If a point P denotes a complex number z=x+iy | vedclass

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(D) Given $z = x + iy$. Then $\frac{z+1}{z+i} = \frac{(x+1) + iy}{x + i(y+1)}$.To make this purely real,we multiply the numerator and denominator by t

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$x+y+1=0, (x, y) 
eq (0, -1)$

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(D) Given $z = x + iy$. Then $\frac{z+1}{z+i} = \frac{(x+1) + iy}{x + i(y+1)}$.To make this purely real,we multiply the numerator and denominator by the conjugate of the denominator: $\frac{(x+1) + iy}{x + i(y+1)} 	imes \frac{x - i(y+1)}{x - i(y+1)}$.The denominator becomes $x^2 + (y+1)^2$,which is real.The numerator is $(x+1)x + y(y+1) + i[xy - (x+1)(y+1)]$.For the expression to be purely real,the imaginary part must be zero:$xy - (x+1)(y+1) = 0$.$xy - (xy + x + y + 1) = 0$.$-x - y - 1 = 0 \Rightarrow x + y + 1 = 0$.Since $z 
eq -i$,the point $(0, -1)$ must be excluded.Thus,the locus is $x+y+1=0, (x, y) 
eq (0, -1)$.

If a point $P$ denotes a complex number $z=x+iy$ in the Argand plane and if $\frac{z+1}{z+i}$ is a purely real number,then the locus of $P$ is

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