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.
