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Question
If the imaginary part of `(2z + 1)/(iz + 1)` is –2, then show that the locus of the point representing z in the argand plane is a straight line.
Solution
Let z = x + iy,
Then `(2z + 1)/(iz + 1) = (2(x + iy) + 1)/(i(x + iy) + 1)`
= `((2x + 1) + i2y)/((1 - y) + ix)`
= `({(2x + 1) + i2y})/({(1 - y) + ix}) xx ({(1 - y) - ix})/({(1 - y) - ix})`
= `((2x + 1 - y) + i(2y - 2y^2 - 2x^2 - x))/(1 + y^2 - 2y + x^2)`
Thus `"Im"((2z + 1)/(iz + 1)) = (2y - 2y^2 - 2x^2 - x)/(1 + y^2 - 2y + x^2)`
But `"Im"((2z + 1)/(iz + 1))` = –2 .....(Given)
So `(2y - 2y^2 - 2x^2 - x)/(1 + y^2 - 2y + x^2)` = –2
⇒ 2y – 2y2 – 2x2 – x = –2 – 2y2 + 4y – 2x2
i.e., x + 2y – 2 = 0, which is the equation of a line.
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