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 +

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ひでたついさし
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1 (pp.14-27) 1

2 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 y 2 + 2iy u = 1 y 2, v = 2y, y ( ) v 2 u = 1 2 2

3 2 vs w = f(z), f(z = x + iy) = u(x, y) + i v(x, y) (1)f 1 (z = x + iy) = z 2 = x 2 y 2 + 2ixy... z (2)f 2 (z = x + iy) = zz = (x + iy)(x iy) = x 2 + y 2... (3)f 3 (z = x + iy) = 2iz + 6z = 2y + 2ix + 6x 6iy = (6x 2y) + i(2x 6y) f(z) = z z (1) OK (2), (3) NG e z = cos z + i sin z cos z = eiz + ie iz 2 3

4 3 3.1 a ( ) {z z a < ρ} {z z a = ρ} ( ) {z z a ρ} Note: z 2 = zz = re iθ re iθ = r 2 = r(cos θ + i sin θ)r(cos θ i sin θ) = (x + iy)(x iy) = x 2 + y 2 z 4

5 3.2 5

6 4... f z = z 0 def z z0 lim f(z) = f(z 0 ) z 0 f(z) f(z 0 ) f(z) f(z 0 ) ϵ > 0 f(z 0 ) f(z) f(z 0 ) < ϵ z 0 z z 0 < δ ϵ > 0 δ > 0 s.t. z z 0 < δ f(z) f(z 0 ) < ϵ 6

7 4.1 f(z) = z 2 z = 0 limz 2 = 0. z 0 z 2 z z f(z + z) f(z) = f (z) z +O( z) 0 as z 0 lim z 0 z2 = 0 0 = 0 + i0 lim x 0 x 2 = 0 z 2 = r 2 e 2iθ. z = re iθ 0 r 0. r 2 0 z 2 0. z < δ Then, z 2 = z 2 < δ 2. So, ϵ > 0 0 < δ < ϵ δ z < δ z z 2 < δ 2 < ϵ z 2 < ϵ z < δ = ϵ 7

8 5 f (z) = lim z 0 f(z + z) f(z) z f(z + z) f(z) z = f (z) + ϵ( z) ϵ( ) lim ϵ( z) = 0 z 0 w = f(z + z) f(z) = f (z) z ϵ( z) z : + ϵ( z) z f(z + z) z... f(z) = f(z 0 ) + f (z 0 )(z z 0 ) + f (n) (z 0 ) (z z 0 ) n n! n=2 f(z + z) f(z) = f (z) z + ϵ( z) z 0 as z 0 y = f (x) x + o( x) o( x) o( x) 0, x 0 dy = f (x)dx dy: 8

9 5.1 (z n ) = nz n 1 (z + z) n z n z n k = = = n k=0 n k=1 = n k=1 n! k!(n k)! n k n k n k z n k ( z) k z n z z n k ( z) k z z n k ( z) k 1 n 1 z n 1 = nz n 1... n 0 = 1 z 0 k = 1 z as 3z 4 z 2 + 6z (i + 2) 12z 3 2z + 6 (cf) = cf, (f + g) = f + g, (fg) = f g + fg, (1/g) = g /g 2 9

10 f(z) = 1/(1 z 4 ) f(z) = 1 (1 z 2 )(1 + z 2 ) = 1 (1 z)(1 + z)(z i)(z + i) z = 1, 1, i, i lim z a f(z). D = {1, 1, i, i} ( ( ) )) f 4z 3 ( ) 1 (z) = (1 z 4 ) 2. = g g g 2 f D def f D D f D 10

11 5.3 f, g f(g(z)) (f(g(z))) = f (g(z))g (z) z g ω f f(g(z + z)) f(g(z)) z f(g(z + z)) f(g(z)) w = g(z + z) g(z) = w = g(z + z) g(z) z f(g(z) + w) f(g(z)) g(z + z) g(z) = w z f (g(z))g (z) as z 0 g(z) so, w = g(z + z) g(z) 0 (z 2 ) 3 = z 6. LHS g(z) = z 2, f(z) = z 3 LHS 3(z 2 ) 2 (2z) = 6z 5 RHS 11

