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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
Solution
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
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