CIII CIII : October 4, 2013 Version : 1.1 A A441 Kawahira, Tomoki TA (Takahiro, Wakasa 3 )

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1 III III : October 4, 203 Version :. Kawahira, Tomoki TA (Takahiro, Wakasa 3 ) pdf (982/03) II (999/06)

2 III II (985/04) (990/0) 5 59 F B80 89 A90 00 S A4 A4 afe David afe David 2:00 3:30 2 () : (2) R: (3) Q: (4) Z: (5) N: (6) : () α: (2) β: (3) γ, Γ: (4) δ, : (5) ϵ: (6) ζ: (7) η: (8) θ, Θ: (9) ι: (0) κ: () λ, Λ: (2) µ: (3) ν: (4) ξ, Ξ: (5) o: (6) π, Π: (7) ρ: (8) σ, Σ: (9) τ: (20) υ, Υ: (2) ϕ, Φ: (22) χ: (23) ψ, Ψ: (24) ω, Ω:

3 III : October 4, 203 Version :.. D f : D 2. f : D α D 3. f : D 4. e z 5. z 2 6. ( ) i α r > 0 α r (α, r) 7. m = (α, r) (z α) m dz = (i, ) z dz = π = (0, 2) 2. () z 2 dz = 2πi z i () (2) f : D D f(z) 300 D 2 f(z) dz

4 III : October, 203 Version :. (0/4). D (domain) : D D 2 D D f : D 2. f : D α D w = f(z) z = α (differentiable) : A f(z) f(α) lim z α z α = A A f α (differential coefficient) A = f (α). z α f(z) f(α) z α A w z A w A z A = re iθ 0 f z = α r = A θ = arg A f(α) 3. f : D w = f(z) D (holomorphic) : { f D f D. w = f(z) D : α D f D w = f(z) z D f (z) f (derivative). f D Goursat 4. e z. f : D w = f(z) z = x + yiw = u + vi f(x + yi) = u + vi (x, y, u, v R) (a) (b) (a) f D (b) u, v D (R) { ux = v y v x = u y

5 III f (z) = u x + iv x (R) the auchy-riemann equation) (R) f xy uv ( ) ( ) ux u y = v x v y e ( ) ( ) z ( = e x+yi ) = e x (cos y + i sin y) ( xy ) x u e uv = x cos y ux u y v e x y = ( ) sin y v x v y e x cos y e x sin y e x sin y e x u, v cos y (R) x, y f f (z) = u x + iv x = e x cos y + ie x sin y = e z 5. z 2 ( ) ( ) ( ) x u x = 2 y 2 ( ) ( ) y v 2xy ux u y 2x 2y = v x v y 2y 2x 6. ( ) i ( ) i := e i log( ) log( ) e z = = e πi = e πi+2mπi (m Z) log( ) = (2m + )πi (m Z) ( ) i := e i log( ) = e i(2m+)πi = e (2m+)π (m Z)..., e 3π, e π, e π, e 3π, e 5π,... α r > 0 α r (α, r) 7. m = (α, r) z(t) = α + re it z (t) = ire it. (z α) m dz = 2π 0 2π (re it ) m (ire it ) dt = ir m+ e i(m+)t dt = 0 [ ir m+ (m + )i ei(m+)t i 2π 0 (z α) m dz ] 2π 0 = 0 (m ) dt = 2πi (m = ). 8.

6 III (auchy s Integral Theorem)D D f : D f(z) dz = 0. (simple closed curve, s.c.c.) D (simply connected) D D D 9. = (i, ) z dz = 7 π 2 { 2πi α z α dz = 0 α ( z ) = 2 2i z 2i z + 2i ( ) z dz = 2 2i z 2i dz z + 2i dz. 2i 2i 2 2i (2πi 0) = π 2 0. (auchy s Integral Formula): D D f : D α ( ) f(α) = f(z) 2πi z α dz. = (0, 2) z 2 dz = 2πi z i

7 III I D = f(z) = z 2, α = i D f D i ( ) f(i) = z 2 dz I = 2πif(i) = 2πi. 2πi z i 2. () () (2) f : D D f(z) 300 D 2 f(z) dz () = 600 (2) z : z = z(t) z (t) = (0 t 2) t z z z (t) t 2 f(z) dz = f(z(t))z (t)dt f(z) z f(z(t))z (t) t 0 t f(z(t))z (t) f(z(t)) z (t) 300 = 300 () f(z(t))z (t) f(z(t))z (t) = () dz (2) z(z 2i) 4z 2 dz (3) + 2z 2 + 3z 2 dz 2-2. z 2 + dz (α, r) = { z(t) = α + re it t [0, 2π] } () = (i, ) (2) = ( i, ) (3) = (0, 2) (4) = (, ) 2-3. a, b E E = {z(t) = a cos t + ib sin t t [0, 2π]} 2π 2π () dz (2) z a 2 cos 2 x + b 2 sin 2 dx = x ab Hint: () 0

8 III : October 5, 203 Version :. (0/) (auchy s Integral Theorem)D D f : D f(z) dz = 0.. x = x(t), y = y(t) R : z = z(t) = x(t) + iy(t) (a t b) (smooth) dz dx dy (t) = (t) + dt dt dt (t)i a < t < b, (piecewise smooth), 2,..., N = + + N (closed curve) : z = z(t) (a t b) z(a) = z(b) : z = z(t) (a t b) (simple closed curve, s.c.c.) a t < t 2 < b z(t ) z(t 2 ) Jordan 3 (interior) int() (exterior) ext() D (simply connected) D int() D D t (orientation) z(a) (initial point)z(b) (terminal point) T T

9 III ,... n = + + n 2,... n 2 0 { } = 0 = n + +. n i i = i 2 { z α dz = 2πi : α int() 0 : α ext(). {α} α z α α D = 0. α r > 0 r := (α, r) n = 7 = = 2πi. r x dx = π 2 i x = 2 tan θ R > R 2 I R = R x dx R := { Re it t [0, π] } [ R, R] R 9 ( ) z dz = 2 2i z 2i dz z + 2i dz. 2i 2i = 2 2i (2πi 0) = π 2 = +I R = π R 2

10 III R 0 (R ) lim R I R = π/ 2 R > 2 z R z + 2 z 2 2 = R 2 2 > z dz R 2 2 l( R) = πr R 2 2 R l( R ) R 0 (R ) D f : D f(x + yi) = u + vi (x, y, u, v R) u, v x, y z z f z := ( 2 x i ) f(x + yi) y f z := ( 2 x + i ) f(x + yi) y z z f z, f z f z(z), f z (z) f(z) = z f z := ( 2 x i y ) (x yi) = 2 { x (x yi) i } (x yi) = {( 0) i(0 i)} = 0. y 2 f z := ( 2 x i ) (x yi) = {( 0) + i(0 i)} =. y 2 f(z) = z n f z (z) = 0f z (z) = nz n (Green s Theorem): D D f : D α int() f(z) dz = 2i int() f z dxdy f(z) = z z dz = 2i dxdy int() 4 α, β α β α + β α + β α = z 2, β = 2

11 III (int() ) = 2i z dz 0 z z f : D w = f(z) u + vi = f(x + yi) f z := ( 2 x + i ) (u + vi) = y 2 {(u x v y ) + i(u y + v x )} D f z 0 u, v f f f z f u+vi = f(x+yi) f(z) dz = (u + vi)(dx + idy) = (u dx v dy) + i (u dy + v dx) () = ( u y v x )dxdy + i ( v y + u x )dxdy = = i int() int() int() int() ( x (iu v) + ) ( u vi) dxdy y ( x + i ) (u + vi) dxdy = 2i y int() f z dxdy.. f : D α D f z (α) = Af z (α) = B z α f α f(z) = f(α) + A (z α) + B (z α) + o( z α ) f z (α) = 0. f(x + yi) = u + vi α = x 0 + y 0 i, f(α) = u 0 + v 0 i x = x x 0 y = y y 0 u = u u 0 v = v v 0 z = z α z = x + i y z = x i y f u, v x, y u = p x + p 2 y + o( x 2 + y 2 ), v = q x + q 2 y + o( x 2 + y 2 ),

