京都大学 情報学研究科 システム科学専攻 2011年8月実施 専門科目 工業数学 Author 犬 (finalized by 祭音Myyura with assistance from GPT 6 Astra)
Description 問題1
(1) ( 1 + i ) 2 (1+i)^2 ( 1 + i ) 2
(2) 領域 Re ( z 2 ) ≤ 1 \operatorname{Re}(z^2)\le1 Re ( z 2 ) ≤ 1
(3) 逆正弦関数 sin − 1 z \sin^{-1}z sin − 1 z
問題2
f ( z ) = ( 1 − e i z ) / z 2 f(z)=(1-e^{iz})/z^2 f ( z ) = ( 1 − e i z ) / z 2
(1) z = 0 z=0 z = 0
(2) C r C_r C r ∣ z ∣ = r |z|=r ∣ z ∣ = r z = − r z=-r z = − r z = r z=r z = r r > 0 r>0 r > 0 lim r → 0 ∫ C r f ( z ) d z \lim_{r\to0}\int_{C_r}f(z)dz lim r → 0 ∫ C r f ( z ) d z
(3) 複素積分を用いて ∫ 0 ∞ ( 1 − cos x ) / x 2 d x \int_0^\infty(1-\cos x)/x^2\,dx ∫ 0 ∞ ( 1 − cos x ) / x 2 d x
問題3
a > 0 a>0 a > 0 D = { z : ∣ z ∣ ≤ a } D=\{z:|z|\le a\} D = { z : ∣ z ∣ ≤ a } f f f D D D
(1) f ( z ) ‾ \overline{f(z)} f ( z ) D D D z z z f f f
(2) ∣ f ( z ) ∣ |f(z)| ∣ f ( z ) ∣ D D D f f f
(3) C ≥ 0 C\ge0 C ≥ 0 ∣ z ∣ = a |z|=a ∣ z ∣ = a ∣ f ( z ) ∣ = C |f(z)|=C ∣ f ( z ) ∣ = C f f f D D D C ≠ 0 C\ne0 C = 0
(4) (3) と同じ条件のもとで、f f f D D D
题目描述
问题1 (1) 求 ( 1 + i ) 2 (1+i)^2 ( 1 + i ) 2 Re ( z 2 ) ≤ 1 \operatorname{Re}(z^2)\le1 Re ( z 2 ) ≤ 1 sin − 1 z \sin^{-1}z sin − 1 z
问题2 设 f ( z ) = ( 1 − e i z ) / z 2 f(z)=(1-e^{iz})/z^2 f ( z ) = ( 1 − e i z ) / z 2 C r C_r C r ∣ z ∣ = r |z|=r ∣ z ∣ = r − r -r − r r r r r → 0 r\to0 r → 0 ∫ C r f ( z ) d z \int_{C_r}f(z)dz ∫ C r f ( z ) d z ∫ 0 ∞ ( 1 − cos x ) / x 2 d x \int_0^\infty(1-\cos x)/x^2dx ∫ 0 ∞ ( 1 − cos x ) / x 2 d x
问题3 设 a > 0 a>0 a > 0 D = { ∣ z ∣ ≤ a } D=\{|z|\le a\} D = { ∣ z ∣ ≤ a } f f f D D D f ( z ) ‾ \overline{f(z)} f ( z ) f f f ∣ f ∣ |f| ∣ f ∣ D D D f f f ∣ f ∣ = C ≥ 0 |f|=C\ge0 ∣ f ∣ = C ≥ 0 f f f C ≠ 0 C\ne0 C = 0 f f f D D D
Kai 問題1
(1) ( 1 + i ) 2 = 2 i (1+i)^2=2i ( 1 + i ) 2 = 2 i 2 \boxed2 2 π / 2 + 2 π k ( k ∈ Z ) \boxed{\pi/2+2\pi k\ (k\in\mathbb Z)} π /2 + 2 πk ( k ∈ Z )
(2) z = x + i y z=x+iy z = x + i y Re z 2 = x 2 − y 2 \operatorname{Re}z^2=x^2-y^2 Re z 2 = x 2 − y 2
− 1 + y 2 ≤ x ≤ 1 + y 2 \boxed{-\sqrt{1+y^2}\le x\le\sqrt{1+y^2}} − 1 + y 2 ≤ x ≤ 1 + y 2 であり、境界を含む。
(3) sin w = z \sin w=z sin w = z u = e i w u=e^{iw} u = e i w u 2 − 2 i z u − 1 = 0 u^2-2izu-1=0 u 2 − 2 i z u − 1 = 0
sin − 1 z = − i log ( i z ± 1 − z 2 ) . \boxed{\sin^{-1}z=-i\log\left(iz\pm\sqrt{1-z^2}\right)}. sin − 1 z = − i log ( i z ± 1 − z 2 ) . 問題2
(1) 指数関数の級数から
f ( z ) = − i z + 1 2 + i z 6 − z 2 24 + ⋯ . \boxed{f(z)=-\frac iz+\frac12+\frac{iz}6-\frac{z^2}{24}+\cdots}. f ( z ) = − z i + 2 1 + 6 i z − 24 z 2 + ⋯ . (2) z = r e i θ z=re^{i\theta} z = r e i θ θ : π → 0 \theta:\pi\to0 θ : π → 0 ( − i / z ) d z = d θ (-i/z)dz=d\theta ( − i / z ) d z = d θ O ( r ) O(r) O ( r ) lim r → 0 ∫ C r f ( z ) d z = − π \boxed{\lim_{r\to0}\int_{C_r}f(z)dz=-\pi} r → 0 lim ∫ C r f ( z ) d z = − π
(3) 上半平面で半径 R R R C r C_r C r ∣ e i z ∣ ≤ 1 |e^{iz}|\le1 ∣ e i z ∣ ≤ 1 O ( R − 1 ) O(R^{-1}) O ( R − 1 )
lim r → 0 , R → ∞ ( ∫ − R − r f ( x ) d x + ∫ r R f ( x ) d x ) = π . \lim_{r\to0,R\to\infty}\left(\int_{-R}^{-r}f(x)dx+\int_r^Rf(x)dx\right)=\pi. r → 0 , R → ∞ lim ( ∫ − R − r f ( x ) d x + ∫ r R f ( x ) d x ) = π . 実部は偶関数なので ∫ 0 ∞ 1 − cos x x 2 d x = π 2 \boxed{\int_0^\infty\frac{1-\cos x}{x^2}dx=\frac\pi2} ∫ 0 ∞ x 2 1 − cos x d x = 2 π
問題3
(1) f = u + i v f=u+iv f = u + i v f f f u x = v y , u y = − v x u_x=v_y,u_y=-v_x u x = v y , u y = − v x f ‾ \overline f f u x = − v y , u y = v x u_x=-v_y,u_y=v_x u x = − v y , u y = v x f f f
(2) ∣ f ∣ = c |f|=c ∣ f ∣ = c c = 0 c=0 c = 0 c > 0 c>0 c > 0 f ‾ = c 2 / f \overline f=c^2/f f = c 2 / f
(3) C = 0 C=0 C = 0 f ≡ 0 f\equiv0 f ≡ 0 C > 0 \boxed{C>0} C > 0
(4) 零点がないと仮定すれば 1 / f 1/f 1/ f D D D f , 1 / f f,1/f f , 1/ f ∣ f ∣ ≤ C |f|\le C ∣ f ∣ ≤ C 1 / ∣ f ∣ ≤ 1 / C 1/|f|\le1/C 1/∣ f ∣ ≤ 1/ C ∣ f ∣ = C |f|=C ∣ f ∣ = C f f f