12 5.4 f(z = z(t)) z d f(z(t + t)) f(z(t)) f(z(t)) = lim dt t 0 t z(t) = x(t) + iy(t) z z(t + t) z(t) (t) = lim t 0 t = lim t 0 x(t + t) x(t) + i(y(t + t) y(t)) t t (e z ) = e z d dt it = i d dt eit = ie it = f (z(t))z (t), = lim t 0 x(t + t) + iy(t + t) (x(t) + iy(t)) t = x (t) + iy (t) ( ddt eit = ddt (cos t + i sin t) = sin t + i cos t = ieit ) θ(t) = t. d dt z(t) = d dt eit = ie it = iz(t) i 12

13 6 z z f(z) = z. z + z z z = z + z z = z x i y = z z x + i y 1 if y = 0 = 1 if x = 0 z 0 z 0 f def f def f 13

14 6.1 f(z = x + iy) = u(x, y) + iv(x, y) f (z) x f (x + iy) lim h 0 u(x + h, y) + iv(x + h, y) u(x, y) iv(x, y) h u(x + h, y) u(x, y) + i lim h 0 v(x + h, y) v(x, y) = lim h h 0 h u, v x f (z) = u x + iv x d dy (f(z = x + iy)) = f (z)i = u y + iv y f (z) = u x + iv x = u y + iv y i = v y iu y f(z = x + iy) = u(x, y) + iv(x, y) u, v u x = v y, v x = u y 14

15 6.2 f(z = x + iy) = z 2 = (x 2 y 2 ) + 2xyi f (z) = 2z. u(x, y) = x 2 y 2, v(x, y) = 2xy u x = 2x = v y, u y = 2y = v x f = u + iv u, v u x, u y, v x, v y : f = u + iv u, v C : f f z 0 f(z 0 + z) = f(z 0 ) + f (z 0 ) z... = f(z 0 + h + z) = f(z 0 + h) + f (z 0 + h) z h h 0. f LHS RHS 1 f (z 0 ) z = lim h 0 f (z 0 + h) z = f (z 0 ) z = lim h 0 f (z 0 + h) z + O( z) s.t. O( z)/ z 0 f (z 0 ) = lim h 0 f (z 0 + h) + O( z)/ z f (z 0 ) = lim h 0 f (z 0 + h) u x = v y, u y = v x u y, v y f u x, v x 15

16 e z = e x+iy = e x (cos y + i sin y) = e x e iy e iy = cos y + i sin y e z 1 ez 2 = ex 1 (cos y 1 + i sin y 1 )e x 2 (cos y 2 + i sin y 2 ) = e x 1+x 2 ( (cos y 1 cos y 2 sin y 1 sin y 2 ) + i(sin y 1 cos y 2 + cos y 1 sin y 2 ) = e x 1+x 2 (cos(y 1 + y 2 ) + i sin(y 1 + y 2 )) = e z 1+z 2 (z 1 + z 2 = (x 1 + x 2 ) + i(y 1 + y 2 )) (e z ) n = e nz, z 1 wz = 1 w w = 1/z (e z ) 1 = 1 e z = 1 e x e iy = (e z ) n = 1/(e z ) n = 1/e nz = e nz e x cos y + i sin y = e x (cos y i sin y) = e x e iy = e z (e z 1 )z 2. e i, (e 1+i ) i u(x, y) = e x cos y, v(x, y) = e x sin y u x = e x cos y, v y = e x cos y, u y = e x sin y, v x = e x sin y u x = v y, u y = v x x y f(z = x + iy) = e x cos y + ie x sin y. x f (z) = e x cos y + ie x sin y = e z y if (z) = u y + iv y = e x sin y + ie x cos y = i(e x cos y + ie x sin y) = ie z f (z) = e z 16 )

17 6.4 f = u + iv u, v C f f n f (n) f = u + iv, f = u x + iv x = v y iu y = n + im n x = u xx = v yx = v xy = m y, n y = u xy = u yx = v xx = m x. f n x = m y, Note: n, m u, v C w def 2 w = w xx + w yy = 0 u xy = u yx u v u x = v y, u y = v x (v x = u y ) u xx = v yx, u yy = v xy. v yx = v xy 2 u = u xx + u yy = 0 n y = m x 17