12 III p = u x (x 0, y 0 )p 2 = u y (x 0, y 0 )q = v x (x 0, y 0 )q 2 = v y (x 0, y 0 ) f(z) f(α) = u + vi = (p x + p 2 y) + (q x + q 2 y)i + o( x 2 + y 2 ) = (p + q i) x + (p 2 + q 2 i) y + o( x 2 + y 2 ) = f z + z (α) + f z z (α) + o( z ) x 2 y 2i = ( ) f (α) i f 2 x y (α) z + ( ) f (α) + i f 2 x y (α) z + o( z ) = A (z α) + B (z α) + o( z α ). (auchy s Integral Formula): D D f : D α int() f(α) = 2πi f(z) z α dz z 2 = (0, 3) dz = 2πi I D = z + f(z) = z 2, α = D f D int() f( ) = 2πi z 2 z i dz I = 2πi( )2 = 2πi. (Pompeiu s Formula): D D f : D α int() f(α) = 2πi f(z) z α dz f z (z) π int() z α dxdy. 2 5 g(z) = f(z)/(z α)d(α, r) := {z z α < r} r D(α, r) int() g g(z) dz g(z) dz = 2i z dxdy ( ) (α,r) () int() D(α,r) z g z = ( ) f(z) = f ( ) z z α z (z) z α + f(z) z z α /(z α) int() D(α, r) 0 5 int() {α}

13 III () (α,r) f(z) z α dz f(z) f(α) = dz f(α) (α,r) z α (α,r) () z α dz z α = r 2 2πif(α) f f z (α) = Af z (α) = B f(z) f(α) = A(z α)+b(z α)+o( z α ) z α = r 0 z α = r f(z) f(α) ( A + B +) z α f(z) f(α) dz ( A + B + )2πr 0 (r 0) (α,r) z α r 0 ( ) I = 0 x 4 + dx () α = e πi/4 z 4 + = 0 ±α, ±α 4 (2) z 4 + = a, b, c, d (3) a z α + b z + α + c z α + d z + α x 2 dx I z, z. D f, g : D () z z f g (f(z)g(z)) = (z) g(z) + f(z) z z (z) f g (f(z)g(z)) = (z) g(z) + f(z) z z (z) Hint: (2) n f(z) = z n f z (z) = nz n (3) f(z) f(z)

14 III : October 25, 203 Version :. (0/8) D f : D f, f, f (3),.... f(x) = x 3/2 (x R) f f (0) 2. n D f : D D α int() n 0 f (n) (α) f (n) (α) = n! 2πi f(z) dz. (z α) n+ f (n) (α) n! = 2πi f(z) (z α) dz (z α) n (z α) n n 2. n = 0 n = n 2 h 0 n = 0 f(α + h) f(α) h = { h 2πi = 2πi f(z) z (α + h) dz f(z) (z α h)(z α) dz } f(z) z α dz h 0 f (α) = f(z) dz 2πi (z α) 2 f(z) (z α h)(z α) dz f(z) (z α) 2 dz = hf(z) (z α) 2 (z α h) dz f(z) M r N r () = z D(z, r) D(z, r) = int((z, r)) r, h 0 α α + h N r () z f(z)

15 III z z α r z α h r h M r 3 l() h 0 0. α D α D α α 2 n f (n) (α) n 0 f (n) D f (n+) f (n) f (n+) 2 α D f (n+2) (α) f (n+) z = α 3 α f (n+) D. = (0, 2) e z I = (z ) 4 dz 2 D = f(z) = e z, α =, n = 3 α int() f (3) () = 3! 2πi f (3) (z) = e z. e z (z ) 4 dz I = 2πif (3) () = eπi 3! 3. f : D D f = int() D f(z) dz = 0. (Morera) f : D D ( )D f(z) dz = 0 f D. ( ) F : D F (z) = f(z) (z D) F D f f F ( ) α D ζ D α ζ D, 2 ( ) 2 + ( ) f(z) dz = 0 = 2+( ) 2 = 2 Goursat 3 g z = α z = α

16 III D α ζ ( ) F (ζ) := f(z) dz = f(z) dz 2 F (ζ) = ζ α f(z) dz F : D f ζ D F (ζ + h) F (ζ) lim = f(ζ) h 0 h ϵ > 0 δ > 0 h δ F (ζ + h) F (ζ) f(ζ) h ϵ f ϵ > 0 δ > 0 z ζ δ h δ F (ζ + h) F (ζ) h f(z) f(ζ) ϵ D { = ζ+h h α ζ α } α ζ ζ ζ + h L : z(t) = ζ + ht (0 t ) F (ζ + h) F (ζ) h L = h dz = { F (ζ + h) F (ζ) h 0 +L } = h L f(z) dz h dt = hl(l) = h L f(ζ) = f(z) dz f(ζ) dz h L h L = h (f(z) f(ζ)) dz h e l(l) = ϵ ζ D ( ) L. f : D F : D F = f F f (primitive function)

17 III F F 2 f F (z) F 2 (z) D f : D ( ) F : D f(z) = z 2 F (z) = z 3 /3 + (z 3 /3) = z 2. f(z) = F (z) = log z + z log z = (0, ) dz = 2πi 0 ( ) z f(z) D = {0} f D f D := (, 0] ( ) F (ζ) = ζ f(z) dz D F log z Im log z < π Liouville s Theorem f : M > 0 f(z) M ( z ) f e z sin z, cos z sin x ( x R) sin z ( z ) z f(z) = + z 2 z f(z) z + z 2 2 f

18 III f : M f(z) M ( z ) α = (α, r) (r > 0) 2 f (α) = f(z) 2πi (z α) 2 dz 2π M r 2 l() = M r. M r > 0 f (α) = 0 α f The Fundamental Theorem in Algebra a n z n + a n z n + + a z + a 0 = 0 (a n 0, n ) x 2 + = 0 Gauss α, α 2,... α n a n z n + a n z n + + a z + a 0 = a n (z α )(z α 2 ) (z α n ) n n x 2 +. f(z) f(z) 0 z g(z) := f(z) z 0 f(z) = z n a n + a n + + a 0 z z n A(z) z r > 0 A(z) a n + + a 0 a n z z n r + + a 0 r n r 0 z r 0 A(z) a n /2 r 0 z r 0 f(z) z n ( a n a n 2 ) rn 0 a n 2 g(z) 2 r0 n a =: M. n E = {z z r} g(z) E M g(z) max{m, M } 4- = (0, 2) e z e z sin z () dz (2) z z 2 dz (3) (z π/2)(z + π) dz

19 III n = (0, 2) e z z + e iz () dz (2) z4 (z ) 2 dz (3) (z + 3) (3z π) 2 dz 4-3 () f : D D f (z) = 0 (2) f : D F, F 2 : D F (z) = F 2 (z) + Hint.

20 III / : November, 203 Version :. (0/25) D D(α, r) := {z z α < r} α r D(α, r) = int((α, r)) g n g g n g g n (z) dz g(z) dz g n g g n g g n : D (n N) D E g n E g : E (uniform convergence) ( ϵ > 0)( N N)( n N)( z E) g n (z) g(z) < ϵ E g n g z 0 g n g (n ) g = lim n g n. 0 < r < g n (z) = + z + + z n, g(z) = z g n g E(r) = {z z r} z < z = + z + + zn +. g n (z) = ( z n+ )/( z) g n (z) g(z) = z n+ z rn+ r 0 (n )

21 III r n+ /( r) z E(r) g n g E(r) = {z z r} r z 4- g n : D (n 0) g : D D g D. ϵ > 0 α D δ > 0 N z α < δ = g(z) g(α) < ϵ ( n N)( z D) g n (z) g(z) < ϵ/3 g n (α) g(α) < ϵ/3 m N g m D z = α δ > 0 z α < δ = g m (z) g m (α) < ϵ/3 z α < δ g(z) g(α) g(z) g m (z) + g m (z) g m (α) + g m (α) g(α) < ϵ/3 + ϵ/3 + ϵ/3 = ϵ 4-2 g n : D (n 0) g : D D D g n (z) dz g(z) dz (n ).. n N ( ϵ > 0)( N N)( n N)( z ) g n (z) g(z) < ϵ. g n (z) dz g(z) dz = (g n (z) g(z)) dz ϵ l() l() lim n g n (z) dz = 4-3 g n : D (n 0) g : D D () g D (2) D g n g g(z) dz

