Question

Number of complex numbers $z$ such that $\left| z \right| = 1$ and $\left| {\dfrac{z}{{\bar z}} + \dfrac{{\bar z}}{z}} \right| = 1$ is $\left( {\arg \left( z \right) \in \left[ {0,2\pi } \right]} \right)$ .
A. $4$
B. $6$
C. $8$
D.More than $8$

Answer 1

Hint : We know that $\left| {z\bar z} \right| = {\left| z \right|^2}$ .So, $\left| {{z^2} + {{\bar z}^2}} \right|$ will be equal to 1. Instead of $z$ take $z = x + iy$ . And then substituting the value of $z$ we will get ${x^2} - {y^2} = \dfrac{1}{2}$ and we also know that ${x^2} + {y^2} = 1$ . Solve that equation then we get the values of $x$ and $y$ .

Complete step-by-step answer :
It is given that $\left| z \right| = 1$ .
Also, given that $\left| {\dfrac{{{z^2} + {{\bar z}^2}}}{{\bar zz}}} \right| = 1$ .........(1)
Since it is known that $z\bar z = {\left| z \right|^2}$
Since, we know that $\left| z \right| = 1$ , so we can say that $z\bar z = 1$ .
Substitute the value $z\bar z = 1$ in the equation (1), we get,
$\left| {{z^2} + {{\bar z}^2}} \right| = 1$ ..........(2)
Since, $z$ is a complex number the value of $z$ as $x + iy$ where $x,y$ are real and $i$ is imaginary number.
We know the formula to find the conjugate of $z$ is $x - iy$ .
Since, the value of z is $z = x + iy$ .
Then if we square on both sides we get,
$\begin{array}{c} {z^2} = {\left( {x + iy} \right)^2}\\\ = \left( {x + iy} \right)\left( {x + iy} \right)\\\ = {x^2} - {y^2} + 2ixy \end{array}$ .........(3)
Since, we know $\bar z = x - iy$ . If we square on both sides for the equation we get,
$\begin{array}{c} {{\bar z}^2} = {\left( {x - iy} \right)^2}\\\ = \left( {x - iy} \right)\left( {x - iy} \right)\\\ = {x^2} - {y^2} - 2ixy \end{array}$ ..............(4)
On substituting (3) and (4) in (2) we get,
$\begin{array}{c} \left| {{z^2} + {{\bar z}^2}} \right| = 1\\\ \left| {{x^2} - {y^2} + 2ixy + {x^2} - {y^2} - 2ixy} \right| = 1 \end{array}$
Now, in the above equation equal but opposite sines $2ixy$ and $- 2ixy$ will get cancelled, we obtain,
$\begin{array}{c} \left| {2{x^2} - 2{y^2}} \right| = 1\\\ {x^2} - {y^2} = \dfrac{1}{2} \end{array}$ ..............(5)
And it is known that ${\left| z \right|^2} = 1$ since $z = x + iy$ we get,
$\begin{array}{l} {\left| {x + iy} \right|^2} = 1\\\ {x^2} + {y^2} = 1 \end{array}$ ............(6)
On equating the equations (5) and (6) we get,
$\begin{array}{c} 2{x^2} = \dfrac{1}{2} + 1\\\ 2{x^2} = \dfrac{3}{2}\\\ {x^2} = \dfrac{3}{4} \end{array}$
If we take the square root on both sides we get,
$x = \pm \dfrac{{\sqrt 3 }}{2}$
On substituting the value of $x$ in (5) we get,
$\begin{array}{c} \dfrac{3}{4} - {y^2} = \dfrac{1}{2}\\\ \dfrac{3}{4} - \dfrac{1}{2} = {y^2}\\\ {y^2} = \dfrac{3}{4} - \dfrac{2}{4} \end{array}$
The value for ${y^2}$ will be calculated as,
${y^2} = \dfrac{1}{4}$
Taking the square root on both sides we get,
$y = \pm \dfrac{1}{2}$
Hence, the value of y is $\dfrac{1}{2}$ and $- \dfrac{1}{2}$ .
The possible complex numbers are $z = \dfrac{{\sqrt 3 }}{2} + i\dfrac{1}{2}$ , $z = \dfrac{{\sqrt 3 }}{2} - i\dfrac{1}{2}$ , $z = - \dfrac{{\sqrt 3 }}{2} + i\dfrac{1}{2}$ and $z = \dfrac{-{\sqrt 3 }}{2} - \dfrac{i}{2}$.
Therefore, the complex numbers $z$ are $4$.
So, the correct answer is “Option A”.

Note : Always the value of $\left| z \right|$ will not be one. In complex numbers if the imaginary part is zero that is $y = 0$ then the complex number is real. In the complex number if the real number is zero that is $x = 0$ the complex number is purely imaginary .