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1 21(2009) (2009) /

2 21(2009)

3

4 1 1 (1.1) z = x + iy x, y i R C i 2 = 1, i = 0. 21(2009) /

5 1 B.C. (N) 1, 2, 3,...; +,, (Z) 0, ±1, ±2,... ; ±,, (Q) q p, p, q p 0; ±,,, (R) ±,,, (C) ±,, 21(2009) /

6 1 (1) ax = b, (2) ax + b = 0, a 0. (1) X = b a, (2) X = b a : (Q). ax 2 + bx + c = 0, a 0 (A + B) 2 = A 2 + 2AB + B 2 X 2 + b a X + c ( a = X + b ) 2 + c 2a ( X + b ) 2 = b2 2a a b2 4a 2 = 0. 4a 2 c a = b2 4ac 4a 2. b 2 4ac. 2a X + b 2a = ± b 2 4ac 4a 2 = ± 21(2009) /

7 1 X = b ± b 2 4ac. 2a b 2 4ac 0 = X (R, X R) b 2 4ac < 0 = ax 3 + bx 2 + cx + d = 0, a 0 Girolamo Cardano, Niccolo Fontana Tartaglia (2009) /

8 1 a > 0 21(2009) /

9 1 (A + B) 3 = A 3 + 3A 2 B + 3AB 2 + B 3 ax 3 + bx 2 + cx + d = 0 a (1.2) X 3 + bx 2 + cx + d = 0 X = Y + u u Y 3 + (3u + b)y 2 + (3u 2 + 2b + c)y + u 3 + bu 2 + d = 0. u = b 3 (1.2) (1.3) Y 3 + cy + d = 0. Y = Z + v Z v (1.3) 21(2009) /

10 1 ( Z + v ) 3 ( + c Z + v ) + d Z Z = Z 3 + 3vZ + 3v 2 Z + v 3 = Z 3 + 3vZ + 3v 2 Z + v 3 = Z 3 + (3v + c) ( Z + v Z Z 3 + c (Z + v Z (Z Z 3 + c + v Z ) + v 3 Z 3 + d. ) + d ) + d v = c/3 (1.2) Z 3 Z 3 + e Z 3 + d = 0. 21(2009) /

11 1 (1.4) (Z 3 ) 2 + d(z 3 ) + e = 0 Z = ( Y = Z + v Z. d ± ) 1/3 d 2 4e, 2 d 2 4e < 0 21(2009) /

12 1 i 2 = 1 i = 0 z = x + iy C Leonhard Euler ( ): e iθ = cos θ + i sin θ Friedrich Gauss ( ) 21(2009) /

13 1 y y i z = x + iy 0 1 x x 21(2009) /

14 1 r α x sin α = y r, y cos α = x r z = x + iy = re iα ( ), w = u + iv = se iβ, z w = (r s)e i(α+β), e iα e iα = 1. 21(2009) /

15 (Gauss) a 0 X n + a 1 X n a n 1 X + a n = 0, a 0 0 Liouville 21(2009) /

16 2 C f (z) z f (z) = c 0 + c 1 z + c 2 z c n z n +, z C. c 1 = c 2 = = 0 C (2009) /

17

18 2 2.3 (Liouville) C f (z) M f (z) M (z C) f (z) c C P(z) = a 0 z n + a 1 z n a n 1 z + a n C f (z) = 1 C P(z) z f (z) 0 C f (z) 21(2009) /

19 2 P(z) a C P(z) = a z C C f (z) = e z = 1 + z 1! + z2 2! + + zn n! + C 0 f (z) f ( z) = e z e z = e (Picard) C f (z) f (z) f (z) Picard Liouville 21(2009) /

20 2 21(2009) /

21 3 3 n z 1, z 2,..., z n f (z 1, z 2,..., z n ) (a 1, a 2,..., a n ) C n f (z 1, z 2,..., z n ) = c ν1...ν n (z 1 a 1 ) ν 1 (z 2 a 2 ) ν2 (z n a n ) νn. ν 1,...,ν n 0 (1) Weierstrass Weierstrass f (z 1, z 2,..., z n ) 19 21(2009) /

22 3 (2) 1 I II. 2. (3) = (4) ( ) = = = = = = 21(2009) /

23 3 (idéaux de domaines indéterminés) idéaux coherent, coherent ideal. H. Cartan J.P. Serre 1 J. Leray, (Oka s Coherence Theorem) (1948) 3 21(2009) /

24 4 4 K. Oka, Sur quelques notions arithmétiques (Bulletin ), Bulletin de la Société Mathématique de France 78 (1950), p P 0 C n (4.1) A 1 (z)f 1 (z) + A 2 (z)f 2 (z) + + A N (z)f N (z) = 0. A i (z) F i (z) 4.2 ( ) (4.1) (F 1,α (z), F 2,α (z),..., F N,α (z)), α = 1, 2,..., M(< ) P 0 Q (4.1) 21(2009) /

25 4 Grauert-Remmert, Coherent Analytic Sheaves (1984) Of great importance in Complex Analysis is the concept of a coherent analytic sheaf. Already in 1944 Cartan had experimented with the notion of coherent system of... He posed the fundamental problem,... In 1948 Oka gave an affirmative answer (2009) /

26 4 VII 21(2009) /

27 4. 21(2009) /

28 4 21(2009) /

29 (2009) /

30 4 (a) v = (z, w) C 2, z 2 + w 2 = 1. = C { } S 1 (b) ( ψ) i t ψ = Hψ. (c) I 1983 A. Vilenkin... 21(2009) /

31 4 A. Vilenkin (4.3) ψ(a, q) = A(a)e is(a)/ ξ(a, q) = ψ 0 (a)ξ(a, q). (4.3) 21(2009) /

32 5 5 : (2009) /

33 Testament :... 21(2009) /

34 5 Testament: (2009) /

35 (2009) /

36

37 5 21(2009) /

e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1, σ,..., σ N ) i σ i i n S n n = 1,,

e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1, σ,..., σ N ) i σ i i n S n n = 1,, 01 10 18 ( ) 1 6 6 1 8 8 1 6 1 0 0 0 0 1 Table 1: 10 0 8 180 1 1 1. ( : 60 60 ) : 1. 1 e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1,

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> > <., vs. > x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D > 0 x (2) D = 0 x (3 13 2 13.0 2 ( ) ( ) 2 13.1 ( ) ax 2 + bx + c > 0 ( a, b, c ) ( ) 275 > > 2 2 13.3 x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D >

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