Question

If $\left|z^2-1\right|={\left|z\right|}^2+1$, then $z$ lies on

• A. ellipse
• B. the imaginary axis
• C. circle
• D. the real axis

Answer 1

The correct option is B

the imaginary axis

Explanation for the correct option.

Find the locus of $z$.

Let $z=x+iy$, then its modulus is given as: $\left|z\right|=\sqrt{x^2+y^2}$.

And the complex number $z^2-1$ is given as:

$\begin{array}{rcl}{z}^2-1& =& {\left(x+iy\right)}^2-1\\ & =& x^2+2xyi+i^2{y}^2-1\\ & =& \left(x^2-y^2-1\right)+2xyi\end{array}$

So the modulus of $\left|z^2-1\right|$ is given as: $\left|\begin{array}{rcl}& & \left(x^2-y^2-1\right)+2xyi\end{array}\right|=\sqrt{{\begin{array}{rcl}& & \left(x^2-y^2-1\right)\end{array}}^2+{\left(2xy\right)}^2}$.

Now, substitute the values of $\left|z\right|=\sqrt{x^2+y^2}$ and $\left|z^2-1\right|=\sqrt{\begin{array}{rcl}& & {\left(x^2-y^2-1\right)}^2+{\left(2xy\right)}^2\end{array}}$ in the equation $\left|z^2-1\right|={\left|z\right|}^2+1$ and simplify and square both sides.

$\begin{array}{rcl}\left|z^2-1\right|& =& {\left|z\right|}^2+1\\ & \Rightarrow & \sqrt{{\left(x^2-y^2-1\right)}^2+{\left(2xy\right)}^2}={\left(\sqrt{x^2+y^2}\right)}^2+1\\ & \Rightarrow & {\left(\sqrt{{\left(x^2-y^2-1\right)}^2+{\left(2xy\right)}^2}\right)}^2={\left(x^2+y^2+1\right)}^2\\ & \Rightarrow & {\left(x^2-y^2-1\right)}^2+{\left(2xy\right)}^2={\left(x^2+y^2+1\right)}^2\\ & \Rightarrow & x^4+y^4+1-2x^2{y}^2+2y^2-2x^2+4x^2{y}^2=x^4+y^4+1+2x^2{y}^2+2x^2+2y^2\\ & \Rightarrow & -4x^2=0\\ & \Rightarrow & x=0\end{array}$

So the locus of $z$ is $x=0$ which represents the imaginary axis on the complex plane.

So the complex number $z$ lies on the imaginary axis.

Hence, the correct option is B.