[UPLIFTJEE]

Let a, b be two real numbers such that $\mathrm{ab}<0$. If the complex number $\frac{1+\mathrm{ai}}{\mathrm{b}+\mathrm{i}}$ is of unit modulus and a + ib lies on the circle $|z-1|=|2 z|$, then a possible value of $\frac{1+[a]}{4 b}$, where [t] is greatest integer function, is [JEE (Main) 01 Feb 2023 S II]
(A) 1
(B) $-\frac{1}{2}$
(C) -1
(D) $\frac{1}{2}$

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[UPLIFTJEE]

Solution

$\left|\frac{1+\mathrm{ai}}{\mathrm{b}+\mathrm{i}}\right|=1 \Rightarrow|1+\mathrm{ai}|=|\mathrm{b}+\mathrm{i}|$

\[ \begin{aligned} & a^{2}+1=b^{2}+1 \Rightarrow a= \pm b \Rightarrow b=-a \text { as } a b<0 \Rightarrow(a, b) \text { lies on }|z-1|=|2 z| \Rightarrow|a+i b-1|=2|a+i b| \\ & (a-1)^{2}+b^{2}=4\left(a^{2}+b^{2}\right) \Rightarrow(a-1)^{2}=a^{2}=4\left(2 a^{2}\right) \Rightarrow 1-2 a=6 a^{2} \Rightarrow 6 a^{2}+2 a-1=0 \Rightarrow a=-\frac{2 \pm \sqrt{28}}{12}=-\frac{1 \pm \sqrt{7}}{6} \end{aligned} \]

Case $\mathrm{I}: \mathrm{a}=\frac{\sqrt{7}-1}{6} \& \mathrm{~b}=\frac{1-\sqrt{7}}{6} \Rightarrow[\mathrm{a}]=0 \therefore \frac{1+[\mathrm{a}]}{4 \mathrm{~b}}=\frac{6}{4(1-\sqrt{7})}=-\left(\frac{1+\sqrt{7}}{4}\right)$
Case II: $\mathrm{a}=-\frac{1-\sqrt{7}}{6}$ then $[\mathrm{a}]=-1 \Rightarrow \frac{1+[\mathrm{a}]}{4 \mathrm{~b}}=0$ No answer is matching

[UPLIFTJEE]

Answer

Bonus

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