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Let $z=x+i y,$ where $x$ and $y$ are real. The points $(x, y)$ in the $X-Y$ plane for which $\frac{z+i}{z-i}$
is purely imaginary, lie on
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is purely imaginary, lie on
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2526 Upvotes
Verified Answer
The correct answer is:
a circle
$\frac{z+i}{z-i}=\frac{(x+j)+i}{(x+i y)-i}$
$=\frac{x+(1+y) i}{x+(y-1) i}$
$=\frac{x+(y+1) i}{x+(y-1) i} \times \frac{x-(y-1) i}{x-(y-1) i}$
$=\frac{x^{2}-x(y-1) i+x(y+1) i-(y+1)(y-1) i^{2}}{x^{2}-(y-1)^{2} i^{2}}$
$=\frac{x^{2}+i[-x y+x+x y+x]+\left(y^{2}-1\right)}{x^{2}+(y-1)^{2}}$
$=\frac{x^{2}+y^{2}-1}{x^{2}+(y-1)^{2}}+\frac{2 x}{x^{2}+(y-1)^{2}} i$
Now, $\frac{z+i}{z-i}$ is purely imaginary.
$\therefore$
$\operatorname{Re}\left(\frac{z+i}{z-i}\right)=0$
$\begin{array}{lr}\Rightarrow & \frac{x^{2}+y^{2}-1}{x^{2}+(y-1)^{2}}=0 \\ \Rightarrow & x^{2}+y^{2}=1\end{array}$
$\therefore(x, y)$ lies on a circle.
$=\frac{x+(1+y) i}{x+(y-1) i}$
$=\frac{x+(y+1) i}{x+(y-1) i} \times \frac{x-(y-1) i}{x-(y-1) i}$
$=\frac{x^{2}-x(y-1) i+x(y+1) i-(y+1)(y-1) i^{2}}{x^{2}-(y-1)^{2} i^{2}}$
$=\frac{x^{2}+i[-x y+x+x y+x]+\left(y^{2}-1\right)}{x^{2}+(y-1)^{2}}$
$=\frac{x^{2}+y^{2}-1}{x^{2}+(y-1)^{2}}+\frac{2 x}{x^{2}+(y-1)^{2}} i$
Now, $\frac{z+i}{z-i}$ is purely imaginary.
$\therefore$
$\operatorname{Re}\left(\frac{z+i}{z-i}\right)=0$
$\begin{array}{lr}\Rightarrow & \frac{x^{2}+y^{2}-1}{x^{2}+(y-1)^{2}}=0 \\ \Rightarrow & x^{2}+y^{2}=1\end{array}$
$\therefore(x, y)$ lies on a circle.
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