An algebra problem by U Z
find the sum of all the possible values of the complex number d such that there exists values of a , b satisfying
1)
{
(
a
+
1
)
(
b
−
1
)
+
(
b
+
1
)
(
a
−
1
)
}
d
+
(
a
−
1
)
(
b
−
1
)
=
0
2)
d
(
a
+
1
)
(
b
+
1
)
−
(
a
−
1
)
(
b
−
1
)
=
0
3) Define
A
=
{
a
−
1
a
+
1
,
b
−
1
b
+
1
}
and
B
=
{
a
+
1
2
a
,
b
+
1
2
b
}
. We have
A
⋂
B
=
∅
.
The answer is 1.
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{ ( a + 1 ) ( b − 1 ) + ( b + 1 ) ( a − 1 ) } d + ( a − 1 ) ( b − 1 ) = 0
= a − 1 a + 1 + b − 1 b + 1 = − d 1 and ( a − 1 ) ( b − 1 ) ( a + 1 ) ( b + 1 ) = d 1
thus
x 2 − d ( − 1 ) x + d 1 = 0 has roots d − 1 d + 1 , b − 1 b + 1
d x 2 + x + 1 = 0
x = a + 1 2 a
a = 2 − x x thus a − 1 a + 1 = x − 1 1
now ,
( x − 1 ) 2 d + x − 1 1 + 1 = 0 is a quadratic equations whose roots are a + 1 2 a , b + 1 2 b
x 2 − x + d = 0
now since intersection is not empty , therefore they must have a common root
if both roots are common then , the curves are parallel
thus 1 d = − 1 1 = d 1 d = − 1
if a root is common then ,
d + 1 x 2 = 1 − d 2 x = − ( d + 1 ) 1 ( can you answer why?)
d = 1 ± i
d = { − 1 , 1 + i , 1 − i }