Question

The complex number z which satisfies the condition $\left|\frac{i+z}{i-z}\right|=1$ lies on

(a) circle x² + y² = 1
(b) the x−axis
(c) the y−axis
(d) the line x + y = 1

Answer 1

$$\begin{gathered} \left|\frac{i+z}{i-z}\right|=1 \\ \Rightarrow {\left|\frac{i+z}{i-z}\right|}^2=1^2 \\ \Rightarrow \left(\frac{i+z}{i-z}\right)\overline{\left(\frac{i+z}{i-z}\right)}=1 \\ \Rightarrow \left(\frac{i+z}{i-z}\right)\left(\frac{-i+\overline{z}}{-i-\overline{z}}\right)=1 \\ \Rightarrow \left(\frac{-i^2-zi+\overline{z}i+z\overline{z}}{-i^2+zi-\overline{z}i+z\overline{z}}\right)=1 \\ \Rightarrow -i^2-zi+\overline{z}i+z\overline{z}=-i^2+zi-\overline{z}i+z\overline{z} \\ \Rightarrow -zi+\overline{z}i=zi-\overline{z}i \\ \Rightarrow \overline{z}i+\overline{z}i=zi+zi \\ \Rightarrow 2\overline{z}i=2zi \\ \Rightarrow \overline{z}=z \\ \Rightarrow z\mathrm{is}\mathrm{purely}\mathrm{real} \end{gathered}$$

Hence, the correct option is (b).