Let integers a, b [-3,3] be such that a+b 0. Then the number of all possible ordered pairs (a, b), for which | z-az+b |=1 and vmatrixz+1& & 2\ &z+ 2&1\ 2&1&z+ vmatrix=1, z C, where and 2 are the roots of x2+x+1=0, is equal to .
Answer
10
Solution
C1 C1+C2+C3 and 1+ + 2=0 ⇒ =z3=1 ⇒ z=1, , 2
[figure]
|z-a|=|z+b| ⇒ z on the bisector of a and -b: Re z= a-b2
z=1: a-b=2 ⇒ 5 pairs, reject (1,-1) (a+b=0) ⇒ 4
z= , 2: a-b=-1 ⇒ 6 pairs
∴ 4+6=10
Answer: 10