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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$4$

Vedclass · Answer

(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

Similar Questions

If $|z + 4| \le 3$,then the maximum value of $|z + 1|$ is

Let $|z_1 - 8 - 2i| \leq 1$ and $|z_2 - 2 + 6i| \leq 2$,where $z_1, z_2 \in \mathbb{C}$. Then the minimum value of $|z_1 - z_2|$ is:

$A$ circle whose radius is $r$ and centre is $z_0$,then the equation of the circle is

If $C$ is a point on the straight line joining the points $A(-2+i)$ and $B(3-4i)$ in the Argand plane and $\frac{AC}{CB}=\frac{1}{2}$,then the argument of $C$ is

Let $A, B, C$ be three sets of complex numbers defined as $A = \{z : \text{Im}(z) \ge 1\}$,$B = \{z : |z - 2 - i| = 3\}$,and $C = \{z : \text{Re}((1 - i)z) = \sqrt{2}\}$. If $z$ is any point in $A \cap B \cap C$,then $|z + 1 - i|^2 + |z - 5 - i|^2$ lies between:

Vedclass Test Series

Exam Paper Generator

Online Exam Module