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ОБРАЗЕЦ РЕШЕНИЯ К СОДЕРЖАНИЮ РАЗДЕЛА
   Восстановить аналитическую в окрестности точки z0 функцию f(z) по известной действительной части u(x, y)или мнимой v(x, y) и значению f(z0).
1u(x, y) = x2 – y2 + xf(0) = 0
2u(x, y) = x3 – 3xy + 1f(0) = 1
3 v (x, y) = ex (y·cosy + x· siny) f(0) = 0
4u (x, y) = x2 – y2 – 2y f(0) = 0
5 f(0) = 2
6 f(0) = 1 + i
7v(x, y) = e-y ·sin x + y f(0) = 1
8v(x, y) = ex· cosyf(0) = 1 + i
9 f(0) = 1
10 f(1) = 2
11u(x, y) = e-y·cosxf(0) = 1
12u(x, y) = y - 2·x·yf(0) = 0
13u(x, y) = x2 – y2 + 2x + 1 f(0) = i
14 u(x, y) = x2 – y2 - 2x + 1 f(0) = 1
15v(x, y) = 3·x2· y– y3 - yf(0) = 0
16 v(x, y) = y + 2·x·yf(0) = 0
17v(x, y) = 3·x2·y– y3f(0) = 1
18u(x, y) = ex (x·cosy - y·siny)f(0) = 0
19v(x, y) = 2·x + 2·x·yf(0) = 0
20u(x, y) = 1 - ex· sinyf(0) = 1 + i
21 f(0) = 2
22 f(0) = 1 + i
23u(x, y) = e-y· cosx + xf(0) = 1
24u(x, y) = e-y·sinxf(0) = 1
25 f(0) = 1
26 f(1) = 2
27v(x, y) = x2 – y2 - xf(0) = 0
28u(x, y) = - 2·x·y – 2·yf(0) = i
29v(x, y) = 2·x·y – 2·yf(0) = 1
30u(x, y) = x3 - 3·x·y2 – xf(0) = 0
31v(x, y) = 2·x·y + xf(0) = 0