If the point P represents the complex number | vedclass

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(B) Given $z=x+iy$, where $P=(x, y)$.Consider the expression $\frac{z+i}{z-1} = \frac{x+i(y+1)}{(x-1)+iy}$.To simplify, multiply the numerator and den

Vedclass · Accepted Answer

$x^2+y^2-x+y=0$ and $(x, y) 
eq (1,0)$

Vedclass · Answer

(B) Given $z=x+iy$, where $P=(x, y)$.Consider the expression $\frac{z+i}{z-1} = \frac{x+i(y+1)}{(x-1)+iy}$.To simplify, multiply the numerator and denominator by the conjugate of the denominator, $(x-1)-iy$:$\frac{x+i(y+1)}{(x-1)+iy} 	imes \frac{(x-1)-iy}{(x-1)-iy} = \frac{x(x-1) - ixy + i(y+1)(x-1) + y(y+1)}{(x-1)^2+y^2}$.Expanding the numerator:$= \frac{x^2-x + y^2+y + i(xy-x+y+1-xy)}{(x-1)^2+y^2} = \frac{(x^2+y^2-x+y) + i(1-x+y)}{(x-1)^2+y^2}$.Since $\frac{z+i}{z-1}$ is purely imaginary, its real part must be zero:$\operatorname{Re}\left(\frac{z+i}{z-1}ight) = 0 \Rightarrow \frac{x^2+y^2-x+y}{(x-1)^2+y^2} = 0$.This implies $x^2+y^2-x+y=0$, provided the denominator $(x-1)^2+y^2 
eq 0$, which means $(x, y) 
eq (1, 0)$.

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

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