The number of solutions of the equation z^2 + | vedclass

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(D) Let $z = x + iy$,so that $\bar{z} = x - iy$. Therefore,the equation becomes:$z^2 + \bar{z} = 0 \iff (x + iy)^2 + (x - iy) = 0$$(x^2 - y^2 + x) + i

Vedclass · Accepted Answer

$4$

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(D) Let $z = x + iy$,so that $\bar{z} = x - iy$. Therefore,the equation becomes:$z^2 + \bar{z} = 0 \iff (x + iy)^2 + (x - iy) = 0$$(x^2 - y^2 + x) + i(2xy - y) = 0$Equating real and imaginary parts to zero,we get:$x^2 - y^2 + x = 0$ .....$(i)$$2xy - y = 0$ .....$(ii)$From $(ii)$,$y(2x - 1) = 0$,which implies $y = 0$ or $x = \frac{1}{2}$.Case $1$: If $y = 0$,then $(i)$ gives $x^2 + x = 0$,so $x(x + 1) = 0$,which gives $x = 0$ or $x = -1$. This provides two solutions: $z = 0$ and $z = -1$.Case $2$: If $x = \frac{1}{2}$,then $(i)$ gives $(\frac{1}{2})^2 - y^2 + \frac{1}{2} = 0$,so $\frac{1}{4} - y^2 + \frac{1}{2} = 0$,which means $y^2 = \frac{3}{4}$,so $y = \pm \frac{\sqrt{3}}{2}$. This provides two solutions: $z = \frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $z = \frac{1}{2} - i\frac{\sqrt{3}}{2}$.Thus,there are $4$ solutions in total.

List-$I$	List-$II$
$(P)$ $\|z\|^2$ is equal to	$(1)$ $12$
$(Q)$ $\|z-\bar{z}\|^2$ is equal to	$(2)$ $4$
$(R)$ $\|z\|^2+\|z+\bar{z}\|^2$ is equal to	$(3)$ $8$
$(S)$ $\|z+1\|^2$ is equal to	$(4)$ $10$
	$(5)$ $7$

The number of solutions of the equation $z^2 + \bar{z} = 0$ is

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