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प्रश्न
If z = x + iy and arg `((z - "i")/(z + 2)) = pi/4`, show that x2 + y3 + 3x – 3y + 2 = 0
उत्तर
Given z = x + iy and arg `((z - "i")/(z + 2)) = pi/4`
Now simplifying `(z - "i")/(z + 2) = (x + iy - "i")/(x + "i"y + 2)`
= `(x + "i"(y - 1))/((x + 2) + "i"y) xx ((x + 2) - "i"y)/((x + 2) - "i"y)`
= `(x(x + 2) + y(y 1))/((x + 2)^2 + y^2) + ("i"[(x + 2)(y - 1) - xy])/((x + 2)^2 + y^2)`
= `(x^2 + y^2 + 2x - y)/((x + 2)^2 + y^2) + "i" ((2y - x - 2))/((x + 2)^2 + y^2)`
Given arg `((z - "i")/(z + 2)) = pi/4`
i.e., `tan^-1 ((2y - x - 2)/(x^2 + y^2 + 2x - y)) = pi/4`
`(2y - x - 2)/(x^2 + y^2 + 2x - y) = tan pi/4` = 1
2y – x – 2 = x2 + 2x + y2 – y
x2 + y2 + 2x + x – y – 2y + 2 = 0
⇒ x² + y² + 3x – 3y + 2 = 0
Hence proved
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