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Step-by-Step Solution
Step 1: Express the Unit Modulus Condition
We are given that the complex number
$ \frac{1 + a i}{b + i} $
is of unit modulus:
$ \left|\frac{1 + a i}{b + i}\right| = 1. $
This directly implies
$ |\,1 + a i\,| = |\,b + i\,|. $
Step 2: Relate a and b from the Modulus Equality
Squaring both sides yields:
$ |\,1 + a i\,|^2 = |\,b + i\,|^2
\ \Longrightarrow\
(1)^2 + (a)^2 = (b)^2 + (1)^2.
$
Simplifying:
$ a^2 = b^2. $
Because $ab < 0,$ one must be positive and the other must be negative. Hence
$ b = - a. $
Step 3: Use the Geometric Condition on the Circle
We also know that $ z = a + i b $ lies on the circle defined by
$ |\,z - 1\,| = |\,2z\,|. $
Substituting $ z = a + i b $:
$ |\,a + i b - 1\,| = 2\,|\,a + i b\,|. $
Squaring both sides:
$ (a - 1)^2 + b^2 = 4(a^2 + b^2). $
With $ b = -a, $ substitute to get:
$ (a - 1)^2 + (-a)^2 = 4\bigl(a^2 + (-a)^2\bigr)
\ \Longrightarrow\
(a - 1)^2 + a^2 = 4\,(a^2 + a^2) = 8\,a^2.
$
Step 4: Simplify and Solve for a
Expand $ (a - 1)^2 = a^2 - 2a + 1,$ so
$ a^2 - 2a + 1 + a^2 = 8\,a^2
\ \Longrightarrow\
2a^2 - 2a + 1 = 8\,a^2.
$
Rearranging gives:
$ 6a^2 + 2a - 1 = 0. $
Solve this quadratic using the quadratic formula:
$
a = \frac{-2 \pm \sqrt{(2)^2 + 4 \cdot 6 \cdot 1}}{2 \cdot 6}
= \frac{-2 \pm \sqrt{4 + 24}}{12}
= \frac{-2 \pm \sqrt{28}}{12}
= \frac{-2 \pm 2\sqrt{7}}{12}
= \frac{-1 \pm \sqrt{7}}{6}.
$
Step 5: Determine Corresponding b and Evaluate the Expression
Since $ b = -a, $ we consider the two possible solutions for $ (a,b). $
$ a = \frac{\sqrt{7} - 1}{6},\ \ b = \frac{1 - \sqrt{7}}{6}. $
Here, $ a $ is positive but less than 1, so $ [\,a\,] = 0.$
Then
$ 1 + [\,a\,] = 1 + 0 = 1.$
Substituting in
$ \frac{1 + [\,a\,]}{4b} = \frac{1}{4 \times \frac{1 - \sqrt{7}}{6}}, $
which does not simplify to $0.$
$ a = \frac{-1 - \sqrt{7}}{6},\ \ b = \frac{1 + \sqrt{7}}{6}. $
Here, $ a $ is negative (less than $-0.5$), thus $ [\,a\,] = -1.$
Hence
$ 1 + [\,a\,] = 1 - 1 = 0.$
Consequently,
$ \frac{1 + [\,a\,]}{4b} = \frac{0}{4 \times \frac{1 + \sqrt{7}}{6}} = 0.$
Therefore, the expression
$ \frac{1 + [\,a\,]}{4b} $
can indeed equal
$0,$
which matches the given correct answer.
