Question

If z = x + iy is a complex number such that Im `((2z + 1)/("i"z + 1))` = 0, show that the locus of z is 2x² + 2y² + x – 2y = 0

Answer 1

Let z = x + iy

Simplyfying `((2z + 1)/("i"z + 1)) = (2(x + "i"y) + 1)/("i"(x + "i"y) + 1)`

= `((2x + 1) + 2"i"y)/((1 - y) + "i"x) xx ((1 - y) - "i"x)/((1 - y) - "i"x)`

= `((2x + 1)(1 - y) - "i"x (2x + 1) + 2"i"y(1 - y) + 2xy)/((1 - y)^2 + x^2`

= `([(2x + 1)(1 - y) + 2xy])/((1 - y)^2 + x^2) + "i" ([2y(1 - y) - x(2x + 1)])/((1 - y)^2 + x^2)`

Given Im `((2z + 1)/("i"z + 1))` = 0

`(2y(1 - y) - x(2x + 1))/((1 - y)^2 + x^2)` = 0

2y(1 – y) – x(2x + 1) = 0

⇒ 2y – 2y² – 2x² – x = 0

∴ The locus is 2x² + 2y² – 2y + x = 0

Hence proved