Question

The number of the solution of the equation, \[{{z}^{2}}+\bar{z}=0\] is

a.1

b.2

c.3

d.4

Answer 1

Hint: Put, \[z=x+iy\] and \[\bar{z}=x-iy\]. Solve the expression by putting these values. Find the real and imaginary part from the expression and find the solutions.

Complete step-by-step answer:
A complex number is a number that can be expressed in the form of \[x+iy\], where x and y are real numbers and i is the solution of the equation \[{{x}^{2}}=-1\]. As no real number satisfies this equation, it is called an imaginary number.

Let us consider, \[z=x+iy\].
\[\therefore \bar{z}=x-iy\].

We have been given that, \[{{z}^{2}}+\bar{z}=0\].

Substitute the value of z and \[\bar{z}\] in the above expression.
\[{{\left( x+iy \right)}^{2}}+\left( x-iy \right)=0\]
\[{{x}^{2}}+2xyi+{{\left( iy \right)}^{2}}+x-iy=0\] \[\left( \because {{i}^{2}}=-1 \right)\]
\[{{x}^{2}}+2ixy-{{y}^{2}}+x-iy=0\]

Now let us rearrange the above expression.
\[\left( {{x}^{2}}-{{y}^{2}}+x \right)+i\left( 2xy-y \right)=0\]

In this above equation, we can find the real part and imaginary part.

Thus the real part \[\Rightarrow {{x}^{2}}-{{y}^{2}}+x=0-(1)\]
Imaginary part \[\Rightarrow 2xy-y=0-(2)\]

Let us consider the equation (2).
\[\begin{align}
  & 2xy-y=0 \\
 & y\left( 2x-1 \right)=0 \\
\end{align}\]
\[\therefore y=0\] and \[2x-1=0\Rightarrow 2x=1\Rightarrow x=\dfrac{1}{2}\].

Thus we got \[x=\dfrac{1}{2}\] and y = 0.

Let us substitute these values in equation (1).

Put y = 0 in equation (1).
\[\begin{align}
  & {{x}^{2}}-{{y}^{2}}+x=0 \\
 & {{x}^{\to 2}}-0+x=0 \\
 & x\left( x+1 \right)=0 \\
\end{align}\]
\[\therefore x=0\] and \[x=-1\].

When y = 0, value of x = 0, -1.
Now put \[x=\dfrac{1}{2}\] in equation (1).
\[\begin{align}
  & {{x}^{2}}-{{y}^{2}}+x=0 \\
 & {{x}^{2}}+x={{y}^{2}} \\
 & {{y}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}}+\dfrac{1}{2} \\
 & y=\sqrt{\dfrac{1}{4}+\dfrac{1}{2}}=\sqrt{\dfrac{3}{4}}=\pm \dfrac{\sqrt{3}}{2} \\
\end{align}\]

Thus when, \[x=\dfrac{1}{2}\], we get the value of \[y=\pm \dfrac{\sqrt{3}}{2}\].

Hence we got the solutions as \[\left( 0,0 \right),\left( -1,0 \right),\left( \dfrac{1}{2},\dfrac{\sqrt{3}}{2} \right)\] and \[\left( \dfrac{1}{2},\dfrac{-\sqrt{3}}{2} \right)\].

Thus we got 4 solutions.
\[\therefore \] Option (d) is the correct answer.

Note:
We can also solve this by, \[{{z}^{2}}+\bar{z}=0\].
\[\therefore {{z}^{2}}=-\bar{z}\]
Taking modulus, \[{{\left| z \right|}^{2}}=\left| z \right|\].
Hence, \[\left| z \right|=0\] or 1
When z = 0, we get z = 0 to satisfy the equation.
In \[{{2}^{nd}}\] case, \[z={{e}^{i\theta }}\] for \[\theta \in \left[ 0,\left. 2\pi \right) \right.\].
\[{{\left| z \right|}^{2}}={{e}^{2i\theta }}={{e}^{\left( 2n+1 \right)i\pi }}.{{e}^{-i\theta }},n\in z\]
\[3\theta =\left( 2n+1 \right)\pi \] and \[\theta \in \left\{ \dfrac{\pi }{2},\pi ,\dfrac{5\pi }{3} \right\}\].
\[\dfrac{\pi }{3}+\dfrac{5\pi }{3}=2\pi \], thus we get the solutions as,
\[\begin{align}
  & z={{e}^{\dfrac{\pi i}{3}}}=\dfrac{1+\sqrt{3}i}{2} \\
 & z={{e}^{\pi i}}=-1 \\
 & z={{e}^{\dfrac{5\pi i}{3}}}={{e}^{\dfrac{-\pi i}{3}}}=\dfrac{1-\sqrt{3}i}{2} \\
\end{align}\]
Thus 4 solutions are, \[z=0,-1,\dfrac{1\pm \sqrt{3}i}{2}\].