22 III () D D D g n D g n (z) dz = g(z) dz = 0 4- g g D (2) E D E D 2 r > 0 α E D(α, r) (α, r) D α E = (α, r) 2 g n(α) = 2πi g n n N g n(α) g (α) 2πi g n (z) (z α) 2 dz g (α) = 2πi g(z) (z α) 2 dz ( ϵ > 0)( N N)( n N)( z D) g n (z) g(z) < ϵ. g n (z) g(z) (z α) 2 dz ϵ 2π r 2 l() = ϵ r. ϵ/r α E g n g E. g n : D g : D E D g n E g E 4-3 g n g D g g n g D (Weiserstrass s Theorem) g n (z) = + z + z z n g(z) = /( z) D(0, ) g (z) = lim n g n(z) ( z) 2 = + 2z + + nzn + sin nx g n (x) = (x R) n g n g(x) = 0 R g n(x) = cos nx g (x) = 0 z g n (z) g(z) 4-4. D f : D α D R > 0 = (α, R) D z int() f(z) = f(α) + f (α)(z α) + f (α) (z α) 2 + 2! g n 2 {e n } E e n D /n { e n(k) } z D E z E E D D D D =

23 III f(z) α (Taylor expansion about α). g(z) = a n (z α) n 3 n=0 f n : D (n = 0,, 2,...) g n (z) := f 0 (z) + f (z) + + f n (z) z D g n (z) g(z) g(z) = f 0 (z) + + f n (z) +, g(z) = f n (z), g(z) = f n (z) n=0 n 0 g : D 4-5 D g n (z) = f 0 (z) + f (z) + + f n (z) g(z) = f n (z) f n (z) dz = n=0 n= f n (z) dz. n=0 g = lim n g n = g(z) dz = lim g n (z) dz = lim n n = 4-2 n f n (z) dz = lim k=0 n n k=0 f n (z) dz = z 0 int() r := z 0 < R z z 0 α z α = r R < f(z) z z 0 = f(z) (z α) (z 0 α) = f(z) z α z 0 α z α = f(z) z α n=0 ( ) n z0 α. z α 3

24 III f n (z) := f(z) z α ( ) n z0 α g n (z) := f 0 (z) + + f n (z)g(z) := f(z) z α z g(z) = lim g n (z) = n=0 z z 0 f n (z) f(z) M = M() z β = z 0 α z α g n (z) g(z) = f n+ (z) + f n+2 (z) + = f(z) z α (βn+ + β n+2 + ) M R β n+ β M β n+ R( β ) β = r/r < z g n g 4 f(z 0 ) = f(z) dz = 2πi z z 0 2πi = 2πi n=0 f n (z) dz n=0 n=0 f n (z) dz 4-5 = ( ) n f(z) 2πi n=0 z α z0 α dz z α ( ) f(z) = dz (z 2πi n=0 (z α) n+ 0 α) n f (n) (α) = (z 0 α) n. n! (/). z e z = + z! + z2 2! z3 + sin z = z 3! + z5 z2 cos z = 5! 2! + z4 4! α = 0 z < R R = (0, R) 4-4. e z e 2 = n! (z 2)n n=0 e z = e 2 e z 2 = e 2 n=0. (z 2) n n! f(z) A f(z) = a n (z α) n B f(z) = b n (z α) n n a n = b n. f(z) = z 2i 4

25 III () z = 0 (2) z = i () f(z) = 2i z < z < 2 2i f(z) = 2i ( z n = 2i) n=0 n=0 z 2i. (2i) n+ zn ( z < 2) (2) () f(z) = (z i) i = i z i i < z i < z i i. f(z) = ( ) n z i = i i i n+ (z i)n ( z i < ) n=0 n=0. a n (z α) n (radius of convergence) z α < R z α > R R () 2(2) z = 0 z 4 () (2) e z2 (3) cos z z + 3 z 5-2 f(z) = z 2 α + () α = 0 (2) α = 2i (3) α = 5-3 g n, g : g n (z) := () g n g R sin nz g(z) := 0 n (2) g n g z D g n g

26 III : November 5, 203 Version :. (/) D D(α, r) α r D(α, r) = int((α, r)) 4-4. D f : D α D R > 0 = (α, R) D z int() f(z) = n=0 f (n) (α) (z α) n n! f(z) α (Taylor expansion about α) 2 (z α) n f (n) (α) n! = 2πi f(z) dz (z α) n+ e z z 2 = z 2 + z + 2! + z 3! + e/z = + z + 2!z + 3!z 3 + z = 0 z 0 z = α D = {z R < z α < R 2 } f : D = (α, R) ( R < R < R 2 ) n a n a n := 2πi f(z) dz (z α) n+ z D f(z) = + a 2 (z α) 2 + a z α + a 0 + a (z α) + a 2 (z α) 2 +.

27 III f(z) α (Laurent expansion about α).. R = 0 R 2 = a n R. {a n } n Z f 0 < z α < R 2 k a k 0 f(z) = a k (z α) k + + a z α + a 0 + a (z α) + a 2 (z α) 2 + f z = α k (pole of order N) ez D = {0 < z < } α = 0 z2 e z z 2 = ) ( z 2 + z + z2 2! + z3 3! = z 2 + z + 2! + z 3! + z = z 2 D = {0 < z < 2} α = 0 (z 2) z 2 (z 2) = z 2 2( z/2) = ) ( 2z 2 z2 z = 2z 2 4z 8 z 6 z = z 0 D 0 := (z 0, ϵ), := (α, r ), 2 = (α, r 2 ) ϵ > 0 R < r < z 0 α < r 2 < R 2 0,, 2 f(z 0 ) = f(z) dz 2πi 0 z z 0 = ( ) 2πi 0 2πi 2 2 = a m 2πi (z 0 α) m, a m = 2πi m = a n (z 0 α) n, a n = 2πi 2 2πi n 0 f(z) dz (z α) m+ f(z) dz (z α) n+

28 III z z α < z 0 α = f(z) 2πi 2πi z 0 α f(z) ( ) n z α dz z 0 α z 0 α z α dz = 2πi z 0 α = 2πi n 0 n 0 ( f(z) (z α) n ) (z 0 α) n+ dz 4-5 = ( ) f(z) dz 2πi (z α) n (z 0 α) n+ = ( ) f(z) dz 2πi (z α) m+ (z 0 α) m. n 0 n+=m 2 5-2D α D f : D {α} D α f(z) = n Z a n(z α) n α D f(z) dz = 2πi a. a z α α., = (α, r) r D r n (z α) n dz = 0n = (z α) n dz = 2πi f(z) dz = ( ) a n (z α) n dz = ) (a n (z α) n dz = 2πi a. n Z n Z =. f(z) a a (z α) f α (residue) Res(f(z), α) 5-2 f(z) dz = 2πi Res(f(z), α) e z. z = 0 z2 e z z 2 = z 2 + ( ) e z + Res z z 2, 0 e z =. = (0, ) dz = 2πi = 2πi. z2

29 III = (0, ) D {0} 2πi 4 = πi 2 z 2 (z 2) = + Res 4z z 2 dz D = { z < 2} (z 2) ( ) z 2 dz = 2πi Res (z 2) z 2 (z 2), 0 ( ) z 2 (z 2), 0 = 4.. g(z) = z 2. g (0) =! 2πi g(z) z 2 dz z 2 (z 2) dz = 2πi g (0) = 2πi 4 = πi 2. {α, α 2,, α n } int() f int() {α, α 2,, α n } ( ) f(z) dz = 2πi n Res(f(z), α k ). k= f. r > 0 k = (α k, r),..., n D = + + n 5-2 = 2πi Res(f(z), α k ) ( k n) k f f(z) = n Z a n(z α) n a n (z α) n dz n Z a n (z α) n dz = n Z 4-5 f(z) f(z) = F (z) + G(z); F (z) := a n (z α) n, G(z) := n=0 n= a n (z α) n 5- D z 0

