Key to Demonstration 3
I. Proof. By Theorem 1.4 e), and the fact that |z|= 1, it is easy to get
¯¯
¯¯
¯¯ Z
C
dz 2z2 + 5
¯¯
¯¯
¯¯≤max
¯¯
¯ 1 2z2+ 5
¯¯
¯
¯¯
¯ Z
C
dz
¯¯
¯=
¯¯
¯ 1
2(−i)2+ 5
¯¯
¯·2π= 2π 3 . II. Key.
Z
C
dz
z2−4 = 1 4
Z
C
dz z−2 −
Z
C
dz z+ 2
.
(a) Z
|z|=1
dz
z2−4 = 1
4(0−0) = 0 ;
(b) Z
|z|=4
dz
z2−4 = 1
4(2πi−2πi) = 0 ;
(c) Z
|z−2|=2
dz
z2−4 = 1
4(2πi−0) = πi 2 . III. Key.
Z
C
dz
z2−1 = 1 2
Z
C
dz z−1 −
Z
C
dz z+ 1
.
(a) Z
|z|=13
dz
z2−1 = 1
2(0−0) = 0 ;
(b) Z
|z|=3
dz
z2−1 = 1
2(2πi−2πi) = 0 ;
(c) Z
|z−1|=1
dz
z2−1 = 1
2(2πi−0) =πi . IV. Key.
Z
C
dz
z(z−1)(z+ 2) =−1 2
Z
C
dz z +1
3 Z
C
dz z−1 +1
6 Z
C
dz z−(−2).
1
(a) Z
|z|=12
dz
z(z−1)(z+ 2) =−1
2·2πi+ 1
3·0 + 1
6·0 = −πi;
(b) Z
|z|=32
dz
z(z−1)(z+ 2) =−1
2 ·2πi+ 1
3·2πi+1
6 ·0 =−π 3i;
(c) Z
|z|=52
dz
z(z−1)(z+ 2) =−1
2·2πi+1
3 ·2πi+ 1
6·2πi= 0 ;.
(d) Z
|z−1|=12
dz
z(z−1)(z+ 2) =−1
2 ·0 + 1
3 ·2πi+1
6 ·0 = 2π 3 i . V. Key. Letz = cosθ+isinθ, dθ = dziz and sinθ = 2i(1z −z), then,
Z2π
0
dθ 3 sinθ+ 5 =
Z
|z|=1
dz
iz(3i2(1z −z) + 5)
= 2 Z
|z|=1
dz
(3z+i)(z+ 3i)
= −i 4
³ Z
|z|=1
dz z+ 3i −
Z
|z|=1
dz z+ 3i
´
= −i
4 ·2πi= π 2.
VI. Key. Let z = cosθ+isinθ,dθ = dziz and cosθ= 12(z+1z), then, Z2π
0
dθ 2 + cosθ =
Z
|z|=1
dz
iz(2 + 12(z+1z)) = 2 i
Z
|z|=1
dz z2+ 4z+ 1
=2 i
Z
|z|=1
dz (z−(−2 +√
3))(z−(−2−√ 3))
=2 i
1 2√
3 Z
|z|=1
dz z−(−2 +√
3)− 1 2√
3 Z
|z|=1
dz z−(−2−√
3)
= 2π
√3.
2
VII. Proof. Letz = cosθ+isinθ,dθ = dziz, cosθ = 12(z+ 1z), then Z2π
0
dθ
1 +acosθ = Z
|z|=1
dz/iz
1 + 12a(z+ 1z) = 2 i
Z
|z|=1
dz az2+ 2z+a
=2 ia
Z
|z|=1
¡ dz
z+1+√a1−a2¢¡
z+ 1−√a1−a2¢
=2 ia
a 2√
1−a2
Z
|z|=1
dz
z+1−√a1−a2 − Z
|z|=1
dz z+ 1+√a1−a2
Since 0 < a <1, we can get −1< −1+√a1−a2 <0, because −1+√a1−a2 <0 and −1+√a1−a2 >
−1 ⇔ 1−a <√
1−a2 ⇔ (1−a2)2 <1−a2 ⇔ a2−a <0 ⇔ a(a−1)<0 ⇔ a <1, thus we can get −1+√a1−a2 ∈I(C) and −1−√a1−a2 ∈E(C), where C ={z : |z|= 1}. So,
Z2π
0
dθ
1 +acosθ = 1 i√
1−a2(2πi−0) = 2π
√1−a2 .
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