推 j0958322080 :感謝，第二題應該是答案錯了 06/13 00:37
※ 引述《j0958322080 (Tidus)》之銘言：
: http://ppt.cc/E9M1
: 第五題
: 我把它換成log展開後就不會算了
z = atan(-0.4+0.2i)
tan(z) = (exp(iz)-exp(-iz))/(exp(iz)+exp(-iz))/i = -0.4+0.2i
令A=exp(iz), B=-0.4+0.2i
(A-1/A)/(A+1/A) = iB
(A^2-1)/(A^2+1) = iB
整理得 A^2 = (1+iB)/(1-iB) = 0.5-0.5i = 2^(-0.5) * exp(-0.25πi)
or 2^(-0.5) * exp(1.75πi)
A = 2^(-0.25) * exp((-0.125+2n)πi) or A = 2^(-0.25) * exp((0.875+2n)πi)
iz = ln(A) = ln(2^(-0.25)) (-0.125+2n)πi or ln(2^(-0.25)) + (0.875+2n)πi
z = (-0.125+2n)π - ln(2^(-0.25))i or (0.875+2n)π - ln(2^(-0.25))i
其中n為任意整數
: http://ppt.cc/t4Ay
: 第三題
: 我有積分出正確虛部的答案，但是實部就錯了。
Γ1: 1+i -> 2+i => z = t+i, dz=dt, t=1~2
f(z) = z + Im{z} = t+i+1
2 |2
∫ f(z) dz = ∫ (t+i+1)dt = (0.5*t^2 +(i+1)t)| = 2.5 + i
Γ1 1 |1
Γ2: 2+i-> 2+2i => z = 2 +ti, dz=idt, t=1~2
f(z) = z + Im{z} = 2 +ti+t
2 |2
∫ f(z) dz = ∫ (2i-t+ti)dt = (2it-0.5t^2+0.5t^2i)| = -1.5 + 3.5i
Γ2 1 |1
∫f(z)dz = ∫ f(z) dz + ∫ f(z) dz = 1 + 4.5 i
Γ Γ1 Γ2
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