30 III (a) F n (z) := (b) G n (z) = n a k (z α) k F (z) k=0 n k= a k G(z) (z α) k (a) 5-3. F (z) = a 0 + a z + + a n z n + z 0 z < z 0 z F (z) 0 < r < z 0 F n (z) = a 0 + a z + + a n z n E(r) := {z z r}. r < z 0 r F (z 0 ) = a 0 + a z a n z0 n + a n z0 n 0 (n ) M > 0 n 0 a n z0 n M n ( ) z r < z 0 z a n z n = a n z0 n z n r z 0 M z 0 { a 0 + a z a n z0 n M + r ( ) n } r z M z 0 r/ z 0. n F (z) F (z) r (< z 0 ) z < z 0 F (z) E(r) z E(r) F (z) F n (z) = a n+ z n+ + a n+2 z n+2 + = lim N a n+z n+ + + a n+n z n+n a n+ z n+ + + a n+n z n+n M { ( ) n+ ( ) } n+n r r + + M (r/ z 0 ) n+ z 0 z 0 r/ z 0. N F (z) F n (z) M (r/ z 0 ) n+. z n r/ z 0 E(r) F (a) (a). 0 < R < R 2 R < z α < R 2 z f(z) = F (z) + G(z) = n Z a n(z α) n z = z 2 F n (z 2 ) = n k=0 a k(z 2 α) k n F (z 2 ) R < r < z 2 α < R 2 r = (α, r) F n (z) F (z) (b) 6- (a) (b) 2 α

31 III D = {z R < z α < R 2 } 5- f(z) = n Z a n(z α) n f D f(z) = m Z b n(z α) m a n = b n n (a) 5-3 (b) = (α, R) R < R < R 2 f(z) = n Z b n(z α) n D a n = 2πi = 2πi = m= f(z) (z α) n+ dz (z α) n+ b n 2πi = { m= b n (z α) m } dz (z α) n m+ dz = b n (b) () F (z) = a 0 + a /z + + a n /z n + z 0 z > z 0 z F (z) r > z 0 F n (z) = a 0 + a /z + + a n /z n E (r) := {z z r} (2) (b) 5-2 (a) = (α, r) 6-2 f(z) = z z 2 + z 2 () D = {z z < } (2) D = {z < z < 2} (3) D = {z z > 2} ( ) z Hint: f(z) = z 2 +z 2 = 3 z + 2 z+2 (2) z = z /z 2 z+2 = +z/2

32 III : November 22, 203 Version :. (/5). f(z) α f(z) = n Z a n(z α) n a (z α) f α (residue) Res(f(z), α) {α, α 2,, α n } int() f int() {α, α 2,, α n } ( ) n f(z) dz = 2πi Res(f(z), α k ). k= f. α f {z 0 < z α < R} 5-() f(z) = n Z a n(z α) n (). m N a m = 0 f(z) = a n (z α) n α (removable singularity) n=0 z 0 f(z) = sin z2 z 2 {0} = {z 0 < z < } f(z) = ( ) z 2 z 2 z6 3! + z0 = z4 5! 3! + z8 5! z = 0 f f(0) := f (2) k N a k 0 a m = 0 m k f(z) = a k (z α) k + + a z α + a 0 + a (z α) + a 2 (z α) 2 + α k (pole of order k) (3). a k 0 k N α f (essential singularity) f(z) = (sin z 2 )/z 2 z = 0 g(z) = ez z cos zi z = 0 sin(sin z) z

33 III e /z = + z + 2!z 2 + 3!z 3 + {0} z = 0 ()(2)(3) (isolated singularity). f(z) = sin(π/z) {0, ±, ±/2, ±/3,...} {±, ±/2, ±/3,...} f 2 z = 0 f I = f(z) dz (i) f(z) α,, α n (ii) Res(f(z), α k ) (iii) I = 2πi (). = (0, 2) I = e iz z 2 dz + 3 f(z) = eiz z 2 {±i} z = ±i + I = f(z) dz = 2πi{Res(f(z), i) + Res(f(z), i)}.. f Res(f(z), i) (z α) eiz z 2 + = { } e iz 2i z i eiz z = i z + i e iz z 2 + = eiz z + i z i = z i f(z) = e /z z 2 sin z 4 a n (z α) n f a

34 III w = z i z = w + i f(z) = e i(w+i) (w + i) 2 + = e e iw = 2ei w { + = 2ei +0 w Res(f(z), i) = /(2ei). 2iw + w 2 = 2ei w + w e iw 2i ( w ) ( + w ) 2 + } 2i 2i ( + iw + (iw)2 2! ) +. 6-f(z) = g(z) g(z) z = α (z α) k Res(f(z), α) = (k )! g(k ) (α). k = Res(f(z), α) = g(α). α k g(α) 0 g k f(z) = g(z) g(z) z = α = (α, r) r (z α) k 2 Res(f(z), α) = f(z) dz = g(z) 2πi 2πi (z α) k dz = (k )! g(k ) (α). 6- Res(f(z), ±i) g(z) = eiz z + i g(i) = e 2i. h(z) = g z = i 6- Res(f(z), i) = eiz z i Res(f(z), i) = h( i) = e 2i.. e iz ( e I = z 2 dz = 2πi + 2i e ) = π(e e). 2i 5 I = 2π cos θ dθ = π 2 Step. z = e iθ (0 θ 2π) z = (0, ) sin θ = z z 2i, cos θ = z + z, dz = ie iθ dθ dθ = dz 2 i z. 5 t = tan θ/2

35 III Step 2. I = z + z 2 dz i z = 2 i Step 3. f(z) f(z) = z = /3 I = 2 ( i 2πi Res f(z), ). 3 3z 2 + 0z + 3 dz 6-2 α = /3g(z) = k = 3(z + 3) ( Res f(z), ) ( = g ) = (z + /3)(z + 3) I = 2 i 2πi 8 = π I = + x 4 dx = π 2 f(z) = + z 4 = + z 4 dz α = eπi/4, β = e 3πi/4 = 2πi {Res(f(z), α) + Res(f(z), β)} f(z) = g(z) Res(f(z), α) = g (z). z = α g(z) = z 4 + ( ) Res + z 4, α = 4α 3 = α 4 = i 4 2. α 4 = ( ) Res + z 4, β = 4β 3 = β 4 = i Mathematica { log ( x 2 2x + ) + log ( x 2 + 2x + ) 2 tan ( 2x ) + 2 tan ( 2x + )} /(4 2)

36 III ( i = 2πi i ) 4 = π. 2 2 I = + J R = R J R := {z = x R x R} { } R := z = Re iθ 0 θ π = J R + R. π R 0 2 R ( ) I = lim = lim = lim = π R J R R R R R 2 z = R > z 4 + z 4 = R 4 > 0 R f(z) R 4 f(z) dz l( R) R R 4 = πr R 4 0 (R ). I = π α h(z) h(α) 0 f(z) = g(z) = (z α)h(z) 6- Res(f(z), α) = h(α). g (z) = h(z) + (z α)h (z) g (α) = h(α) e z ( () (2) 2z2 + z πi (z ) 2 (3) z 2 exp ) z 7-2. z () dz (2) dz (3) (0,) sin z (0,3) (z )(z + 2) e z (4) z 2 (z 2 dz (5) 4) z 2 (z 2 dz (6) 4) (,2) (,2) (0,2) (i,) (4) z 2 sin z e z (z )(z 3) dz z 4 dz

37 III : November 29, 203 Version :. (/22). f(z) = I = cos x + x 2 dx = π e eiz 2 + z2 = f(z) dz = R + J R, R := { z = Re it 0 t π }, J R := {z = t R t R}. z = i = 2πi Res(f(z), i). f(z) = z i e iz z + i eiz z + i z = i 6- ei i Res(f(z), i) = i + i = e 2i. = 2πi e = π 2i e. 2 R 0 z = R > z 2 + R z 2 = R 2 > 0 z = x + yi R y 0 e iz = e i(x+yi) = e y R f(z) = e iz + z 2 R 2 f(z) dz l( R) R R 2 = πr R 2 0 (R 0). e ix R + x 2 dx = R R = J R R 0 J R cos x R + x 2 dx + i R sin x sin x dx + x2 + x 2 ( ) I = lim = lim = lim = π R J R R R R R e..

