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The complex number z which satisfies the condition \left |\frac{i+z}{i-z} \right |=1 lies on
A. circle x^2 + y^2 = 1
B. the x-axis
C. the y-axis
D. the line x + y = 1

Answers (1)

The answer is the option (b).

\begin{array}{c} \text { Let } z=x+i y \quad\left|\frac{i+x+i y}{i-x-y i}\right|=1 \\\\ \left|\frac{x-(y+1) i}{-x-(y-1) i}\right|=1 \\\\ |x+(y+1) i|=|-x-(y-1) i| \\\\ \sqrt{x^{2}+(y+1)^{2}}=\sqrt{x^{2}+(y-1)^{2}} \\\\ x^{2}+(y+1)^{2}=x^{2}+(y-1)^{2} \\\\ y^{2}+2 y+1=y^{2}-2 y+1 \\ 2 y=-2 y \\\\ y=0 ; x-a x i s \end{array}

Hence, b is correct

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