z 1 and z 2 are two constant complex numbers such that ∣ z 1 − z 2 ∣ = 5 . A complex number, z , satisfies the following equation.
∣ z + z 2 + z 1 z 2 − z z 1 − z z 2 − 2 z 1 + z 2 ∣ + ∣ z − z 2 + z 1 z 2 − z z 1 − z z 2 − 2 z 1 + z 2 ∣ = 1 0
The locus of another complex number
z
3
is given as :
For the equation
tan
(
arg
(
z
−
z
3
)
)
=
tan
ϕ
, there exists exactly one value of
z
for two different values of
ϕ
,
(
ϕ
1
and
ϕ
2
)
such that
∣
ϕ
1
−
ϕ
2
∣
=
m
π
+
2
π
;
m
∈
Z
.
( The value of the complex number
z
is not same for the two values of
ϕ
but the number of solutions is unity for the two values. )
The value ∣ 2 z 3 − z 1 − z 2 ∣ is found to be constant. Enter the answer as the square of the constant value.
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