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(D) We have,$\frac{z-1}{z+i} = \frac{x+iy-1}{x+i(y+1)}$.Multiplying the numerator and denominator by the conjugate of the denominator,$x-i(y+1)$,we ge

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$(2016, -2017)$

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(D) We have,$\frac{z-1}{z+i} = \frac{x+iy-1}{x+i(y+1)}$.Multiplying the numerator and denominator by the conjugate of the denominator,$x-i(y+1)$,we get:$\frac{(x-1)+iy}{x+i(y+1)} 	imes \frac{x-i(y+1)}{x-i(y+1)} = \frac{x(x-1) - i(x-1)(y+1) + ixy + y(y+1)}{x^2+(y+1)^2}$.The real part is $\frac{x(x-1)+y(y+1)}{x^2+(y+1)^2}$.Given that the real part is $1$,we have:$\frac{x^2-x+y^2+y}{x^2+(y+1)^2} = 1$.$x^2-x+y^2+y = x^2+y^2+2y+1$.$-x+y = 2y+1$.$x+y+1 = 0$.Checking the options,for $(2016, -2017)$,we have $2016 + (-2017) + 1 = 2016 - 2017 + 1 = 0$.Thus,the point $(2016, -2017)$ lies on the locus.

Let $z=x+iy$ and a point $P$ represent $z$ in the Argand plane. If the real part of $\frac{z-1}{z+i}$ is $1$,then a point that lies on the locus of $P$ is

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