# The complex number z which satisfies the condition

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

Solution:

$\left|\frac{i+z}{i-z}\right|=1$

$\Rightarrow\left|\frac{i+z}{i-z}\right|^{2}=1^{2}$

$\Rightarrow\left(\frac{i+z}{i-z}\right) \overline{\left(\frac{i+z}{i-z}\right)}=1$

$\Rightarrow\left(\frac{i+z}{i-z}\right)\left(\frac{-i+\bar{z}}{-i-\bar{z}}\right)=1$

$\Rightarrow\left(\frac{-i^{2}-z i+\bar{z} i+z \bar{z}}{-i^{2}+z i-\bar{z} i+z \bar{z}}\right)=1$

$\Rightarrow-i^{2}-z i+\bar{z} i+z \bar{z}=-i^{2}+z i-\bar{z} i+z \bar{z}$

$\Rightarrow-z i+\bar{z} i=z i-\bar{z} i$

$\Rightarrow \bar{z} i+\bar{z} i=z i+z i$

$\Rightarrow 2 \bar{z} i=2 z i$

$\Rightarrow \bar{z}=z$

$\Rightarrow z$ is purely real

Hence, the correct option is (b).