Explanation:
C The number of solution of equation
\(z^2+\bar{z}=0\)
Let, \(z=x+\) iy
\((x+i y)^2+(x-i y)=0\)
\(x^2+i^2 y^2+i 2 x y+x-i y=0\)
\(x^2-y^2+x+i 2 x y-i y=0\)
\(x^2-y^2+x+i(2 x y-y)=0\)
On comparing both real \& imaginary parts,
\(x^2+x-y^2=0\)
\(2 x y-y=0\)
\(y(2 x-1)=0\)
From equation (ii)
\(y(2 x-1)=0\)
\(y=0\)
\(x=\frac{1}{2}\)
Putting \(y=0\) in equation (i) we get,
\(x^2+x=0\)
\(x(x+1)=0 \Rightarrow x=0,-1\)
Now putting in equation (i) we get,
\(\frac{1}{4}+\frac{1}{2}-y^2=0\)
\(y^2=\frac{3}{4}\)
\(y= \pm \frac{\sqrt{3}}{2}\)
Thus, \((\mathrm{x}, \mathrm{y})=(0,0)(-1,0),\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}, \frac{-\sqrt{3}}{2}\right)\)
Hence, the number of solution of given equation is 4 .