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

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Question

(d) Let $z = x + iy,$ so that $\overline z = x - iy,$ therefore
${z^2} + \overline z = 0\, \Leftrightarrow ({x^2} - {y^2} + x) + i\,(2xy - y) = 0$
E

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(d) Let $z = x + iy,$ so that $\overline z = x - iy,$ therefore
${z^2} + \overline z = 0\, \Leftrightarrow ({x^2} - {y^2} + x) + i\,(2xy - y) = 0$
Equating real and imaginary parts, we get
${x^2} - {y^2} + x = 0$ .....$(i)$
and $2xy - y = 0$ ==> $y = 0$or $x = \frac{1}{2}$
If $y = 0$, then $(i)$  gives ${x^2} + x = 0\,\, \Rightarrow x = 0$or $x = - 1$
If $x = \frac{1}{2},$ Then ${x^2} - {y^2} + x = 0$

==>${y^2} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$==>$y = \pm \frac{{\sqrt 3 }}{2}$
Hence, there are four solutions in all.

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

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