Q103 / 170 · Manipal (MET) 2018 · Mathematics

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

  • A
    a circle $x^2+y^2=1$
  • Correct answer
    B
    the $x$-axis
  • C
    the $y$-axis
  • D
    the line $x+y=1$

Solution

Let $z=x+i y$ $\begin{aligned} & \therefore & i+z & =x+i(1+y) \\ & \text { and } & i-z & =-x+i(1-y) \\ & \because & & \left|\frac{i+z}{i-z}\right|\end{aligned}$ $\begin{array}{cc} \Rightarrow & x^2+(1+y)^2=x^2+(1-y)^2 \\ \Rightarrow & x^2+y^2+2 y+1=x^2+y^2+1-2 y \\ \Rightarrow & 4 y=0 \\ \Rightarrow & y=0 \end{array}$ i.e. $x$-axis.

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

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