Question

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

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

Answer 1

Answer: (B)

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