Конформные отображения. Мицик М.Ф. - 6 стр.

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; 2) , 
, .
.
. 
,0,
1111
11
11
+
+
= cbda
dzc
bza
ς
(5)
,0,
2222
22
22
+
+
= cbda
dzc
bza
w
(6)
 (5) (6), 
d
cz
baz
w
+
+
=
, (7)

( )( )
,0
22221111
=
cbdacbdabcad
 ,  (7)
.
  .   (1) 
z
,

,0,
+
= bcad
a
cw
bdw
z
(8)
   (1).  (8)    
     . 
.
.    .
, 
( )
( )
1,
1
+== zz
zw ς
,

( )( ) ( )( ) ( )( ) ()( )
.,1
1
,
1
1
zwzw
zw
zw ςςςς +=
+
=
5
.   
zz
 

Cl
, : 1) 
z
z

l
,
2)
( )
(
)
lzlz ;;
ρ
.
6
.   
z

z
 
 ( 
0
z

R
), 
, 
0
z

2
00
Rzzzz =
.
  (6) ,   
:
zRz
2
=
, , 
1
=
R
,
zz 1=
.
                                                                                        ; 2)                      ,
                                      ,                                                        .
                                                                                    .
                                                          .
                                    a1 z + b1
                                  ς=          , a1d1 − b1c1 ≠ 0,                                                          (5)
                                    c1 z + d1
                                    a z + b2
                                  w= 2         , a2 d 2 − b2 c2 ≠ 0,                                                      (6)
                                    c2 z + d 2
                         (5) (6),
                                az + b
                                         ,  w=                                                                            (7)
                                cz + d
      ad − bc = (a1d1 − b1c1 )(a2 d 2 − b2 c2 ) ≠ 0,                                                      ,                     (7)
                                                  .
                                                              .                                    (1)                          z,

                                           dw − b
                                  z=               , ad − bc ≠ 0 ,                                                        (8)
                                          − cw + a
                                           (1).                      (8)
                                                                                                                      .
             .
                              .                                                                                                     .
                 ,
                                            w( z ) =
                                                , ς (z ) = z + 1 ,
                                              1
                                              z
                         w(ς (z )) =      , ς (w(z )) = + 1, w(ς ( z )) ≠ ς (w( z )) .
                                       1                1
                                     z +1               z
         5.                                                                     z z∗
                                          l ∈C ,              : 1)              z − z∗                        l                 ,
   2) ρ (z; l ) = ρ z ; l .  ( )  ∗


         6.                                                                     z       z∗
                                                      (                    z0                      R ),
                     ,                                        z0         z − z0 ⋅ z ∗ − z0 = R 2 .
                                              (6)                    ,
                                                                                ∗
                                                                            : z = R z,                                     R = 1,
                                                                                         2
                                                                                                                  ,
z∗ = 1 z .