Question

The number of solutions of the equation $z^2+\bar{z}=0$, where $z \in C$ are

• A. 1
• B. 4
• C. 5
• D. 6

Answer 1

The Correct Option is B

The correct option is(B): 4.

Let \(z = x + iy\)
\(\Rightarrow z^{2} =x^{2} -y^{2} + 2ixy\)
\(\because z^{2} + \bar{z} = 0\) (given) 
\(\therefore x^{2} -y^{2} + 2ixy+x-iy=0\)
\(\Rightarrow \left(x^{2} + x - y^{2}\right) + i \left(2.xy - y\right) = 0\) 
Equating the real and imaginary parts, we get 
\(x^{2} +x - y^{2} =0\,\,\,\,\,\dots(i)\) 
and \(2xy - y = 0\,\,\,\,\,\,\dots(ii)\)
By E(ii), we gety 
\(\left(2x - 1\right) = 0\)
\(\Rightarrow y=0\) or \(x = \frac{1}{2}\) 

so at y=0,

x²+x=0

x=0,-1.

again for \(x = \frac{1}{2}\) from 2 we get

y²= \(x = \frac{-1}{y}\)+\( \frac{1}{2}\)=\( \frac{3}{4}\)

\(y =\frac{+√2}{2}\)

therefor the solution for the given equation are: (a,y), are (0,0), (-1,0), (±\(\frac{√3}{2}\). \(\frac{1}{2}\).)

hence correct answer is 4.