38 III I = 0 sin x x dx = π f(z) = eiz z = + J J 2 := { z = Re it 0 t π } J := {z = t R t r} { } 2 := z = re i(π t) 0 t π J 2 := {z = t r t R} r < R r 0, R int() f f(z) dz = 0 + J J 2 = 0. 2 i R r sin x x dx = (R ) : z = Re it (0 t π) dz = ire it dt = iz dt dz/z = idt f(z) dz = e iz z dz π = e ireit idt 0 π = e ir(cos t+i sin t) π dt = e R sin t e ir cos t dt 2 0 sin x x dx = 0 π ( ) e R sin t dt 0 0. (R 0) 3 sin x/x x = 0 I = lim r +0 R sin x/x 2I = lim r +0 R ( r R sin x x dx + R r 0 sin x x dx ) R r sin x dx x

39 III ( ) π 0 e R sin t e ir cos t dt t e R sin t e ir cos t 0 π e R sin t e ir cos t = πi (r 0) 2 π 0 e R sin t dt 7- α f {z 0 < z α < R} α r := { z = α + re it 0 t π } lim f(z) dz = πi Res(f(z), α). r 0 r. r = (α, r) r 0 f(z) dz = 2πi Res(f(z), α) r r α f(z) = a + g(z) g(z) z = α z α a f(z) dz = r z α dz + g(z) dz r r π 0 r a re it ireit dt = πia = πi Res(f(z), α).. 2 g(z) z = α r r g(z) g(α) r g(z) g(α) + g(z) dz r = ( g(α) + ) l( r) = ( g(α) + ) πr 0. (r +0) f(z) = e iz /z, α = 0 0 f(z) = ) ( + iz + (iz)2 + = +0 z 2! z Res(f(z), 0) = r = 2 lim r +0 f(z) dz = lim r +0 2 f(z) dz = πi Res(f(z), 0) = πi. r r +0R 2 i 0 sin x x dx = 0 + ( πi) I = i 2 πi = π 2.

40 III D D f D (meromorphic) P := {α, α 2, } D f D P α k f 4 P = f D. f(z) = ez f P = {0, } P z + z2. f(z) = f P = Z Z sin πz. f(z) f(α) = 0 α f ( zero) f z = α α k f(z) = a k (z α) k + a k+ (z α) k+ + k α (order) α k f(z) = a k (z α) k + + a z α + c 0 + a (z α) + k α (order) α k f(z) = 0 α k f(z) = 5 7-2D D f D f (z) dz =. 2πi f(z). 4 α k P α k P D lim z αk f(z) = 5

41 III int() int() α,..., α N β,..., β M r > 0 n := (α n, r), D n := D(α n, r) ( n N) m := (β m, r), D m := D(β m, r) ( m M) r int() f (z)/f(z) D := int() N D n n= 6 f (z) 2πi f(z) dz = N n= f (z) 2πi n f(z) dz () M m= + D m M m= f (z) 2πi f(z) dz. m (2) (). n =,..., N α n k n D n g = g n g(α n ) 0 f(z) = (z α n ) k n g(z) f g f (z) f(z) = k n(z α n ) kn g(z) + (z α n ) kn g (z) (z α n ) k = k n + g (z) n g(z) z α n g(z). g(α n ) 0 g r D n g(z) 0 g (z)/g(z) n D n 02-3 f (z) 2πi n f(z) dz = 2πi n () = k n dz + g (z) z α n 2πi n g(z) dz = k n + 0 = k n. N k n = n= 6 D f f D f /f D f /f D

42 III (2). () m =,..., M β m l m D m h = h m h(β m ) 0 h(z) f(z) = (z β m ) l m f h f (z) = l m (z β m ) lm+ h(z) + (z β m ) h (z) lm f (z) f(z) = l m + h (z) z β m h(z). () g h(β m ) 0 h (z)/h(z) m D m f (z) 2πi f(z) dz = l m dz + h (z) 2πi m z β m m 2πi h(z) dz = l m + 0 = l m. m M (2) = ( l m ) = m= () 4 2 i (2) 4 (a) π 0 R r e R sin t dt = 2 sin x x dx = + 2 π/2 0 e R sin t dt (b) t [0, π/2] sin t 2t/π 8-2. () I = (2) I = π 0 e R sin t dt 0 (R 0) x 6 dx (Hint x sin x (x 2 + ) 2 dx = π zeiz (Hint 3 f(z) = 4e (z 2 + ) (HintE := int() int() E

43 III : December 6, 203 Version :. (/29). D f D (meromorphic) P := {α, α 2, } D f D P α k f P = f D 7-2D D f D f (z) dz =. 2πi f(z) 7.2 w = f(z) f() arg w w = f(z) dw = f (z)dz f (z) 2πi f(z) dz = 2πi f() w dw w f() w = 0 f f() {0} w f() w = e x+yi (x, y R) f() : w(t) = : z = z(t)f() : w = f(z(t)) t

44 III e x(t)+i y(t) (0 t ) x(t)y(t) t arg w(t) = y(t) t t x := x(t + t) x(t) y := y(t + t) y(t) w(t + t) = e x(t+ t)+i y(t+ t) = e (x(t)+i y(t))+ x+i y = w(t) e x e i y arg w(t + t) = arg w(t) + y w = e x+yi dt dw = e x+iy dx + e x+iy i dy dw w 2πi f() = dx + i dy w dw = + 2πi dx() 2π dy(2) w(t) = e x(t)+i y(t) f() x(t) + i y(t) () 0 0 dx = x (t) dt = x() x(0) 0 f() w(0) = w() e x() = e x(0) x() = x(0) (2) () dy = y (t) dt = y() y(0) 0 e i y() = e i y(0) y() = y(0) + 2mπ (m Z) 2mπ w(t) f() y(t) = arg w(t) 2 m = m = 0 y(t) f (z) 2πi f(z) dz = 2πi w dw = m f() : w = 0 f() 2 y = arg w dy = d arg w 2π 2π f()

45 III RouchéD f, g : D D int() D f(z) > g(z) ( f(z) + g(z) ) = ( f(z) ) f f g f > g. z 3 +3z + = 0 f(z) = z 3 g(z) = 3z + z = 2 f(z) = 8 g(z) = 3z + 3 z + = 7 = (0, 2) z < 2 f(z) + g(z) z < 2 f(z) f(z) = z 3 3 z = 0 z < 2 z 3 + 3z + = 0 3 z 3 + 3z + = 0 3 D(0, 2). z = g(z) = 3z + 3 z = 2 f(z) = = (0, ) z < g(z) + f(z) z < g(z) g(z) = 3z + z = /3 z < z 3 + 3z + = 0 2 z < w = h t (z) := f(z) + tg(z) (0 t ) z 2 h t w t = 0, 0.2, 0.4 t = 0.6, 0.8,.0

46 III z 00 5z 3 + = 0 f(z) = 5z 3 g(z) = z 00 + z = f(z) = 5 g(z) z 00 + = 2 f(z) = D f n f f g n (z) := f(z) f n (z) f n (z) = f(z) + g n (z) g n (z) 0. z ( f(z) > g(z) 0 f(z) f(z) + g(z) = f(z) + g(z) ) h(z) = + g(z) f(z) f(z) {f(z)h(z)} dz = f (z) 2πi f(z)h(z) 2πi f(z) dz = 2πi ( f (z) f(z) + h (z) ) dz = + h (z) h(z) 2πi h(z) dz 0 w = h(z) (z ) w = h(z) = g(z) f(z) < h h() w w = w /w 3 h (z) 2πi h(z) dz = dw = 0. 2πi h() w f(z)g(z) n n + α,, α n+ f(α k ) = g(α k ) (k =,..., n + ) z f(z) = g(z) h(z) := f(z) g(z) n h(α k ) = 0 (k =,..., n + ) h(z) = (z α )(z α 2 ) (z α n+ )Q(z) 3 h()

47 III Q(z) Q(z) n + Q(z) = 0 Identity Theorem D f(z)g(z) α D {z n } n=0 D {α} z n α (n ) f(z n ) = g(z n ) (n =, 2,...) z D f(z) = g(z) f :, x + yi f(x + yi) ( x R) f(x) = e x f f(x + yi) = e x+2yi, f(x + yi) = e x ( + yi), f(x + yi) = e x cos y, f ( z ) f(z) = e z f(x + yi) = e x+yi g(z) = e z, z n = /n R, α = 0 D D(α, r) f g r > 0 r = D f g D f(z) = z3 (z )(z 2 4) (z 2 + ) 2 r = f (z) /2, 3/2, 5/2 dz f(z) (0,r) 4

