SATHEE: Complex Numbers 4 Question 7

Complex Numbers 4 Question 7

7. Let $a, b \in R$ and $a^{2} + b^{2} \neq 0$ .

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

(2016 Adv.)

(a) the circle with radius $\frac{1}{2 a}$ and centre $\frac{1}{2 a}, 0$ for $a > 0, b \neq 0$

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

(d) the $Y$ -axis for $a = 0, b \neq 0$

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Answer:

Correct Answer: 7. (d)

Solution:

Here, $x + i y = \frac{1}{a + i b t} \times \frac{a - i b t}{a - i b t}$

$∴ x + i y = \frac{a - i b t}{a^{2} + b^{2} t^{2}}$

Let $a \neq 0, b \neq 0$

$∴ x = \frac{a}{a^{2} + b^{2} t^{2}}$ and $y = \frac{- b t}{a^{2} + b^{2} t^{2}}$

$\Rightarrow \frac{y}{x} = \frac{- b t}{a} \Rightarrow t = \frac{a y}{b x}$

On putting $x = \frac{a}{a^{2} + b^{2} t^{2}}$ , we get

$x a^{2} + b^{2} \cdot \frac{a^{2} y^{2}}{b^{2} x^{2}} = a \Rightarrow a^{2} (x^{2} + y^{2}) = a x$

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

$x - {\frac{1}{2 a}}^{2} + y^{2} = \frac{1}{4 a^{2}}$

$∴$ Option (a) is correct.

For

$\begin{matrix} a \neq 0 and b = 0, \\ x + i y = \frac{1}{a} \Rightarrow x = \frac{1}{a}, y = 0 \end{matrix}$

$\Rightarrow z$ lies on $X$ -axis.

$∴$ Option (c) is correct.

For $a = 0$ and $b \neq 0, x + i y = \frac{1}{i b t} \Rightarrow x = 0, y = - \frac{1}{b t}$

$\Rightarrow z$ lies on $Y$ -axis.

$∴$ Option $(d)$ is correct.

Complex Numbers 4 Question 7

7. Let $a, b \in R$ and $a^{2} + b^{2} \neq 0$ .

Answer:

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Complex Numbers 4 Question 7

7. Let a,b∈R and a2+b2≠0.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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7. Let $a, b \in R$ and $a^{2} + b^{2} \neq 0$ .