Let a, b ∈ ℝ and a ^ 2 + b ^ 2 ≠ 0 . Suppose S = \ z ∈ ℂ: z = (1)/(a + ibt), t ∈ ℝ, t ≠ 0\ , where i = - 1 . If z = x + iy and z ∈ S, then (x, y) lies on

Correct answer: A. The circle with radius (1)/(2a) and centre ((1)/(2a), 0) for a > 0, b ≠ 0 x + iy = (1)/(a + ibt) (Multiply & divide by conjugate a + ibt ) x + iy = a - ibt a ^ 2 + b ^ 2 t ^ 2 Let a ≠ 0 and b ≠ 0 x = a a ^ 2 + b ^ 2 t ^ 2 ......(i) y = - bt a ^ 2 + b ^ 2 t ^ 2 ......(ii) (y)/(x) = (- bt)/(a) ⇒ t = - (ay)/(bx) Put in (i) x\ a ^ 2 + b ^ 2 . a ^ 2 y ^ 2 b ^ 2 x ^ 2 \ = a ⇒ a ^ 2 ( x ^ 2 + y ^ 2 ) = ax x ^ 2 + y ^ 2 - (1)/(a)x = 0 (x - (1)/(2a)) ^ 2 + y ^ 2 = 1 4 a ^ 2 Circle with centre ( (1)/(2a) , 0 ) , radius = (1)/(2| a |) array ll a > 0;(1)/(2a) a For a ≠ 0, b = 0 x + iy = (1)/(a) x = (1)/(a), y = 0 ⇒ z lies on x - axis For a = 0, b ≠ 0 x + iy = (1)/(ibt) y = - (1)/(bt), x = 0 ⇒ z lies on y - axis.

JEE Advanced 2016Paper 2

Exam: JEE Advanced
Year: 2016
Paper: Paper 2
Subject: Mathematics
Chapter: Complex Number

JEE Advanced 2016 · Paper 2 · Mathematics · Complex Number

Let a, b ∈ ℝ and a ^ 2 + b ^ 2 ≠ 0 . Suppose S = \ z ∈ ℂ: z =...

Let

a, b \in \mathbb{R} and

{a}^{2} + {b}^{2} \ne 0

. Suppose

S = \left\{z \in \mathbb{C}: z = \dfrac{1}{a + ibt}, t \in \mathbb{R}, t \ne 0\right\},

where

i = \sqrt{- 1} .

If z = x + iy and

z \in S,

then \left(x, y\right)

lies on

Options

AThe circle with radius $\dfrac{1}{2a}$ and centre $\left(\dfrac{1}{2a}, 0\right)$ for $a > 0, b \ne 0$
BThe circle with radius $- \dfrac{1}{2a}$ and centre $\left(- \dfrac{1}{2a}, 0\right)$ for $a < 0, b \ne 0$
CThe x - axis for $a \ne 0, b = 0$
DThe y - axis for $a = 0, b \ne 0$

Correct answer

A. The circle with radius $\dfrac{1}{2a}$ and centre $\left(\dfrac{1}{2a}, 0\right)$ for $a > 0, b \ne 0$

Step-by-step solution

x + iy = \dfrac{1}{a + ibt}

(Multiply & divide by conjugate

a + ibt

)

x + iy = \dfrac{a - ibt}{{a}^{2} + {b}^{2}{t}^{2}}

Let

a \ne 0

and

b \ne 0

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

......(i)

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

......(ii)

\dfrac{y}{x} = \dfrac{- bt}{a} ⇒ t = - \dfrac{ay}{bx}

Put in (i)

x\left\{{a}^{2} + {b}^{2} . \dfrac{{a}^{2}{y}^{2}}{{b}^{2}{x}^{2}}\right\} = a

⇒

{a}^{2}\left({x}^{2} + {y}^{2}\right) = ax

{x}^{2} + {y}^{2} - \dfrac{1}{a}x = 0

{\left(x - \dfrac{1}{2a}\right)}^{2} + {y}^{2} = \dfrac{1}{4{a}^{2}}

Circle with centre

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

, radius

= \dfrac{1}{2| a |}

{

\begin{array}{ll}a > 0\text{;}\dfrac{1}{2a} \\ a < 0; \dfrac{- 1}{2a}\end{array}

<br />For

a \ne 0, b = 0

<br />

x + iy = \dfrac{1}{a}

<br />

x = \dfrac{1}{a}, y = 0 ⇒ z

lies on x - axis <br />For

a = 0, b \ne 0

<br />

x + iy = \dfrac{1}{ibt}

<br />

y = - \dfrac{1}{bt}, x = 0

<br />

⇒$ z lies on y - axis.

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