Q88 / 155 · WBJEE 2017 · Mathematics

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

  • A
    a straight line
  • B
    an ellipse
  • C
    a hyperbola
  • Correct answer
    D
    a circle

Solution

$\frac{z+i}{z-i}=\frac{(x+j)+i}{(x+i y)-i}$ $=\frac{x+(1+y) i}{x+(y-1) i}$ $=\frac{x+(y+1) i}{x+(y-1) i} \times \frac{x-(y-1) i}{x-(y-1) i}$ $=\frac{x^{2}-x(y-1) i+x(y+1) i-(y+1)(y-1) i^{2}}{x^{2}-(y-1)^{2} i^{2}}$ $=\frac{x^{2}+i[-x y+x+x y+x]+\left(y^{2}-1\right)}{x^{2}+(y-1)^{2}}$ $=\frac{x^{2}+y^{2}-1}{x^{2}+(y-1)^{2}}+\frac{2 x}{x^{2}+(y-1)^{2}} i$ 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.

Hence, the correct answer is option: 'D'.

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