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 .