《复变函数论》张锦豪邱维元版高等教育出版社课后答案【khdaw_lxywyl】.pdf

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\j\R 1.B z = x + iy, 0) → aei (π+ϕ). (3) (−4 − 3i)−3 → 5−3ei (π−3 arctan(3/4)). (4) 2.bFWEFS4 3.^UA3 p, q,V>`℄ z!fU z∗E [] z∗ = p + (z − p) 4.^h C(a, r),V>`℄ z!fh z∗E [] z∗ = a + → 2√2ei 19π/12 (√3i − 1)2 1 − i q − p q − p r2 z − a 课后答案网 www.khdaw.com

2 . z − a 1 − ¯az   < 1, = 1, > 1, if |a| < 1; if |a| = 1; if |a| > 1. |z − a|2 − |1 − az|2 = (1 − |a|2)(|z|2 − 1), 5.B |z| < 1,8 F[` 5d".9Y5 |z1| < 1. |z2| < 1,8 7.B [F]1b 0 < c ≤ a ≤ b,2 a − c 9.:7AhU;7F/ Riemann;7A h [F];7F h#U℄ |z − p| = λ,0!f ξ1, ξ2, ξ3 UXk8 ξ1, ξ2, ξ3 f℄ :7AC6ei ;7AaÆ9G h ≤ |z1| + |z2| 1 + |z1||z2| a b − c ≤ b ≤ |z1| − |z2| 1 − |z1||z2| ≤ ξ3 = |z|2 − 1 |z|2 + 1 z + z |z|2 + 1 , 1 i z − z |z|2 + 1 , 1 + ξ3 1 − ξ3 . ξ1 + iξ2 1 − ξ3 , z1 + z2 1 + z1z2 a + c b + c . |z − q| ξ1 = ξ2 = z = |z|2 = 课后答案网 www.khdaw.com

10.B z, z′;7F (ξ1, ξ2, ξ3)g (ξ′1, ξ′2, ξ′3). d(z, z′) = q(ξ1 − ξ′1)2 + (ξ2 − ξ′2)2 + (ξ3 − ξ′3)2. 8 [F] f ξ1, ξ2, ξ3E0 E=T?_5 (ξ1 − ξ′1)2 + (ξ2 − ξ′2)2 + (ξ3 − ξ′3)2 = 2(1 − ξ1ξ′1 − ξ2ξ′2 − ξ3ξ′3), p(|z|2 + 1)(|z′|2 + 1) 2|z − z′| 2 . p1 + |z|2 d(z, z′) = lim z ′→∞ 3 d(z, z′) = . 课后答案网 www.khdaw.com

2. T (z) = , z1 = 1, zn = T (zn−1) , n ≥ 2.V lim zn ℄ (1 + i)/√2. [? ] c = (1 + i)/√2, wn = zn− c.V wn → 0 (n → +∞).g| wn+1 + c = zn+1 = ." c2 = i,ÆÆ wn + c + i wn + c + 1 zn + i zn + 1 z + i z + 1 = 1 wn+1 = wn + c + i wn + c + 1 − c wn + c + i − cwn − c2 − c wn + c + 1 (1 − c)wn wn + c + 1 . = = \" 4 c = (1 + i)/√2, c2 = i, wn = zn − c, z1 = 1. 1 + i √2 (√2 − 1)2 + 1 |w1|2 = |1 − c|2 = 1 − √2 − 1 − i = √2 1 + 2 − 2√2 + 1 = = 2 2 2 √2 < 1, = 2 − 2 |c + 1|2 = 1 + 1 + i √2 2 = √2 + 1 + i √2 2 课后答案网 www.khdaw.com

2 + 2√2 + 1 + 1 2 = = 2 + √2. |w1|2 = |1 − c|2 = 2 − √2 < 1, |c + 1|2 = 2 + √2. |c+1|−|w1| = |c+1|−|1−c|. (|c + 1| − |1 − c|)2 = q2 + √2 − q2 − √22 2 2 = (√2 + 1)2 + 1 wn = zn−c, z1 = 1wT U 1 A wn+1 = |wn| ≤ |1 − c|n.XR /Y7 |w2| = |1 − c| |w1| ≤ |1 − c||w1| ≤ |1 − c|2. √2 − 2q2 + √2q2 − √2 = 2 + √2 + 2 − = 4 − 2√2 > 1, , |w1| = |1 − c|.ÆÆ |c + 1| − |1 − c| > 1. (1 − c)wn wn + c + 1 |w1 + c + 1| ≤ |1 − c| |w1| |c + 1| − |w1| |wn| |1 − c|n |wn + c + 1| ≤ |1 − c| |c + 1| − |wn| |c + 1| − |1 − c|n ≤ |1 − c|n+1 1 1 |c + 1| − |1 − c| |wn+1| = |1 − c| ≤ |1 − c|n+1 ≤ |1 − c|n+1. 课后答案网 www.khdaw.com

f |1 − c| < 1 wn → 0 (n → +∞). [?] c = (1 + i)/√2, wn = zn− c.V wn → 0 (n → +∞).g| wn+1 + c = zn+1 = ." c2 = i,ÆÆ wn + c + i wn + c + 1 zn + i zn + 1 = 3 wn + c + i wn + c + 1 − c wn + c + i − cwn − c2 − c wn + c + 1 = = (1 − c)wn wn + c + 1 . . ζn = (n ∈ N ), 1 wn wn+1 = _i 1 wn+1 ζn+1 = 1 + c 1 − c ζn + 1 wn + 1 1 − c = 1 + c 1 − c 1 1 − c . 课后答案网 www.khdaw.com

ζn+1 = 4 1 + c 1 − c ζn + 1 1 − c .ze = = = = ζn+1 + 1 2c ζn+1 + 1 2c = 1 + c 1 − c ζn + 1 + c 1 − c 1 + c 1 − c 1 + c 1 − c 1 + c 1 − c 1 2c ζn + ζn + + 1 1 1 − c 2c 2c + 1 − c (1 − c)2c 1 + c (1 − c)2c , ζn + ζn + 1 2c ζn−1 + 1 2c . 1 2c ζn + 1 2c 1 + c 1 − c n ζ1 + = ζn+1 + 1 2c 1 + c 1 − c = 1 + c 1 − c = 1 + c 1 − c Æ limn→∞ |ζn| = +∞. ζn = 1 + c 1 − c ,8 limn→+∞ wn = 0. > 1, = p2 + √2 p2 − √2 1 wn 课后答案网 www.khdaw.com

5 = 1 − c 1 + c zn−1 − c zn−1 + c zn − c zn + c = T (z) − c T (z) + c = 1 − c 1 + c z − c z + c . n−1 z1 − c z1 + c T (zn−1) − c T (zn−1) + c [?a] c = (1 + i)/√2.g|"e .ze Æ limn→∞ zn − c = 0. [?j]g|" 4 x1 = 1, y1 = 0,ES zn ~ \/YV , 2 ._ n = 1. z1 = 1, z2 = arg i = π xn−1 + i(yn−1 + 1) xn−1 + 1 + iyn−1 n−1 + y2 x2 = 1 − c 1 + c = 1 − c 1 + c (xn−1 + 1)2 + y2 zn−1 + i zn−1 + 1 ∀n. 1 + i |z2n−1| = 1, n → 0. arg z2n = π 4 , = = zn = n−1 + xn−1 + yn−1 + i(xn−1 + yn−1 + 1) , n−1 2 |z1| = 1, arg z2 = π 4 . 课后答案网 www.khdaw.com

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《复变函数论》张锦豪邱维元版高等教育出版社课后答案【khdaw_lxywyl】.pdf

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