《复变函数+》完整第四版习题解答第5章.pdfVIP

1 1 ( )1 z z2 +1 2 sin z 2 z3 3 1 z3 ? z2 ? z +1 4 ln(z + 1) z z 5 (1+ z2 )(1+ eπ z ) 1 6 e1?z 1 7 z 2 (ez ?1) 1 f (z) = z ( 1 )z2 +1 2 8 z 2n 1+ zn 1 9 sin z2 . ( )z z2 +1 2 =0 z=0 z =±i z=0 z = ±i 2 lim z→0 sin z z3 = ∞ z=0 a. z = 0 sin z z = 0 z3 sin z z=0 z3 b. lim z→0 z2 sin z z3 = lim z→0 sin z z =1≠ 0 z=0 3 = (z2 1 ? 1)( z ? 1) = (z 1 ?1)2 (z + 1) z =1 z = ?1 ∑4 a ln(1 + z) = ∞ (?1)n zn+1 n=0 n + 1 b lim z→0 ln(1 + z z ) = lim z→0 (1 1 + z ) = 1 0 | z | 1 z=0 ∑ln(1+ z) = ∞ (?1)n zn z n=0 n + 1 z=0 ( )5 1 + z2 = 0 z = ± i 1 + z2 ( )1+ eπ z = 0 zk = (2k +1) i (k = 0,±1,±2,) 1+ ez ( )1+ eπ z ′ = π eπ zk = ?π ≠ 0 zk zk = (2k +1)i (k = 1, ±2,) ( )zk 1+ ez z = ±i ∑6 1∞ e1?z = (?1)n n=0 n!(z ?1)n z =1 ∑7 ∞ ez ?1= z zn = z(1+ z + z2 +) n=0 (n +1)! 2 3! z = 2kπ i, (k = ±1, ±2,) z = 0 z2 (ez ?1) 1 z2 (ez ?1) ( )8 z n + 1 = 0 zn = ?1 (2k +1)π i zk = e n k = 0,1,n ?1 9 sin z2 = 0 ? z = ± kπ , z = ±i kπ k = 0,1, 2, -1- ( )sin z2 ′ |z2 =kπ = 2z cos z2 |z2 =kπ = ?0 ??≠ 0 k =0 k≠0 (sin z2 ) = 2 z=0 z=0 1 sin z2 z = ± kπ z = ±i kπ k = 1, 2,3, 1 sin z2 2 z0 f (z) m m 1 z0 f (z) m ?1 f (z) = (z ? z0 )m?(z) ?(z0 ) ≠ 0 f (z) = m(z ? z0 )m?1?(z)+ (z ? z0 )m? (z) = (z ? z0 )m?1[m?(z) + (z ? z0 )?(z)] z0 f (z) m-1 3 z = π i ch z 2 ch π i = cos π = 0 22 (ch z) z= πi 2 = sh π i 2 = isin π 2 =i z = π i ch z 2 4 z=0 (sin z + sh z ? 2z)?2 (sin z + sh z ? 2z) = 0, (sin z + sh z ? 2z) = (cos z + ch z ? 2) = 0 z=0 z=0 z=0 (sin z + sh z ? 2z) = (? sin z + sh z) = 0, (sin z + sh z ? 2z) = (? cos z + ch z) = 0 z=0 z=0 z=0 z=0 (sin z + sh z ? 2z)(4) = (sin z + sh z) = 0, (sin z + sh z ? 2z)(5) = (cos z + ch z) = 2 z=0 z=0 z=0 z=0 z = 0 sin z + sh z ? 2z (sin z + sh z ? 2z)?2 5 f (z) g(z) z0 lim f (z) = lim f (z) ( z→z0 g (z) z→z0 g ( z) ∞) f (z) g(z) z0 f (z) = (z ? z0 )?(z) g(z) = (z ? z0 )ψ (z) ?(z),ψ (z) f (z) = (z ? z0 )?(z) = ?(z) f (z) = ?(z) + (z ? z0 )? (z) g(z) (z ? z0 )ψ (z) ψ (z) g (z) ψ (z) + (z ? z0 )ψ (z) lim f (z) = lim ?(z) lim f (z) = lim ?(z) + (z ? z0 )? (z) = lim ?(z)