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

Step-by-step solution

x + i y = 1 a + i b t (Multiply & divide by conjugate a + i b t ) x + i y = a - i b t a 2 + b 2 t 2 Let a ≠ 0 and b ≠ 0 x = a a 2 + b 2 t 2 ......(i) y = - b t a 2 + b 2 t 2 ......(ii) y x = - b t a ⇒ t = - a y b x Put in (i) x a 2 + b 2 . a 2 y 2 b 2 x 2 = a ⇒ a 2 x 2 + y 2 = a x x 2 + y 2 - 1 a x = 0 x - 1 2 a 2 + y 2 = 1 4 a 2 Circle with centre ( 1 2 a , 0 ) , radius = 1 2 | a | a > 0 ; 1 2 a a < 0 ; − 1 2 a For a ≠ 0 , b = 0 x + i y = 1 a x = 1 a , y = 0 ⇒ z lies on x - axis For a = 0 , b ≠ 0 x + i y = 1 i b t