Let a, b be two real numbers such that ab < 0 . If the complex...

Given,

$a, b$ be two real numbers such that $ab < 0$, \left|\dfrac{1 + ai}{b + i}\right| = 1$,$a + ib$lies on the circle \left|z - 1\right| = \left|2z\right|$ and $ab < 0$

Now using \left|\dfrac{1 + ai}{b + i}\right| = 1$we get,</p><p>$⇒ \left|1 + ai\right| = \left|b + i\right|$</p><p>Now squaring both side we get,</p><p>$⇒ {\left|1 + ai\right|}^{2} = {\left|b + i\right|}^{2}$</p><p>$⇒ \left(1 + ai\right)\left(1 - ai\right) = $\left(b + i\right)\left(b - i\right)\left\{{\left|z\right|}^{2} = z\overset{¯}{z}\right\}$

$⇒ 1 + {a}^{2} = {b}^{2} + 1$

$⇒ a = \pm b ⇒ b = - a$ as $ab < 0$

Now given,

\left(a, b\right)$lies on \left|z - 1\right| = \left|2z\right|$

So, \left|a + ib - 1\right| = 2\left|a + ib\right|$</p><p>$⇒ {\left|a - 1 + ib\right|}^{2} = 4{\left|a + ib\right|}^{2}$</p><p>$⇒ {\left(a - 1\right)}^{2} + {b}^{2} = 4\left({a}^{2} + {b}^{2}\right)$</p><p>$⇒ {\left(a - 1\right)}^{2} + {a}^{2} = 4\left(2{a}^{2}\right)\left\{\text{as}b = - a\right\}$</p><p>$⇒ 6{a}^{2} + 2a - 1 = 0$</p><p>$⇒ a = \dfrac{- 2 \pm \sqrt{28}}{12} = \dfrac{- 1 \pm \sqrt{7}}{6}$</p><p>Taking positive sign we get,</p><p>$a = \dfrac{\sqrt{7} - 1}{6} \text{ and } b = \dfrac{1 - \sqrt{7}}{6}$\left\{\text{as}b = - a\right\}$

Now the value of \left[a\right] = 0$as$0 < \dfrac{\sqrt{7} - 1}{6} < 1$</p><p>$∴ \dfrac{1 + \left[a\right]}{4b} = \dfrac{6}{4\left(1 - \sqrt{7}\right)} = - \left(\dfrac{1 + \sqrt{7}}{4}\right)$</p><p>Now taking negative sign we get,$a = \dfrac{- 1 - \sqrt{7}}{6} \text{ and } b = \dfrac{\sqrt{7} + 1}{6}$</p><p>Now \left[a\right] = - 1\left\{\text{as} - 1 < \dfrac{- 1 - \sqrt{7}}{6} < 0\right\}$

So, the value of $\dfrac{1 + \left[a\right]}{4b} = \dfrac{1 + \left(- 1\right)}{4b} = \dfrac{0}{4b} = 0$

Note- Official answer key was drop by NTA, here we have modified the option.