Комплексные числа. Понятие функции комплексного переменного. Аксентьева Е.П. - 15 стр.

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3) z
k
= π/2 + 2kπ − i ln(3 ± 2
√
2), 4) z
k
=
1
2
[(2k + 1)π − arctg
1
2
] +
i
4
ln 5.
k = 0, ±1, ±2, ...
1) 0 < |w| < 1, 0 < arg w < π, 2) 1 < |w| < ∞, 0 < arg w < π/2.
1) x
1
= −1, x
2
= −3, P (x) = (x + 1)(x + 3), 2) x
1
= 1/2, x
2
= 2,
P (x) = 2(x − 2)(x − 1/2), 3) x
1,2
= −1 ± 3i, 4)x
1
= 2, x
2,3
= −1 ±
√
3i,
P (x) = (x − 2)(x
2
+ 2x + 4), 5) x
1
= x
2
= x
3
= 0, x
4
= x
5
= 3,
P (x) = x
3
(x − 3)
2
, 6) x
1
= x
2
= i, x
3
= x
4
= −i, P (x) = (x
2
+ 1)
2
,
7) x
1
= x
2
= 1, x
3
= −1, P (x) = (x − 1)
2
(x + 1), 8) x
1
= 0, x
2
= x
3
= 2i,
x
4
= x
5
= −2i, P (x) = x(x
2
+ 4)
2
, 9) x
1,2
= ±1, x
3,4
= ±i,
P (x) = (x − 1)(x + 1)(x
2
+ 1), 10) x
1,2
= ±i, x
3,4
= ±
√
3i,
P (x) = (x
2
+ 3)(x
2
+ 1), 11) x
1,2
= 1 ± i, x
3,4
= −1 ± i,
P (x) = (x
2
− 2x + 2)(x
2
+ 2x + 2), 12) x
1
= 0, x
2,3
= ±1, x
4,5
= ±3,
P (x) = x(x − 1)(x + 1)(x − 3)(x + 3), 13) x
1
= x
2
= 1, x
3
= x
4
= −1,
x
5
= x
6
= i, x
7
= x
8
= −i, P (x) = (x − 1)
2
(x + 1)
2
(x
2
+ 1)
2
,
14) x
1,2
= ±1, x
3,4
= ±i, x
5,6
=
√
2
2
(1 ± i), x
7,8
=
√
2
2
(−1 ± i),
P (x) = (x − 1)(x + 1)(x
2
+ 1)(x
2
−
√
2x + 1)(x
2
+
√
2x + 1), 15) x
1
= 2,
x
2,3
= (−1 ±
√
3i)/2, P (x) = (x − 2)(x
2
+ x + 1), 16) x
1
= −2,
x
2,3
= (5 ±
√
15i)/4, P (x) = (x + 2)(2x
2
− 5x + 5).
                              √
3) zk = π/2 + 2kπ − i ln(3 ± 2 2), 4) zk = 12 [(2k + 1)π − arctg 12 ] + 4i ln 5.
      Çäåñü âåçäå k = 0, ±1, ±2, ...
      19 1) 0 < |w| < 1, 0 < arg w < π, 2) 1 < |w| < ∞, 0 < arg w < π/2.
      20 1) x1 = −1, x2 = −3, P (x) = (x + 1)(x + 3), 2) x1 = 1/2, x2 = 2,
                                                                          √
P (x) = 2(x − 2)(x − 1/2), 3) x1,2 = −1 ± 3i, 4)x1 = 2, x2,3 = −1 ±           3i,
P (x) = (x − 2)(x2 + 2x + 4), 5) x1 = x2 = x3 = 0, x4 = x5 = 3,
P (x) = x3 (x − 3)2 , 6) x1 = x2 = i, x3 = x4 = −i, P (x) = (x2 + 1)2 ,
7) x1 = x2 = 1, x3 = −1, P (x) = (x − 1)2 (x + 1), 8) x1 = 0, x2 = x3 = 2i,
x4 = x5 = −2i, P (x) = x(x2 + 4)2 , 9) x1,2 = ±1, x3,4 = ±i,
                                                       √
P (x) = (x − 1)(x + 1)(x2 + 1), 10) x1,2 = ±i, x3,4 = ± 3i,
P (x) = (x2 + 3)(x2 + 1), 11) x1,2 = 1 ± i, x3,4 = −1 ± i,
P (x) = (x2 − 2x + 2)(x2 + 2x + 2), 12) x1 = 0, x2,3 = ±1, x4,5 = ±3,
P (x) = x(x − 1)(x + 1)(x − 3)(x + 3), 13) x1 = x2 = 1, x3 = x4 = −1,
x5 = x6 = i, x7 = x8 = −i, P (x) = (x − 1)2 (x + 1)2 (x2 + 1)2 ,
                                   √                     √
                                     2                    2
14) x1,2 = ±1, x3,4 = ±i, x5,6 =   2 (1 ±   i), x7,8 =   2 (−1   ± i),
                                       √                     √
P (x) = (x − 1)(x + 1)(x2 + 1)(x − 2x + 1)(x2 + 2x + 1), 15) x1 = 2,
                                   2
             √
x2,3 = (−1 ± 3i)/2, P (x) = (x − 2)(x2 + x + 1), 16) x1 = −2,
           √
x2,3 = (5 ± 15i)/4, P (x) = (x + 2)(2x2 − 5x + 5).




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