Question

Let $a, b$ be two real numbers such that $a b<0$. If the complex number $\frac{1+ai}{b+i}$ is of unit modulus and $a$ $+i b$ lies on the circle $|z-I|=|2 z|$, then a possible value of $\frac{1+[a ]}{4 b}$, where $[t]$ is greatest integer function, is :

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $1$
• D. $-1$

Answer 1

Answer: (B)

$ ab <0\left|\frac{1+ ai }{ b + i }\right|=1 $
$|1+ ai |=| b + i |$
$ a^2+1=b^2+1 \Rightarrow a=\pm b \Rightarrow b=-a \,\,\,\,a s\,\, a b<0$
$ (a, b) \text { lies on }|z-1|=|2 z| $
$ l a+i b-1|=2 | a+i b|$
$(a-1)^2+b^2=4\left(a^2+b^2\right) $
$(a-1)^2=a^2=4\left(2 a^2\right) $
$ 1-2 a=6 a^2 \Rightarrow 6 a^2+2 a-1=0 $
$ a=\frac{-2 \pm \sqrt{28}}{12}=\frac{-1 \pm \sqrt{7}}{6}$
$ a=\frac{\sqrt{7}-1}{6} \& b=\frac{1-\sqrt{7}}{6}$
$[a]=0$
$\therefore \frac{1+[ a ]}{4 b }=\frac{6}{4(1-\sqrt{7})}=-\left(\frac{1+\sqrt{7}}{4}\right)$
or $[ a ]=0$
Similarly it is not matching with $a =\frac{-1-\sqrt{7}}{6}$
No answer is matching.