Question

What is the number of complex numbers z satisfying ${z^3} = \overline z $?
$
  {\text{A}}{\text{. 1}} \\
  {\text{B}}{\text{. 2}} \\
  {\text{C}}{\text{. 4}} \\
  {\text{D}}{\text{. 5}} \\
 $

Answer 1

Hint- Here, we will proceed by assuming the complex number which satisfies the given equation as $z = a + ib$ and then we will apply the formula ${\left( {x + y} \right)^3} = {x^3} + {y^3} + 3{x^2}y + 3x{y^2}$ and \[\overline z = \overline {a + ib} = a - ib\] in the equation.

Complete Step-by-Step solution:
Let us suppose that any complex number $z = a + ib$ satisfies the equation ${z^3} = \overline z {\text{ }} \to (1{\text{)}}$
By putting $z = a + ib$ in equation (1), we get
${\left( {a + ib} \right)^3} = \overline {a + ib} {\text{ }} \to {\text{(2)}}$
Using the formula ${\left( {x + y} \right)^3} = {x^3} + {y^3} + 3{x^2}y + 3x{y^2}$ in the equation (2), we have
\[
   \Rightarrow {a^3} + {\left( {ib} \right)^3} + 3{a^2}\left( {ib} \right) + 3a{\left( {ib} \right)^2} = \overline {a + ib} \\
   \Rightarrow {a^3} + {i^3}{b^3} + 3i{a^2}b + 3a\left( {{i^2}} \right)\left( {{b^2}} \right) = \overline {a + ib} \\
   \Rightarrow {a^3} + \left( {{i^2}} \right)\left( i \right){b^3} + 3i{a^2}b + 3a\left( {{i^2}} \right)\left( {{b^2}} \right) = \overline {a + ib} \\
 \]
Using $
  i = \sqrt { - 1} \\
   \Rightarrow {i^2} = - 1 \\
 $ in the above equation, we get
\[
   \Rightarrow {a^3} + \left( { - 1} \right)\left( i \right){b^3} + 3i{a^2}b + 3a\left( { - 1} \right)\left( {{b^2}} \right) = \overline {a + ib} \\
   \Rightarrow {a^3} - i{b^3} + 3i{a^2}b - 3a{b^2} = \overline {a + ib} {\text{ }} \to {\text{(3)}} \\
 \]
As we know that the conjugate of any complex number $z = a + ib$ is given by
\[\overline z = \overline {a + ib} = a - ib{\text{ }}\]
Using the above formula in equation (3), we get
\[
   \Rightarrow {a^3} - i{b^3} + 3i{a^2}b - 3a{b^2} = a - ib \\
   \Rightarrow \left( {{a^3} - 3a{b^2}} \right) - i\left( {{b^3} - 3{a^2}b} \right) = a - ib \\
 \]
By comparing the real and imaginary parts of the complex numbers given on the LHS and the RHS of the above equation, we get
\[
  {a^3} - 3a{b^2} = a \\
   \Rightarrow {a^3} - 3a{b^2} - a = 0 \\
   \Rightarrow a\left( {{a^2} - 3{b^2} - 1} \right) = 0 \\
 \]
Either \[a = 0\] or \[
  {a^2} - 3{b^2} - 1 = 0 \\
   \Rightarrow {a^2} = 3{b^2} + 1{\text{ }} \to {\text{(4)}} \\
 \]
and \[
  {b^3} - 3{a^2}b = b \\
   \Rightarrow {b^3} - 3{a^2}b - b = 0 \\
   \Rightarrow b\left( {{b^2} - 3{a^2} - 1} \right) = 0 \\
 \]
Either \[b = 0\] or \[{b^2} - 3{a^2} - 1 = 0{\text{ }} \to {\text{(5)}}\]
By substituting equation (4) in equation (5), we get
\[
   \Rightarrow {b^2} - 3\left( {3{b^2} + 1} \right) - 1 = 0 \\
   \Rightarrow {b^2} - 9{b^2} - 3 - 1 = 0 \\
   \Rightarrow - 8{b^2} - 4 = 0 \\
   \Rightarrow 8{b^2} = - 4 \\
   \Rightarrow {b^2} = \dfrac{{ - 4}}{8} = \dfrac{{ - 1}}{2} \\
   \Rightarrow b = \pm \sqrt {\dfrac{{ - 1}}{2}} = \pm \dfrac{{\sqrt { - 1} }}{{\sqrt 2 }} \\
   \Rightarrow b = \pm \dfrac{i}{{\sqrt 2 }} \\
 \]
Put \[b = \dfrac{i}{{\sqrt 2 }}\] in equation (4), we get
\[
   \Rightarrow {a^2} = 3{\left( {\dfrac{i}{{\sqrt 2 }}} \right)^2} + 1 = 3\left[ {\dfrac{{{i^2}}}{{{{\left( {\sqrt 2 } \right)}^2}}}} \right] + 1 = 3\left[ {\dfrac{{ - 1}}{2}} \right] + 1 = - \dfrac{3}{2} + 1 = \dfrac{{ - 3 + 2}}{2} = - \dfrac{1}{2} \\
