Number of solutions of equation {z^2} + bar z = 0 is

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4-1.Complex numbers

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

$1$

$2$

$3$

$4$

Solution

(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.

Standard 11

Mathematics

Let $\alpha$ and $\beta$ be the sum and the product of all the non-zero solutions of the equation $(\bar{z})^2+|z|=0, z \in C$. Then $4\left(\alpha^2+\beta^2\right)$ is equal to :

hard

(JEE MAIN-2024)

Let $z_1$ and $z_2$ be any two non-zero complex numbers such that $3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|$. If $z = \frac{{3{z_1}}}{{2{z_2}}} + \frac{{2{z_2}}}{{3{z_1}}}$ then

hard

(JEE MAIN-2019)

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

Solution

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