im C C C C f(z)f (z)dz f(d) =f( D) u = w + w, v = w w i idudv = dwd w 14. f : z z = x iy f = u + iv u x =1, u y =0; v x =0, v y = 1 [ ] C

16 IV (1) =1,b= 1; f(z) =z ; f (i) =i. 16. () = 6,b=,c=0,p=6,q =0;f(z) =z 3 ; f (i) = p = 6,q = 5,r =3,s =10 f(z) =u(x, y) +iv(x, y) = (5 + 3i)z f (5 3i) = { C if 0 T (C) =, S(C \{ d/c}) =C \{/c} {b} if = 0 λ := d bc 1 := / λ, b := b/ λ, c := c/ λ, d := d/ λ z z + b S c z + d d b c =1 19. T =0 0 T 1 C T 1 (w) = w b,w C. T 1 C S 1 C \{/c} S 1 (w) = dw + b,w C \{ d/c}. cw dt 1 dw = 1, ds 1 dw = 1 cw d). 0. z = x + iy u(x, y) +iv(x, y) z u = x + y,v =0 x = y =0 z z u = x + y,v =0 x = y =0 u x = x x + y, u y = v x = 0, v y = 0 y x + y ;

${ C if 0 T (C) =, S(C \{ d/c}) =C \{/c} {b} if = 0 λ := d bc 1 := / λ, b := b/ λ, c := c/ λ, d := d/ λ z z + b S c z + d d b c =1 19.$

17 IV 17 z z 3 u =(x + y ) 3,v =0 x = y =0 u x = 3x x + y, u y = 3y x + y ; v x = 0, v y = 0 x = y =0 [ ] z z 3 h 0 lim = lim h = 0; h 0 h h 0 h 3 0 lim h 0 h 0 = lim h 0 h h =0 1. z z( 0) z = x + iy w = z = u + iv u = x y, v =xy c 1,c 0 x = c 1 y u = c 1 y, v =c 1 y y ( ) v u = c 1. c 1 y = c x u = x c, v =xc x ( ) v u = c c.

18 IV 18 (c 1,c ) ( c, c 1 ), (c 1, c ) 0 ( ) ( ). () G f () z G z 0 G z = z 0.G ε>0 D(z 0,ε) G. D(z,ε) G z D(z,ε) z z 0 = z z = z z <ε z D(z 0,ε) z G z G D(z,ε) G G G z1,z G z 1, z G (b) f (z) G f (z) z = f( z) z f = f( z) z =0 3. f f(z) = {f(z)f(z)} =4 z =4 z =4 f z f (z) ( z {f(z)f(z)} =4 f f z z ( f f ) =4 f f z z z +4f f z z =4 f f z z =4 ) f + f z

19 IV 19 u = u(x, y),v(x, y) u u (u, v) (x, y) := x y v v = u v x y u v y x x y ( ) ( ) u v + = f (z) 0 x x 4. f C {zf(z)} =4 z f =0 z {zf(z)} =4 f {z z z } =4 f z + z f {zf(z)} =0 f z =0. 5. C u (u )=4 z z (u )=4 ( u u ) z z =8 u u z z +u u =8 u u z z +8u z u u ( ) u + x u u z z =0 ( ) u =0 y ( ) u z u u u ū (z )

20 IV 0 6. (1) () z G ε>0 z G ε G ε S[1,z] G ε.g ε G (3) z G G S[1,z] 7. (1) () {f,z} =0 g := f /f g (z) = 1 g(z) 1 g(z) = 1 z + c 1, c 1 C f (z) f (z) =. z +c 1 1 c log f (z) = log(z +c 1 )+c f (z) = c 3,c 4 f(z) = c 3 z +c 1 + c 4 c 3 (z +c 1 ) d bc 1, b, c, d C z + b cz + d 8. {ϕ,ζ} = ϕ (ζ) ϕ (ζ) 3 ( ) ϕ (ζ) ϕ (ζ)

21 IV 1 ϕ (ζ) = f (z)g (ζ) ϕ (ζ) = f (z)g (ζ)+f (z)g (ζ) ϕ (ζ) = f (z)g 3 (ζ)+3f (z)g (ζ)g (ζ)+f (z)g (ζ) {f g, ζ} = {f,z}g (ζ) + {g, ζ}

22 IV [] O (0) z 0 C 1 f f z 0 f z 0 f z 0 f f z 0 ( ) z 0 z 0 z 0 (43 ) ( ), /0 = f(ζ) F (ζ)

1 : f(z = re iθ ) = u(r, θ) + iv(r, θ). (re iθ ) 2 = r 2 e 2iθ = r 2 cos 2θ + ir 2 sin 2θ r f(z = x + iy) = u(x, y) + iv(x, y). (x + iy) 2 = x 2 y 2 +

1.3 1.4. (pp.14-27) 1 1 : f(z = re iθ ) = u(r, θ) + iv(r, θ). (re iθ ) 2 = r 2 e 2iθ = r 2 cos 2θ + ir 2 sin 2θ r f(z = x + iy) = u(x, y) + iv(x, y). (x + iy) 2 = x 2 y 2 + i2xy x = 1 y (1 + iy) 2 = 1