   \Rightarrow a = \pm \sqrt {\dfrac{{ - 1}}{2}} = \pm \dfrac{{\sqrt { - 1} }}{{\sqrt 2 }} = \pm \dfrac{i}{{\sqrt 2 }} \\
 \]
Put \[b = - \dfrac{i}{{\sqrt 2 }}\] in equation (4), we get
\[
   \Rightarrow {a^2} = 3{\left( { - \dfrac{i}{{\sqrt 2 }}} \right)^2} + 1 = 3\left[ {\dfrac{{{i^2}}}{{{{\left( {\sqrt 2 } \right)}^2}}}} \right] + 1 = 3\left[ {\dfrac{{ - 1}}{2}} \right] + 1 = - \dfrac{3}{2} + 1 = \dfrac{{ - 3 + 2}}{2} = - \dfrac{1}{2} \\
   \Rightarrow a = \pm \sqrt {\dfrac{{ - 1}}{2}} = \pm \dfrac{{\sqrt { - 1} }}{{\sqrt 2 }} = \pm \dfrac{i}{{\sqrt 2 }} \\
 \]
Therefore, a = 0, b = 0, a = \[ \pm \dfrac{i}{{\sqrt 2 }}\] and b = \[ \pm \dfrac{i}{{\sqrt 2 }}\]
Corresponding to a = 0 and b = 0, the complex number is $
  z = a + ib = 0 + i\left( 0 \right) \\
   \Rightarrow z = 0{\text{ }} \to {\text{(5)}} \\
 $
Corresponding to a = \[ \pm \dfrac{i}{{\sqrt 2 }}\] and b = \[ \pm \dfrac{i}{{\sqrt 2 }}\], the complex numbers are $z = a + ib = \pm \dfrac{i}{{\sqrt 2 }} + i\left( { \pm \dfrac{i}{{\sqrt 2 }}} \right)$ which includes the following complex numbers
 $
  z = \dfrac{i}{{\sqrt 2 }} + i\left( {\dfrac{i}{{\sqrt 2 }}} \right) = \dfrac{i}{{\sqrt 2 }} + \dfrac{{{i^2}}}{{\sqrt 2 }} = \dfrac{i}{{\sqrt 2 }} + \dfrac{{ - 1}}{{\sqrt 2 }} \\
   \Rightarrow z = \dfrac{{ - 1}}{{\sqrt 2 }} + \dfrac{i}{{\sqrt 2 }}{\text{ }} \to {\text{(6)}} \\
 $
or $
  z = - \dfrac{i}{{\sqrt 2 }} + i\left( { - \dfrac{i}{{\sqrt 2 }}} \right) = - \dfrac{i}{{\sqrt 2 }} - \dfrac{{{i^2}}}{{\sqrt 2 }} = - \dfrac{i}{{\sqrt 2 }} - \left( {\dfrac{{ - 1}}{{\sqrt 2 }}} \right) = - \dfrac{i}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} \\
   \Rightarrow z = \dfrac{1}{{\sqrt 2 }} - \dfrac{i}{{\sqrt 2 }}{\text{ }} \to {\text{(7)}} \\
 $
or \[
  z = \dfrac{i}{{\sqrt 2 }} + i\left( { - \dfrac{i}{{\sqrt 2 }}} \right) = \dfrac{i}{{\sqrt 2 }} - \dfrac{{{i^2}}}{{\sqrt 2 }} = \dfrac{i}{{\sqrt 2 }} - \left( {\dfrac{{ - 1}}{{\sqrt 2 }}} \right) = \dfrac{i}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} \\
   \Rightarrow z = \dfrac{1}{{\sqrt 2 }} + \dfrac{i}{{\sqrt 2 }}{\text{ }} \to {\text{(8)}} \\
 \]
or $
  z = - \dfrac{i}{{\sqrt 2 }} + i\left( {\dfrac{i}{{\sqrt 2 }}} \right) = - \dfrac{i}{{\sqrt 2 }} + \dfrac{{{i^2}}}{{\sqrt 2 }} = - \dfrac{i}{{\sqrt 2 }} + \dfrac{{ - 1}}{{\sqrt 2 }} = - \dfrac{i}{{\sqrt 2 }} - \dfrac{1}{{\sqrt 2 }} \\
   \Rightarrow z = - \dfrac{1}{{\sqrt 2 }} - \dfrac{i}{{\sqrt 2 }}{\text{ }} \to {\text{(9)}} \\
 $
Therefore, z = 0, $z = \dfrac{{ - 1}}{{\sqrt 2 }} + \dfrac{i}{{\sqrt 2 }}$, $z = \dfrac{1}{{\sqrt 2 }} - \dfrac{i}{{\sqrt 2 }}$, \[z = \dfrac{1}{{\sqrt 2 }} + \dfrac{i}{{\sqrt 2 }}\] and $z = - \dfrac{1}{{\sqrt 2 }} - \dfrac{i}{{\sqrt 2 }}$ satisfies the equation ${z^3} = \overline z $ which means there are total 5 complex numbers satisfying the given equation.
Hence, option D is correct.

Note- In this particular problem, we have compared two complex numbers which are equated together i.e., $a + ib = c + id$ this gives that both the real part and the imaginary part of the complex numbers given on the LHS and the RHS of the equation $a + ib = c + id$ will always be equal i.e., a = c and b = d.