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Complex Numbers Part 2

Section: Mathematics

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Question : 76 of 106

Marks: +1, -0

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+ib

lies on the circle

|z-1|=|2z|

, then a possible value of

\;\frac{1+[a]}{4b}

[t]

is greatest integer function, is :

[1-Feb-2023 Shift 2]

$-\;\frac{1}{2}$
$-1$
1
$\;\frac{1}{2}$
$0$

Solution:

\left|\;\frac{1+ai}{b+i}\right|=1

|1+ia|=|b+i|

a^2+1=b^2+1 \Rightarrow a=\pm b \Rightarrow b=-a \;\;\;\text{ as }\; ab<0

(a+ib) \;\text{ lies on }\; |z-1|=|2z|

|a+ib-1|=2|a+ib|

(a-1)^2+b^2=4(a^2+b^2)

(a-1)^2=a^2=4(2a^2)

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}

and

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)

Similarly when

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

and

b=\;\frac{1+\sqrt{7}}{6}

then

[a]=-1

\therefore \; \frac{1+[a]}{4b} = \frac{1-1}{4 \times \frac{1+\sqrt{7}}{6}} = 0

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