Let a , b and a 2 + b 2 0 . Suppose S = { z C : z = 1 a + i b t , t + , t0 } , where i = - 1 . If z = x + i y and z S , then ( x , y ) lies on [2016]

Question

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]

Smart Achievers · Accepted Answer

(1, 3, 4)

$z = \frac{1}{a + i b t} = x + i y$

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

$\Rightarrow x^{2} + y^{2} = \frac{1}{a^{2} + b^{2} t^{2}} = \frac{x}{a} \Rightarrow x^{2} + y^{2} - \frac{x}{a} = 0$

$∴$ $Locus of z is a circle with centre (\frac{1}{2 a}, 0) and radius \frac{1}{2 | a |}$

$irrespective of' a' being + v e or - v e$

$Also for b = 0, a \neq 0, we get y = 0$

$∴$ $locus is x-axis$

$and for a = 0, b \neq 0, we get x = 0$

$∴$ $locus is y-axis.$

$Hence, 1, 3, 4 are the correct options.$

Solution

Q.
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]

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

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

3 the $x$ -axis for $a \neq 0, b = 0$

4 the $y$ -axis for $a = 0, b \neq 0$

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Solution

Q. Let a,b∈ℝ and a2+b2≠0. Suppose S={z∈C:z=1a+ibt, t∈ℝ+, t≠0}, where i=-1. If z=x+iy and z∈S, then (x,y) lies on [2016]

1 the circle with radius 12a and centre (12a,0) for a>0, b≠0

2 the circle with radius -12a and centre (-12a,0) for a<0, b≠0

3 the x-axis for a≠0, b=0