48 III z 00 5z 3 + = z <.02 log 0.02 = , log 0 7 = g(z), h(z) z = α g(α) 0 f(z) = g(z) g(α) z = α Res(f(z), α) = h(z) h (α) Hint 6-2

49 III : December 3, 203 Version :. (2/6) Identity Theorem D f(z)g(z) α D {z n } n=0 D {α} z n α (n ) f(z n ) = g(z n ) (n =, 2,...) z D f(z) = g(z) h(z) = f(z) g(z) n z n α( n N) f(z n ) = 0 = ( z D) f(z) = 0 E z f(z) = 0 E f 0 9- α D n = 0,, 2, f (n) (α) = 0 D f(z) 0. β D D α β : z = z(t), (0 t ); α 0 := α, t 0 := 0 R 0 (0, ] α 0 D D = R 0 = D(α, ) := D(α 0, R 0 ) f z D(α 0, R 0 ), f(z) = n= f (n) (α 0 ) (z α 0 ) n n! α 0 = α f (n) (α) = 0 D(α 0, R 0 ) f 0 D = D z(0) = α, z() = β k 0 D(α k, R k ) f 0 α k+ α k+ D(α k, R k ) α k+ = z(t k+ ) 0 t k < t k+ < R k+ > 0 α k+ D D(α k+, R k+ ) z D(α k+, R k+ ), f(z) = E 0 n= f (n) (α k+ ) (z α k+ ) n n!

50 III α k+ D(α k, R k ) D(α k, R k ) f 0 n = 0,, 2, f (n) (α k+ ) = 0 D(α k+, R k+ ) f 0 t k N N β D(α N, R N ) N R k 0 (k ) D f(β) = 0 β ( 9-) f D f(α) = f( lim n z n) = lim n f(z n) = 0. D f 0 9- k N f (k) (α) 0 k α f(z) = (z α) k g(z) g(z) α g(α) 0 z = z n z n α g(z) 0 = f(z n ) = (z n α) k g(z n ) g(z n ) = 0 g(α) = g( lim n z n) = lim n g(z n) = 0. g(α) 0 D f 0 f : D D f α r > 0 z D(α, r) {α}, f(z) 0. D(α, r) α z n α < /n, f(z n ) = 0 z n Maximum Principlef : D α D ( z D), f(α) f(z) f D Maximum Modulus Principle 9-2f : D E D z f(z) E α E ( z E), f(α) f(z)

51 III f(z) = z 2 x (, ) 0 < f(x) f(0) = 2 (, ) x = 0 f(x) E = D(0, ) = {z z } f(z) + z 2 2 z = ±i f(z) = 2 ±i E f(z) = z zz + f(z) = z z z = f f z 2, sin z, e z 2 α D r > 0 = (α, r) int() D z f(z) z α = f(z) f(α) r r f(α) = f(z) 2πi z α dz 2π f(α) l() = f(α) r f(z) = f(α)

52 III r r < r int() f(z) = f(α) 2 int() f f(z) = f(α)f D 9-2 z f(z) D E D E E E = E E, E E = α E f(α) E α D f D f(z) = f(α)f D E 3 D = D(0, ) f : D D f(0) = 0 f(z) z f (0) z D f(z) = e iθ z (θ R) f 00 f (0) f (0) > f(z) = az + O(z 2 ) (a = f (0) ) g(z) = f(z)/z = a + O(z) z r < 9-2 z 0 z 0 = r g(z) g(z 0 ) = f(z 0 ) / z 0 /rr < z D g(z) f(z 0 ) = z 0 z 0 D g(z 0 ) = f (0) = g(0) = f (0) g(0) = g(z) g(z) = f(z)/z = e iθ D, D 2 f : D f 2 : D 2 D D 2 f = f 2 f 2 f (analytic continuation) f f f (z) = + z + z 2 + D = D(0, ) f 2 (z) = z D 2 = {} f D D 2 D f = f 2 s = σ + it (σ, t R) ζ (s) := + 2 s + 3 s + 2 f(x + yi) = u + vi f(z) 2 = u 2 + v 2 2uu x + 2vv x = 2uu y + 2vv y = 0 u x = v x = u y = v y = f f 2 f f 2 5

53 III n s := e s log n log n > 0 ζ D := {s Re s = σ > } n s e s log n ζ : D ζ ζ : {} ζ s = ζ n Nζ( 2n) = 0 ζ (Riemann s zeta function) 6 00 (The Riemann Hypothesis) ζ(α) = 0 = α = 2n (n N) Re α = / D f, g : D D f int() f α, α 2,, α N α n ( n N) k n g(z) f (z) 2πi f(z) dz = k g(α ) + + k N g(α N ). (g(z) = 0-2 f(z) = z n + a n z n + + a z + a 0 () R ( z R = f(z) R n /2 Hint: f(z) z n a n + + a z z n + a ) 0 z n (2) f(z) g(z) = /f(z) g(z) 2/R n R () g(z) = 0 6 Γ( s) ( z) s ζ(s) = 2πi e z dz Γ ( z) s = e (s ) log( z) Im log( z) < π

54 III203 - : December 20, 203 Version :. (2/3) () (2) () w = zn w = z /n w = log z D, D 2 f : D f 2 : D 2 D D 2 f = f 2 f 2 f (analytic continuation) f f xy R 2 p r D(p, r) ( ) 2 ( ) D R 2 F : D R 2 x u F : p = p 0 D y v det ( u x v x u y v y ) p=p0 0 q 0 = F (p 0 ) r > 0 G : D(q 0, r) R 2 () G(q 0 ) = p 0. (2) q D(q 0, r), F G(q) = q. (3) p G(D(q 0, r)), G F (p) = p. ( ) ( ) u x (4) G : F v y ( x u y u x v y v ) ( u x v x u y v y ) ( ) 0 =. 0 () (2) 2 f f 2 3

55 III203-2 D f : D f : z w z 0 D f (z 0 ) 0 w 0 = f(z 0 ) r > 0 g : D(w 0, r) () g(w 0 ) = z 0. (2) w D(w 0, r), f g(w) = w. (3) z g(d(w 0, r)), g f(z) = z. (4) g : w z f dg dw df dz =. ( z ( = x+yiw = u+vi f(x+yi) = u+vi x u 2 F : f y) v) z 0 = x 0 + y 0 i ( ( ) ux u det y ux v = det x = u v x v y) 2 v z=z x u x + v 2 x = u x + v x i 2 z=z0 = f (z 0 ) 2 0 x 0 z=z z=z0 0 ( ( ) x x0 = z = z y) y 0 2 () 0 (3) r > 0 D(w 0, r) g : D(w 0, r), g(u + vi) = x + yi () (3) (4) g(u + vi) = x + yi 2 (4) ( ) ( ) ( ) ( ) xu x v ux u = y ux v = x ux v = x y u y v v x v y v x u x u 2 x + vx 2 v x u x x u = y v, x v = y u g w 2 = z dg dw = g u = x u + y u i = u 2 x + vx 2 (u x v x i) = u x + v x i = f (z) = df/dz. z = w 2 = f(w) w 0 z 0 z = re iθ (r > 0, 0 θ < π) w = ± re iθ/2 w z /2 = e (log z)/2 w = z z = w 2 dz dw = 2w 0 w 0 0 z 0 = w0 2 z = f(w) w = g (z) w 0 z 0 = ( w 0 ) 2 f w = g 2 (z) g 2 (z) = g (z)

56 III203-3 z 0 = g g 2 g () =, g 2 () = = (0, ) z g (z) g 2 (z) z g 2 (z) g (z) : z w g g () = g 2 g 2 () = Mathematica z 2 w S =z 2 w 4 z z := {0} 2 4

57 III203-4 S 5 f g g 2 S w w w = 0 S z = 0 S w = z (Riemann surface) z = 0 S (branched point) z S (branch) w = g(z) S g z g w w = z /n w = z /n z z = w n =: f(w) n w z = 0 n w n z n w = log z w = log z z 0 z = e w =: f(w) w w z w = df dw = d(ew ) dw = ew 0 z k Z k z = w = 2kπi f = exp g k g k (, 0] w S k := {w (2k )π < Im w < (2k + )π} g k g k± w = log z 5

