Complex Number Question 11

Question 11 - 01 February - Shift 2

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 $a+i b$ 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 :

(1) $-\frac{1}{2}$

(2) -1

(3) 1

(4) $\frac{1}{2}$

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Answer: No answe matching

Solution:

Formula: Properties of modulus, Equation of a circle, Algebra of complex numbers

Given that $ab<0 $ and $ | \frac{1+ai}{b+i}|=1$

$|1+ai|=|b+i|$

$a^{2}+1=b^{2}+1 \Rightarrow a= \pm b \Rightarrow b=-a \quad$ as $a b<0$

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

$|a+i b-1|=2|a+i b|$

$(a-1)^{2}+b^{2}=4(a^{2}+b^{2})$

$(a-1)^{2}+a^{2}=4(a^{2} + a^{2})= 4(2 a^{2})$

$1-2 a=6 a^{2} \Rightarrow 6 a^{2}+2 a-1=0$

$a=\frac{-2 \pm \sqrt{28}}{12}=\frac{-1 \pm \sqrt{7}}{6}$

$a=\frac{\sqrt{7}-1}{6} $

$ b=\frac{1-\sqrt{7}}{6}$

$[a]=0$

$\therefore \frac{1+[a]}{4 b}=\frac{6}{4(1-\sqrt{7})}=-(\frac{1+\sqrt{7}}{4})$

or $[a]=0$

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

No answer is matching.