minaminoさん 2025/12/23 9:02
1) |z|=|√3+i|=√(3+1)=2
|1/z|=1/|z|=1/2
iz=-1+i√3
=2{-1/2+i(√3/2)}
=2{cos(2π/3) + i sin(2π/3)}
arg(iz)=2π/3
2) x²/4+(y-1)²=1
両辺をⅹで微分
x/2+2(y-1)y´=0
(√3,3/2)を代入
√3/2+2(3/2-1)y´=0
y´=-√3/2
y=-√3/2(x-√3)+3/2
y=-(√3/2)x+3
x²/4+(y-1)²=1, y=-(√3/2)x+3
https://ja.wolframalpha.com/input?i=x%C2%B2%2F4%2B%28y%EF%BC%8D1%29%C2%B2%EF%BC%9D1%2C+y%3D%EF%BC%8D%28%E2%88%9A3%2F2%29x%2B3
3) ∫[π/6,π/3](1-cos2θ)/2dx
=(1/2)π/6-1/4[sin2θ][π/6,π/3]
=π/12
https://ja.wolframalpha.com/input?i=%E2%88%AB%5B%CF%80%2F6%2C%CF%80%2F3%5D%281%EF%BC%8Dcos2%CE%B8%29%2F2d%CE%B8
x=2sinθ と置換
∫[1,√3]x²/√(4-x²)dx
=∫[π/6,π/3]4sin²θ/(2cosθ)*(2cosθ)dθ
=4∫[π/6,π/3]sin²θdθ
=4×π/12
=π/3
4) Σ[k=1~n]1/(2k-1)(2k+1)
=1/2 Σ[k=1~n]{1/(2k-1)-1/(2k+1)}
=1/2{(1-1/3)+(1/3-1/5)+…+1/(2n-1)-1/(2n+1)}
=1/2{(1-1/(2n+1)} → 1/2
∫[1,n]dx/(2x-1)(2x+1)
=1/2∫[1,n]{1/(2x-1)-1/(2x+1)}dx
=1/4[log|2x-1|-log|2x+1|][1,n]
=1/4[log|(2x-1)/(2x+1)][1,n]
=1/4{log|(2n-1)/(2n+1)-log(1/3)}
→ 1/4(log1+log3)=1/4 log3