The number of solutions of the equation z^{2} | vedclass

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(B) Let $z = x + iy$. Then $z^{2} = x^{2} - y^{2} + 2ixy$. Given $z^{2} + \overline{z} = 0$,we have $(x^{2} - y^{2} + 2ixy) + (x - iy) = 0$. Grouping

Vedclass · Accepted Answer

$4$

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(B) Let $z = x + iy$. Then $z^{2} = x^{2} - y^{2} + 2ixy$. Given $z^{2} + \overline{z} = 0$,we have $(x^{2} - y^{2} + 2ixy) + (x - iy) = 0$. Grouping real and imaginary parts: $(x^{2} + x - y^{2}) + i(2xy - y) = 0$. Equating real and imaginary parts to zero: $x^{2} + x - y^{2} = 0$ $(i)$ $y(2x - 1) = 0$ (ii) From (ii),either $y = 0$ or $x = 1/2$. Case $1$: If $y = 0$,then $x^{2} + x = 0 \Rightarrow x(x + 1) = 0$,so $x = 0$ or $x = -1$. This gives solutions $z = 0$ and $z = -1$. Case $2$: If $x = 1/2$,then $(1/2)^{2} + 1/2 - y^{2} = 0$ $\Rightarrow 1/4 + 1/2 = y^{2}$ $\Rightarrow y^{2} = 3/4$ $\Rightarrow y = \pm \sqrt{3}/2$. This gives solutions $z = 1/2 + i\sqrt{3}/2$ and $z = 1/2 - i\sqrt{3}/2$. Thus,there are $4$ solutions in total.

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

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