Let $a, b \in \mathbb{R}$ and $a^{2}+b^{2} \neq 0$. Suppose $S=\left\{z \in \mathbb{C}: z=\frac{1}{a+i b t}, t \in \mathbb{R}, t \neq 0\right\}$, where $i=\sqrt{-1}$. If $z=x+i y$ and $z \in S$, then $(x, y)$ lies on

(A) the circle with radius $\frac{1}{2 a}$ and centre $\left(\frac{1}{2 a}, 0\right)$ for $a>0, b \neq 0$; (C) the $x$-axis for $a \neq 0, b=0$; (D) the $y$-axis for $a=0, b \neq 0$

(B) the circle with radius $-\frac{1}{2 a}$ and centre $\left(-\frac{1}{2 a}, 0\right)$ for $a<0, b \neq 0$

Complex Numbers — JEE Advanced · 2016 · 22 May 2016 · Paper 2 Question with Solution

Let

a, b \in R

and

a^{2} + b^{2} \neq = 0

. Suppose

S = {z \in C : z = \frac{1}{a + ib t}, t \in R, t \neq = 0}

, where

i = - 1

. If

z = x + i y

and

z \in S

, then

(x, y)

lies on

(A)the circle with radius $\frac{1}{2 a}$ and centre $(\frac{1}{2 a}, 0)$ for $a > 0, b \neq = 0$ ✓ Correct
(B)the circle with radius $- \frac{1}{2 a}$ and centre $(- \frac{1}{2 a}, 0)$ for $a < 0, b \neq = 0$
(C)the $x$ -axis for $a \neq = 0, b = 0$ ✓ Correct
(D)the $y$ -axis for $a = 0, b \neq = 0$ ✓ Correct

Answer: (A), (C), (D)

Solution

Write

z = \frac{1}{a + ib t} = \frac{a - ib t}{a ^{2} + b ^{2} t ^{2}}

, so

x = \frac{a}{a ^{2} + b ^{2} t ^{2}}

and

y = \frac{- b t}{a ^{2} + b ^{2} t ^{2}}

. Eliminating

t

x^{2} + y^{2} = \frac{a ^{2} + b ^{2} t ^{2}}{( a ^{2} + b ^{2} t ^{2} ) ^{2}} = \frac{1}{a ^{2} + b ^{2} t ^{2}} = \frac{x}{a}

, giving

x^{2} + y^{2} - \frac{x}{a} = 0

. For

a > 0, b \neq = 0

, this is a circle with centre

(\frac{1}{2 a}, 0)

and radius

\frac{1}{2 a}

, confirming (A). For

a < 0, b \neq = 0

, the centre is

(\frac{1}{2 a}, 0)

(which is negative) and radius

∣ \frac{1}{2 a} ∣ = - \frac{1}{2 a}

, but option (B) states centre

(- \frac{1}{2 a}, 0)

which is positive—this is consistent, confirming (B) is actually correct as stated. For

b = 0

z = \frac{1}{a}

(constant), so

(x, y) = (\frac{1}{a}, 0)

lies on the

x

-axis, confirming (C). For

a = 0

z = \frac{1}{ib t} = \frac{- i}{b t}

, so

x = 0

and

y = - \frac{1}{b t}

, lying on the

y

-axis, confirming (D).

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · One or more correct options

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Complex Numbers

Solution

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