58 III203-5 w = log z 6 (, 0] k f = exp g k : (, 0] S k df dw dg k dz = dg k dz = df/dw = e w = z. g k /z k /z (, 0] g k (z) g k () = z ζ dζ g k(z) = z dζ + 2kπi ζ z (, 0] Log z g 0 (z) = Log z g k (z) = Log z + 2kπi Log ( + z) = z z2 2 + z3 3 z4 + ( z < ) 4 (, 0] D(, ) z < Log ( + z) = +z ζ dζ + z + η = ζ z < Log ( + z) = z 0 z + η dη = 0 ( η) n dη = n=0 z n=0 0 ( η) n dη = ( ) n z n+ n=0 n + 6 w = Log z

59 III203-6 = ( η) n η < z < ( + z) α z < α 7 n=0 ( + z) α αlog (+z) := e e αlog (+z) = e α(z z2 /2+z 3 /3 ) = + α(z z2 2 + z3 3 ) + α2 (z z2 2 + z3 3 )2 2! 2 + ( + z) α α(α ) = + αz + (z α) 2 + 2! ( ) α α(α ) (α n + ) := n n! z < α ( ) ( + z) α α = z n n n= w = z 2 S S w Hint: z 2 x > 7 ( + z) α = e α log(+z)

60 III : January 0, 204 Version :. (2/20) XY Z R 3 S := { (X, Y, Z) R 3 X 2 + Y 2 + Z 2 = } (0, 0, ) N XY Z = 0 { (x, y, 0) R 3 x, y R } = {x + yi x, y R} (x, y, 0) x + yi XY N x + yi (x, y, 0) S N P = (X, Y, Z) P = (X, Y, Z) x + yi S N - z = x + yi P = (X, Y, Z) X = 2x + z 2, Y = 2y + z 2, Z = + z 2 + z 2 x = X Z, y = Y Z

61 III P = (X, Y, Z) S N z = x + yi S N z z P z N N point at infinity N = (0, 0, ) S N S { } S { } (the Riemann sphere) Ĉ 2 Ĉ Ĉ S R3 - Ĉ z, z 2 d(z, z 2 ) := 2 z z 2 + z 2 + z 2 2, d(z, ) := 2 + z 2. d(, ) := 0 Ĉ z Ĉ r > 0 } B(z, r) := {w Ĉ d(w, z) < r 3 () z n, α n z n α 0 d(z n, α) 0 (2) U U U (Ĉ, d) (3) U Ĉ R > 0 {z z > R} U. z n, α Ĉ d(z n, α) 0 (n ) {z n } α z n α (n ) lim n z n = α z w n {w n } lim (z + w n) = lim (w n + z) = lim w n n n n z + = + z = z z + = + z = 2 P 3

62 III z z = z =, z = 0, z 0 =., ±, /, { αzz + βz + β z + γ = 0 α, γ R, β, β 2 αγ > 0 ( ) α 0 ( ) z + β α 2 = β2 αγ α 2 (> 0) β/α ( β 2 αγ)/α 2 z 0 r > 0 z z 0 = r (z z 0 )(z z 0 ) = r 2 zz z 0 z z 0 z + z 0 z 0 r 2 = 0 α =, β = z 0, γ = z 0 z 0 r 2 β 2 αγ = r 2 > 0 ( ) α = 0 β = p + qi ( ) βz + β z + γ = 0 2Re ( βz) + γ = 0 2(px + qy) + γ = 0 β 2 αγ = β 2 > 0 (p, q) (0, 0) ( ) ax + by + c = 0 (a, b) (0, 0) α = 0, β = (a + bi)/2, γ = c β 2 αγ = (a 2 + b 2 )/4 > 0 ( ) -3z ( ) S (X, Y, Z) Π Π : 2pX + 2qY + (α γ)z + (α + γ) = 0. ( ) β = p + qi β 2 αγ > 0 ( S (0, 0, 0) Π) < - z = X + Y i Z ( ) X2 + Y 2 = Z 2 ( ) ((0, 0, 0) Π) < α + γ (2p)2 + (2q) 2 + (α γ) 2 < (α + γ)2 < 2β 2 + (α γ) 2 4( β 2 αγ) > 0. R 3 Π S N α = 0 N Π

63 III Ĉ Ĉ a, b, c, d, ad bc 0 T (z) = az + b cz + d (Möbius transformation) (fractional linear transformation) X f : X f(x) = Y (homeomorphism, topological map) f f f -5 T (z) = az + b cz + d T ( ) = a ( c, T d ) = c Ĉ Ĉ c = 0 T (z) = αz + β (α 0) T () = T ( ) = (complex affine transformation) 4 T Ĉ 4

64 III T (z) = 2z = 2z z + 0, T (z) = z + = z + 0 z + τ(z) = z = 0 z + T z + 0 τ z = ± (±, 0, 0) S w = T (z) = (az + b)/(cz + d) z = S(w) = (dw b)/( cw + a) S( ) = d/cs(a/c) = () w ĈT S(w) = w (2) z ĈS T (z) = z (3) T, S Ĉ S = T T -6 Ĉ Ĉ w = T (z) = (az + b)/(cz + d) c = 0 w = a d z + b d c 0 w = a c z a d z =: z z + b d = w. + bc ad c 2 z + d/c z z + d c =: z z =: z 2 bc ad c 2 z 2 =: z 3 a c + z 3 = w. T 3 () z λz (λ 0) (2) z z + l (l 0) (3) z z 3 () w = T (z) = λz (λ 0) ( ) Ĉ ( ) λ λ α(λz)(λz) + β λλz + βλ λz + λλγ = 0 αw w + βλw + βλ w + λ 2 γ = 0 5 z /z S 2 P P PP PP

65 III α = α, β = βλ, γ = λ 2 γ β 2 α γ = λ 2 ( β 2 αγ) > 0 ( ) w = T (z) { α w w + β w + β w + γ = 0 α, γ R, β, β 2 α γ ( > 0 ) Ĉ Ĉ (2)(3) f : D, w = f(z)d z 0 A 0 := f (z 0 ) 0 z z 0 f(z) f(z 0 ) = A 0 (z z 0 )( + o( z z 0 )) w w 0 A 0 (z z 0 ) z 0 f A 0 A 0 = A 0 e i arg A 0 f z = z 0 A 0 arg A 0 z = z 0 2 f 2 w = w 0 z = z 0 2 f w = w 0-7 3, 2, 3 3 0, 2, 3 3 4, 2, 3 2 α T (z) = 3 z α k = T ( k) (k =, 2, 3), 2 Ĉ T ( 0) T ( 4) T ( ) = : 2 3

66 III : January 24, 204 Version :. (/0) D f D f D f Ĉ α D w = f(z) k N w = f(z) = a k (z α) k + a k+ (z α) k+ + ( = a k (z α) k + a ) k+ (z α) + a k a k 0 ϵ > 0 z = α + ϵe it (0 t 2π) w = f(z) = a k ϵ k e ikt { + O(ϵ)} a k ϵ k e ikt z α f(z) 0 k Ĉ R3 w 0 k Ĉ k α D w = f(z) k N a k w = f(z) = (z α) k + a k+ (z α) k + + a 0 + = a ( k (z α) k + a ) k+ (z α) + a k a k 0 ϵ > 0 z = α + ϵe it (0 t 2π) w = f(z) = a k ϵ k e ikt { + O(ϵ)} a k ϵ k e ikt 2 a k /ϵ k e ikt t z α w = f(z) k f(z) a k ϵ k e ikt O(ϵ) 2 f(z) (a k /ϵ k ) e ikt O(ϵ)

