Let bar{z} denote the complex conjugate of a | vedclass

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Question

Given,

$\bar{z}-z^2=i\left(\bar{z}+z^2\right)$

$\Rightarrow(1-i) \bar{z}=(1+i) z^2$

$\Rightarrow \frac{(1-i)}{(1+i)} \bar{z}=z^2$

$\Righta

Vedclass · Accepted Answer

$4$

Vedclass · Answer

Given,

$\bar{z}-z^2=i\left(\bar{z}+z^2ight)$

$\Rightarrow(1-i) \bar{z}=(1+i) z^2$

$\Rightarrow \frac{(1-i)}{(1+i)} \bar{z}=z^2$

$\Rightarrow\left(-\frac{2 i}{2}ight) \bar{z}=z^2$

$	herefore z^2=-i \bar{z}$

Let $z = x +i y$,

$	herefore\left( x ^2- y ^2ight)+i(2 xy )=-i( x -i y )$

$	ext { so, } x ^2- y ^2+ y =0$

$	ext { and }(2 y +1) x =0$

$\Rightarrow x =0 	ext { or } y =-\frac{1}{2}$

Case $I$ : When $x =0$

$	herefore(1) \Rightarrow y(1-y)=0 \Rightarrow y=0,1$

$	herefore(0,0),(0,1)$

Case $II$ : When $y =-\frac{1}{2}$

$	herefore(1) \Rightarrow x^2-\frac{1}{4}-\frac{1}{2}=0 \Rightarrow x^2=\frac{3}{4} \Rightarrow x= \pm \frac{\sqrt{3}}{2}$

$	herefore\left(\frac{\sqrt{3}}{2},-\frac{1}{2}ight),\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}ight)$

$\Rightarrow$ Number of distinct ' $z$ ' is equal to $4$ .

Let $\bar{z}$ denote the complex conjugate of a complex number $z$ and let $i=\sqrt{-1}$. In the set of complex numbers, the number of distinct roots of the equation

$\bar{z}-z^2=i\left(\bar{z}+z^2\right)$ is. . . . . .

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