The area of the polygon,whose vertices are th | vedclass

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

Question

(C) Let $z = x + iy$,where $x, y \in \mathbb{R}$.Given the equation $\bar{z} = i z^{2}$.Substituting $z = x + iy$,we get $x - iy = i(x + iy)^{2}$.$x -

Vedclass · Accepted Answer

$\frac{3 \sqrt{3}}{4}$

Vedclass · Answer

(C) Let $z = x + iy$,where $x, y \in \mathbb{R}$.Given the equation $\bar{z} = i z^{2}$.Substituting $z = x + iy$,we get $x - iy = i(x + iy)^{2}$.$x - iy = i(x^{2} - y^{2} + 2xyi) = i(x^{2} - y^{2}) - 2xy$.Equating real and imaginary parts:$x = -2xy$ $\Rightarrow x(1 + 2y) = 0$ $\Rightarrow x = 0$ or $y = -\frac{1}{2}$.$-y = x^{2} - y^{2}$.Case $1$: If $x = 0$,then $-y = -y^{2}$ $\Rightarrow y^{2} - y = 0$ $\Rightarrow y(y - 1) = 0$. So $y = 0$ or $y = 1$.The roots are $z = 0$ and $z = i$. Since the question asks for non-real roots,we consider $z = i$.Case $2$: If $y = -\frac{1}{2}$,then $-(-\frac{1}{2}) = x^{2} - (-\frac{1}{2})^{2}$ $\Rightarrow \frac{1}{2} = x^{2} - \frac{1}{4}$ $\Rightarrow x^{2} = \frac{3}{4}$ $\Rightarrow x = \pm \frac{\sqrt{3}}{2}$.The roots are $z = \frac{\sqrt{3}}{2} - \frac{1}{2}i$ and $z = -\frac{\sqrt{3}}{2} - \frac{1}{2}i$.The vertices of the polygon are $(0, 1)$,$(\frac{\sqrt{3}}{2}, -\frac{1}{2})$,and $(-\frac{\sqrt{3}}{2}, -\frac{1}{2})$.This forms a triangle with base $b = \frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2}) = \sqrt{3}$ and height $h = 1 - (-\frac{1}{2}) = \frac{3}{2}$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \sqrt{3} 	imes \frac{3}{2} = \frac{3\sqrt{3}}{4}$.

The area of the polygon,whose vertices are the non-real roots of the equation $\bar{z} = i z^{2}$ is

Similar Questions

If $\mu = \frac{2z + 5i}{z - 3}$ and $|\mu| = 2$,then the locus of $z$ is:

The vertices $B$ and $D$ of a parallelogram are $1 - 2i$ and $4 + 2i$. If the diagonals are at right angles and $AC = 2BD$,the complex number representing $A$ is

If $a$ and $b$ are real numbers between $0$ and $1$ such that the points $z_1 = a + i$,$z_2 = 1 + bi$,and $z_3 = 0$ form an equilateral triangle,then

The points in the set $\{z \in \mathbb{C} : \arg \left(\frac{z-2}{z-6i}\right) = \frac{\pi}{2}\}$ (where $\mathbb{C}$ denotes the set of all complex numbers) lie on the curve which is a

The locus of the point $z$ satisfying $arg\left( \frac{z - 1}{z + 1} \right) = k$ (where $k$ is non-zero) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module