The number of complex numbers z satisfying $\bar{z}=i z^2$ is
The number of complex numbers z satisfying $\bar{z}=i z^2$ is
3
4
2
5
Solution
Given equation is $\overline{\mathrm{Z}}=\mathrm{iz}^2...(i)$
Let $\mathrm{Z}=\mathrm{x}+\mathrm{iy} \Rightarrow \overline{\mathrm{Z}}=\mathrm{x}-\mathrm{iy}$
Now, from equation (i)
$
\begin{aligned}
& x-i y=(x+i y)^2 \\
& x-i y=i\left(x^2-y^2+2 i x y\right) \\
& x-i y=i x^2-i y^2-2 x y \\
& \Rightarrow 2+2 x y=0 \text { and } y=y^2-x^2
\end{aligned}
$
On solving get $(0,0)(0,1)\left(\frac{-\sqrt{3}}{2}, \frac{-1}{2}\right)\left(\frac{\sqrt{3}}{2}, \frac{-1}{2}\right)$ are four solutions