Re: [理工] 工數 複變
※ 引述《bboycookie (閉關BBOY餅乾)》之銘言：
: SIN(Z)=3
: z=??
: 沒答案
: 域超凡上課的筆記看到的
: 附變
以下是不負責任的解答...
Let z = x+iy,
then sin(z) = sin(x) cosh(y) + i cos(x) sinh(y) = 3.
╭ sin(x) cosh(y) = 3 ......(1)
=> │
╰ cos(x) sinh(y) = 0 ......(2)
由(2)式可知， cos(x) = 0 or sinh(y) = 0.
e^(y) - e^(-y)
If sinh(y) = 0 => ──────── = 0 => e^(y) = e^(-y) => y = 0 ,
2
e^(y) + e^(-y) By (1)式
then cosh(y) = ──────── = 1 => sin(x) = 3.
2
但 sin(x)的值 介於 -1～1 之間，所以 sin(x) = 3 不合. (矛盾)
π 1
∴ cos(x) = 0. => x = ── + 2 nπ = ( 2n + ── )π , n 屬於 Z .
2 2
=> sin(x) = ±1 ( -1不合，因為 cosh(y)恆大於零 ) => sin(x) = 1
由(1)式： sin(x) = 1 => cosh(y) = 3.
e^(y) + e^(-y)
cosh(y) = ──────── = 3
2
=> e^(y) + e^(-y) = 6.
同乘以e^(y)
y 2 y
=> ( e ) - 6 e + 1 = 0
利用公式解
y 6 ±√ 36 - 4 ×1 ×1
=> e = ───────────── = 3 ±2√2
2
=> y = ln ( 3 ±2√2 ) .
1
因為 ln ( 3 - 2√2 ) = ln ( ───── ) = - ln ( 3 + 2√2 ) .
3 + 2√2
所以 y = ±ln ( 3 + 2√2 ) .
將 x , y 的值代回 z = x + iy ,
1
可解出 z = ( 2n + ── )π ± i ln ( 3 + 2√2 ) , n 屬於 Z .
2
不負責任的解答，有錯再請大家指正＠＠
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