FOR ANY COMPOSITE MODULUS, REAL OR COMPLEX. 
293 
Now 
therefore 
also 
therefore 
therefore 
therefore, if 
(4) (4) (2) 
3 + 2i = {\. 2. 0), 
(3 + 20^= (2. 0. 0); 
3 = (3. 3. 0), 
3(3 + (1. 3. 0), 
e = {S. 1 . 0 ), 
i = {a. h. c) 
and therefore 
Examj^le 5. 
This is 
therefore 
Let 
Then 
From the tables 
3« = 3 (mod L)'"! a = I 
3b = 1 (mod 4) >b = 3 
3c = 0 (mod 2)^ c = 0 
^=(1. 3. 0) 
= 5 + (mod 10), 
a; = (3 + 2i) (5 + 6i) (mod 10 3 + 2i) 
— 3 + 28i 
= 103 + (mod 10.3+ 2i). 
4x^ = 14 + (mod 1 5 + lOi). 
4x^ = 7 (2 + i) (mod 2 ^^ 4 ), 
4 = 7 (mod 8 + 1). 
.r = (2 + i) t 
4 (2 + iY^^ = 7 (mod 8 + i). 
(12) (4) 
7 = (11. 2), 
(2 + i) = 59 (mod 8 + i), 
s(ll. 3), 
therefore 
(2 + 2 + = ( 8 . 0 ), 
4 = (2. 0), 
