Question

If \[\left| {{z^2} - 1} \right| = \left| {{z^2}} \right| + 1\], then \[z\] lies on
A. An ellipse
B. The imaginary axis
C. A circle
D. The real axis

Answer 1

Hint: In this question, we will find the locus of \[z\]. For this, we will first consider \[z = x + iy\] and its modulus as \[\left| z \right| = \sqrt {{x^2} + {y^2}} \]. We will now find out the value of \[{z^2} - 1\] and its modulus of \[\left| {{z^2} - 1} \right|\] by substituting the value of \[z\].
After that we will compute the value of given expression \[\left| {{z^2} - 1} \right| = \left| {{z^2}} \right| + 1\] by putting the value of \[\left| z \right|\] and \[\left| {{z^2} - 1} \right|\]. In the end we calculate the value of \[x\].

Formula Used: The modulus is denoted by the symbol “z” which is a non-negative real number that is equal to
\[\sqrt {{a^2} + {b^2}} \] for the complex number \[z = a + ib\].

Complete step by step solution: We have the expression \[\left| {{z^2} - 1} \right| = \left| {{z^2}} \right| + 1\]
Let us consider \[z = x + iy\] and its modulus is represented as \[\left| z \right| = \sqrt {{x^2} + {y^2}} \].
First, we will find the value of \[{z^2} - 1\] by substituting the value of \[z\].
\[{z^2} - 1 = {\left( {x + iy} \right)^2} - 1\]
\[{z^2} - 1 = {x^2} + 2xyi + {i^2}{y^2} - 1\]
\[{z^2} - 1 = \left( {{x^2} - {y^2} - 1} \right) + 2xyi\]
Similarly, we will find the value of \[\left| {{z^2} - 1} \right|\] by substituting the value of \[z\].
\[\left| {\left( {{x^2} - {y^2} - 1} \right) + 2xyi} \right| = \sqrt {{{\left( {{x^2} - {y^2} - 1} \right)}^2} + {{\left( {2xyi} \right)}^2}} \]
We will now substitute the values of \[\left| z \right| = \sqrt {{x^2} + {y^2}} \] and \[\left| {{z^2} - 1} \right| = \sqrt {{{\left( {{x^2} - {y^2} - 1} \right)}^2} + {{\left( {2xyi} \right)}^2}} \] in the given expression \[\left| {{z^2} - 1} \right| = \left| {{z^2}} \right| + 1\].
\[\begin{array}{l}\left| {{z^2} - 1} \right| = \left| {{z^2}} \right| + 1\\ \Rightarrow \sqrt {{{\left( {{x^2} - {y^2} - 1} \right)}^2} + {{\left( {2xyi} \right)}^2}} = \sqrt {{x^2} + {y^2}} + 1\end{array}\]
Further, we will square on both sides to get.
\[\begin{array}{l} \Rightarrow {\left( {\sqrt {{{\left( {{x^2} - {y^2} - 1} \right)}^2} + {{\left( {2xyi} \right)}^2}} } \right)^2} = {\left( {\sqrt {{x^2} + {y^2}} } \right)^2} + 1\\ \Rightarrow {\left( {\sqrt {{{\left( {{x^2} - {y^2} - 1} \right)}^2} + {{\left( {2xyi} \right)}^2}} } \right)^2} = {x^2} + {y^2} + 1\\ \Rightarrow {x^4} + {y^4} + 1 - 2{x^2}{y^2} + 2{y^2} - 2{x^2} + 4{x^2}{y^2} = {x^4} + {y^4} + 1 + 2{x^2}{y^2} + 2{y^2} + 2{x^2}\end{array}\]
After further calculations, we get
\[\begin{array}{l} \Rightarrow 2{x^2}{y^2} = 2{x^2}{y^2} + 4{x^2}\\ \Rightarrow x = 0\end{array}\]
Then, we get
\[z = x + iy = 0 + iy = iy\]
As a result, the locus of \[z\] is \[x = 0\] which represents that imaginary axis on the complex plane.
Thus, the complex number \[z\] lies on the imaginary axis.

Option ‘B’ is correct

Note: Many students make the error of starting their writing with the positive value of the modulus, which is incorrect because the value of the modulus depends on whether the value contained within the modulus is positive or negative, and we should present two examples to illustrate this.