KCET · Maths · Complex Number
The number of solutions of the equation \(\mathrm{z}^{2}+\overline{\mathrm{z}}=0\), where \(\mathrm{z} \in \mathrm{C}\), are
- A 1
- B 4
- C 5
- D 6
Answer & Solution
Correct Answer
(B) 4
Step-by-step Solution
Detailed explanation
Let \(z=x+i y\)
\[
\begin{aligned}
&\Rightarrow \quad z^{2}=x^{2}-y^{2}+2 i x y \\
&\because \quad z^{2}+\bar{z}=0 \\
&\therefore \quad x^{2}-y^{2}+2 i x y+x-i y=0 \\
&\Rightarrow \quad\left(x^{2}+x-y^{2}\right)+i(2 x y-y)=0
\end{aligned}
\]
Equating the real and imaginary parts, we get
\[
x^{2}+x-y^{2}=0
\]
and \(\quad 2 x y-y=0 \quad \ldots\) (ii)
By Eq. (ii), we get
\(y(2 x-1)=0\)
\(\Rightarrow \quad \mathrm{y}=0 \quad\) or \(\quad \mathrm{x}=\frac{1}{2}\)
On solving these two with first equation, we get four solutions \((0,0),(-1,0)\) and \(\left(\frac{1}{2}, \pm \frac{\sqrt{3}}{2}\right)\).
\[
\begin{aligned}
&\Rightarrow \quad z^{2}=x^{2}-y^{2}+2 i x y \\
&\because \quad z^{2}+\bar{z}=0 \\
&\therefore \quad x^{2}-y^{2}+2 i x y+x-i y=0 \\
&\Rightarrow \quad\left(x^{2}+x-y^{2}\right)+i(2 x y-y)=0
\end{aligned}
\]
Equating the real and imaginary parts, we get
\[
x^{2}+x-y^{2}=0
\]
and \(\quad 2 x y-y=0 \quad \ldots\) (ii)
By Eq. (ii), we get
\(y(2 x-1)=0\)
\(\Rightarrow \quad \mathrm{y}=0 \quad\) or \(\quad \mathrm{x}=\frac{1}{2}\)
On solving these two with first equation, we get four solutions \((0,0),(-1,0)\) and \(\left(\frac{1}{2}, \pm \frac{\sqrt{3}}{2}\right)\).
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