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  • Let a, b be two real numbers such that  IF the complex number <img alt="\frac{1+a i}{b+i}" src="

Let a, b be two real numbers such that a b<0 IF the complex number \frac{1+a i}{b+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 :

Option: 1

-\frac{1}{2}


Option: 2

-1


Option: 3

1


Option: 4

\frac{1}{2}


Answers (1)

best_answer

\begin{aligned} & a b<0\left|\frac{1+a i}{b+i}\right|=1 \\ & |1+i a|=|b+i| \\ & a^2+1=b^2+1 \Rightarrow a= \pm b \Rightarrow b=-a \quad \text { as } a b<0 \\ & \text { (a,b) lies on }|z-1|=|2 z| \end{aligned}

\begin{aligned} & |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 \end{aligned}

\begin{aligned} & 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} \end{aligned}

\begin{aligned} & \therefore \frac{1+[a]}{4 b}=\frac{6}{4(1-\sqrt{7})}=-\left(\frac{1+\sqrt{7}}{4}\right) \\ & \text { or }[a]=0 \end{aligned}

Similarly it is not matching with a=\frac{-1-\sqrt{7}}{6}

No answer is matching.

Posted by

Kuldeep Maurya

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