If a complex number $z$ satisfies $|z|^2+1=\mid z^2-1$, then the locus of $z$ is

If a complex number $z$ satisfies $|z|^2+1=\mid z^2-1$, then the locus of $z$ is
  1. a circle
  2. the real axis
  3. the imaginary axis
  4. the straight line $y=x$

Solution

$\begin{aligned} & \text { Let } z=x+i y \text {, so } \\ & \qquad|z|^2+1=\left|z^2-1\right|\end{aligned}$ $ \begin{aligned} & \Rightarrow x^2+y^2+1=\sqrt{\left(x^2-y^2-1\right)^2+4 x^2 y^2} \\ & \Rightarrow\left(x^2+y^2+1\right)^2=\left(x^2-y^2-1\right)^2+4 x^2 y^2 \\ & \Rightarrow\left(x^2+y^2+1\right)^2-\left(x^2-y^2-1\right)^2=4 x^2 y^2 \\ & \Rightarrow \quad\left[\left(x^2+y^2+1\right)+\left(x^2-y^2-1\right)\right] \\ & \quad\left[\left(x^2+y^2+1\right)-\left(x^2-y^2-1\right)\right]=4 x^2 y^2 \\ & \Rightarrow\left(2 x^2\right)\left(2 y^2+2\right)=4 x^2 y^2 \\ & \Rightarrow \quad x^2 y^2+2 x^2=x^2 y^2 \Rightarrow x^2=0 \Rightarrow x=0, \end{aligned} $ so locus is a imaginary axis

Asked in: AP EAMCET 2018 (23 Apr Shift 2)

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