If z is a complex number satisfying the equation z - 1 + i 2 = 2 and ω = 2 z z ≠ 0 , then the locus of ω is

Question

If z is a complex number satisfying the equation z - 1 + i 2 = 2 and ω = 2 z z ≠ 0 , then the locus of ω is, x - y - 1 = 0, x + y - 1 = 0, x - y + z = 0, x + 2 y + 1 = 0

MARKS App · Accepted Answer

We have, z - 1 + i 2 = 2 ⇒ x - 1 2 + y - 1 2 = 2 (Putting z = x + i y ) ⇒ x 2 + y 2 = 2 x + y … … . . i Let, ω = h + i k = 2 z = 2 x + i y = 2 x - i y x 2 + y 2 , so h = 2 x x 2 + y 2 , k = - 2 y x 2 + y 2 ⇒ h - k = 2 x + y x 2 + y 2 = 1 (from equation i ) ∴ Locus of the point ω h , k will be x - y = 1 .

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