18 6.4.1 f (t) = 0 on (x 0 h, x 0 + h) f(x 0 + t) = f(x 0 ) + f (x 0 )t + f (c + ) t 2 2 x 0 < c + < x 0 + t f(x 0 t) = f(x 0 ) f (x 0 )t + f (c ) t 2 2 x 0 t < c < x 0 f(x 0 t) + f(x 0 + t) = 2f(x 0 ) x0 +h f(t)dt = x0 +h 2f(x 0 )dt = 2hf(x 0 ) x 0 h x 0 f(x 0 ) = 1 x0 +h 2h f(t)dt... x 0 h 2 u = u xx + u yy = 0 u(x, y) = 1 u(x, y) dxdy πr 2 U (x,y) (r) u (*1) (*1) 187 page. a = (a 1, a 2 ) 2r (x, y) = (a 1 + rt, a 2 + rt) where 1 t 1 2 f(a, b) = ( fxx (a, b) f yx (a, b) f xy (a, b) f yy (a, b) ) 18

19 6.4.2 f(z = x + iy) = u(x, y) + iv(x, y) u x + iy s.t. u(x, y) = d v x + iy s.t. v(x, y) = c u x = v y, u x u y u = u x + iu y,, v x v y v = v x + iv y v x = u y i u = i(u x + iu y ) = u y + iu x = v x + iv y = v u v u v 19

20 6.5 f(z) = c f(z + i h), f(z + h) f (z) = 0 f (z) = 0 f f(z) = f(z 0 ) + f (z 0 )(z z 0 ) + f (2) (z 0 ) (z z 0 ) f (z 0 ) = f (2) (z 0 ) = = 0 f(z) = f(z 0 ) k f(z) = k f(x + iy) = u(x, y) + iv(x, y) 2 = u(x, y) 2 + v(x, y) 2 = k 2 20

21 6.6 Cauchy-Riemann f(z = re iθ ) = u(r, θ) + iv(r, θ). Then u r = v θ r, v r = u θ r r : d dr f(reiθ ) = e iθ f (re iθ ) = u r + iv r θ : d dθ f(reiθ ) = rie iθ f (re iθ ) = u θ + iv θ e iθ f (re iθ ) = u r + iv r = r 1 i(u θ + iv θ ) = r 1 v θ r 1 iu θ z(t) = te iθ, z(t) = re it parameter t θ z = (r + r)e iθ re iθ = re iθ f = f(z + z) f(z) = u(r + r, θ) + iv(r + r, θ) (u(r, θ) + iv(r, θ)) f u(r + r, θ) u(r, θ) + i(v(r + r, θ) + iv(r, θ)) = z z = re iθ u r + iv r = e iθ (u e iθ r + iv r ) = f (z) r z = re iθ+i θ re iθ = re iθ (e i θ 1) f z = f θ θ re iθ (e i θ 1) ( (e (u θ + iv θ )r 1 e ) ) iθ 1 iθ θ=0 ( ) e iθ = (cos θ + i sin θ) = sin θ + i cos θ = ie iθ... θ f (z) = (u θ + iv θ )r 1 e iθ ( i) = r 1 e iθ (v θ iu θ ) hence, u r = v θ r, v r = u θ r 21

22 f(z) = Arg z f(z = re iθ ) = Arg z = (u(r, θ) = θ) + i(v(r, θ) = 0) π < θ π u, v C 1 ow f u r = v θ r, v r = u θ r f(z = x + iy) = u(x, y). (v(x, y) = 0) v x = u y = 0 u(x, y) = h(x) u x = v y = 0 u(x, y) = h(x) = c c f(z = x + iy) = c f(z = x + iy) = u(x, y) + iv(x, y) arctan y x 0... case 1 x π/2 x = 0, y > 0... case 2 f(z = x+iy) = Arg z = π/2 x = 0, y < 0... case 3 x = y = 0 case 1: u(x, y) = arctan y, v(x, y) = 0. x 1/x u y = 1 + (y/x) 0. v 2 x = 0. case 2 case 3: z = yi case1 case1 22

x (x, ) x y (, y) iy x y z = x + iy (x, y) (r, θ) r = x + y, θ = tan ( y ), π < θ π x r = z, θ = arg z z = x + iy = r cos θ + ir sin θ = r(cos θ + i s

x (x, ) x y (, y) iy x y z = x + iy (x, y) (r, θ) r = x + y, θ = tan ( y ), π < θ π x r = z, θ = arg z z = x + iy = r cos θ + ir sin θ = r(cos θ + i s ... x, y z = x + iy x z y z x = Rez, y = Imz z = x + iy x iy z z () z + z = (z + z )() z z = (z z )(3) z z = ( z z )(4)z z = z z = x + y z = x + iy ()Rez = (z + z), Imz = (z z) i () z z z + z z + z.. z