67 III Ĉ k τ(z) = /z 3 τ f(z) = (z α)k = f(z) = a k (z α)k a k ( + a k+ ) (z α) + a k ) ( a k+ a k (z α) + = (z α) k a k+ (z α) k+ + a k a 2 k k f k k D Ĉ D f : D Ĉ (holomorphic map) α D α f(α) α f α f(α) = α τ f α = f(α) τ(α) = 0 f τ α = f(α) = τ(α) = 0 τ f τ D f D α D f(α) = f : D Ĉ f : D Ĉ f : D f ({ }) T : Ĉ Ĉ, T (z) = (az + b)/(cz + d) (ad bc 0) Ĉ (automorphism) R(z) = a mz m + + a z + a 0 b n z n + + b z + b 0 (rational functioin) R : Ĉ Ĉ (rational map) 4 R(z) = lim z 3z5 z 2 R(±) = R( ) = 3z 3 = R : Ĉ Ĉ /z2 3 ( + x) α = + αx + 4 R(z)

68 III () f : Ĉ Ĉ f : Ĉ Ĉ (2) f : Ĉ Ĉ f : Ĉ Ĉ (3) f : Ĉ = f (3) f f(ĉ) f() f f () = = /f(z) f( ) f P Ĉ P z 0 z 0 = g = τ f τ g g 0 z 0 g = τ f z 0 g 0 α,, α n α j k j f(z) = a kj (z α j ) kj + + a z α j + a 0 + a (z α j ) + a kj 0 h j (z) h j (z) f(z) {h (z) + + h n (z)} Ĉ (3) f 5 (2) = () = () f 2 2 w = z w = ( z 2 )( k 2 z) (0 < k < ) 5

69 III X (separable) E X E = X Ĉ X d 0 X = d 0 (z, w) = z w X = Ĉd 0 = d X 6 D Ĉ {f n : D X} E D f : E X ϵ > 0, N N, n N, z E, d 0 (f(z), f n (z)) < ϵ {f n : D X} f : D X D f n f f n (z) = nz z + n f n D = Ĉ f = id D Ĉ D X = Ĉ 0 (D, X) F 0 (D, X) (normal family) F {f n } n N F D {f n(k) } k N {f n(k) } f = lim k f n(k) F 0 (D, X) F F 6

70 III : January 3, 204 Version :. (/24). X d 0 X = d 0 (z, w) = z w X = Ĉd 0 = d X D Ĉ D D X = Ĉ 0 (D, X) 0 (D, X) F 0 (D, X) (normal family) F {f n } n N F D {f n(k) } k N 0 (D, X) F F (the Ascoli-Arzelà theorem, AAT) F 0 (D, X) { () z D F (2) z D F () : ( z D)( ϵ > 0)( δ = δ(z) > 0)( f F), w D d(z, w) < δ = d 0 (f(z), f(w)) < ϵ δ z f F D F (2) : ( z D)( E = E(z) X)( f F), {f(z) f F} E E z f F D F AAT AAT 2 3-(X = Ĉ ) F 0 (D, Ĉ) () z D F 2

71 III z D E(z) = Ĉ AAT (2) AAT D X D, X R n ( =) () (2) 5 A = (Q + Qi) D A A = D Ĉ A A = {a, a 2, a 3, } F {f n } n N a k A (2) {f n (a k )} n N k = N = {, 2, 3, 4, } N = {n, n 2, n 3, n 4, } {f n (a )} n N lim j f nj (a ) N N 2 = {n 2, n 22, n 23, n 24, } N {f n (a 2 )} n N2 N N N 2 N 3 N k k {f n (a k )} n Nk N k = {n k, n k2, n k3, n k4, } M = {n, n 22, n 33, n 44, } N k {f n (a k )} n M 3 M {f n } n M = {f nkk } k N D E D ϵ > 0 () 3 M = {0, 200, 3000,...} ( ζ E)( δ = δ(ζ) > 0)( f F), z D d(z, ζ) < δ(ζ) = d 0 (f(z), f(ζ)) < ϵ/6 N = {, 2, 3, 4,...} N = {0, 20, 30, 40,...} N 2 = {00, 200, 300, 400,...} N 3 = {000, 2000, 3000, 4000,...}

72 III E ζ, ζ 2,..., ζ l z E j {, 2,..., l} z B(ζ j, δ(ζ j )) 4 B(α, r) α r z, w B(ζ j, δ(ζ j )) f F d 0 (f(z), f(w)) d 0 (f(z), f(ζ j )) + d 0 (f(ζ j ), f(w)) ϵ/6 + ϵ/6 = ϵ/3 A D a j B(ζ j, δ(ζ j )) (j =, 2,..., l) {f n } n M = {f nkk } k N g k := f nkk k 0 µ, ν k 0 j =, 2,..., l d 0 (g µ (a j ), g ν (a j )) ϵ/3 z E z B(ζ j, δ(ζ j )) j {, 2,..., l} d 0 (g µ (z), g ν (z)) d 0 (g µ (z), g µ (a j )) + d 0 (g µ (a j ), g ν (a j )) + d 0 (g ν (a j ), g ν (z)) g µ F d 0 (g µ (z), g µ (a j )) d 0 (g µ (z), g µ (ζ j )) + d 0 (g µ (ζ j ), g µ (a j )) < ϵ/6 + ϵ/6 < ϵ/3 3 2 d 0 (g µ (a j ), g ν (a j )) < ϵ/3 d 0 (g µ (z), g ν (z)) < ϵ/3 + ϵ/3 + ϵ/3 < ϵ {g µ (z)} µ N X z E {f n } n M = {g µ (z)} µ N ( =) (= ) ()(2) F E ϵ > 0 z n, z n E f n F d(z n, z n) < /n d 0 (f n (z n ), f n (z n)) ϵ E {z n } {z n} α {f n } f : E X E d 0 (f n (z n ), f n (z n)) d 0 (f n (z n ), f(z n )) + d 0 (f(z n ), f(z n)) + d 0 (f(z n ), f n (z n)) 3 n ϵ/3 f f n E 2 n ϵ/3 () (2) z D U = {f(z) f F} {w n } U f n F d 0 (f n (z), w n ) < /n {f n } { } f n(k) { fn(k) (z) } lim w n(k) U 4 E {B(ζ, δ(ζ)) ζ E} {B(ζ j, δ(ζ j ))} (= )

73 III X = F (Montel s Theorem F 0 (D, ) F (2) E D F (2) f F K = K(E) f F f(e) K AAT (2) AAT ()(2) (2) (2) (2) α D F α D r > 0 D(α, r) D f F ζ.ζ D(α, r/2) = (α, r) f(ζ) f(ζ ) = 2πi f(z) z ζ dz 2πi f(z) z ζ dz = 2π f(z)(ζ ζ ) (z ζ)(z ζ ) dz (2) f M > 0 f(z) < M z z ζ r/2 z ζ r/2 f(ζ) f(ζ ) 2π M ζ ζ (r/2)(r/2) l() = 4M r ζ ζ. l() 2πr r M f F ζ = α F D D int() D D := {z z < } (the Riemann Mapping Theorem, RMT) D ψ : D D

74 III D D D ψ (z) 0 RMT z 0 D r > 0 T (z) = r(z z 0 ) D T (D) D 0 D D RMT F ()(3) f : D f(d) () f(d) D (2) f(0) = 0 (3) f : D f(d) f 0 (z) = z 3 f 0 F z D f F f(z) D D F F M := sup{ f (0) f F} f 0 F M sup M n M (n ) f n F M n f n(0) < M F {f n } D ψ ψ (0) = M 5 ψ F 6 z D f n (z) ψ(z) ψ(z) < () (2) ψ(0) = lim f n (0) = 0 (3) a, b D ψ(a) = ψ(b) =: p D F n (z) := f n (z) pψ(z) := ψ(z) p D int() a b Ψ Ψ(a) = Ψ(b) = 0 Ψ (z) 2πi Ψ(z) dz 2 F n F n F n(z) 2πi F n (z) dz F n F n Ψ Ψ F n(z) Ψ (z) dz dz 2πi F n (z) 2πi Ψ(z) ψ(d) = D α D T α : w w α D D 7 αw α D ψ(d) ψ(z) α ϕ(z) := (= T α (ψ(z))) αψ(z) ψ(0) = 0 α 0 0 ϕ : D ϕ(d) D ϕ(z) ϕ(0) Φ(z) := ϕ(0)ϕ(z) = ϕ(z) α (= T ϕ(0) (ϕ(z))) αϕ(z) Φ F α = 0, ( ) Φ (0) = α + ψ (0) > ψ (0) = M 2 α M 5 6 7