Question

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

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

Answer 1

The correct option is B

the imaginary axis

Explanation for the correct option.

Step 1. Assume a complex number $z$ and find the value of $\left|z^2-1\right|$.

Let $z=x+iy$ be a complex number. Its modulus is given as: $\left|z\right|=\sqrt{x^2+y^2}$.

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

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

Now, the $\left|z^2-1\right|$ is given as:

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

Step 2. Form the equation using the given information and find the locus of $z$.

In the given equation $\left|z^2-1\right|={\left|z\right|}^2+1$ substitute the values of $\left|z^2-1\right|$ and $\left|z\right|$ and simplify by squaring both sides.

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

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

So $z$ lies on the imaginary axis.

Hence, the correct option is B.