If A = {z : |frac{z - 2}{z + 2}| = 3, z in C} | vedclass

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

Question

(B) The equation $|\frac{z - 2}{z + 2}| = 3$ represents a circle in the complex plane.Squaring both sides,$|z - 2|^2 = 9|z + 2|^2$.Let $z = x + iy$. T

Vedclass · Accepted Answer

$\frac{9}{2}$

Vedclass · Answer

(B) The equation $|\frac{z - 2}{z + 2}| = 3$ represents a circle in the complex plane.Squaring both sides,$|z - 2|^2 = 9|z + 2|^2$.Let $z = x + iy$. Then $(x - 2)^2 + y^2 = 9((x + 2)^2 + y^2)$.$x^2 - 4x + 4 + y^2 = 9(x^2 + 4x + 4 + y^2) = 9x^2 + 36x + 36 + 9y^2$.$8x^2 + 8y^2 + 40x + 32 = 0 \Rightarrow x^2 + y^2 + 5x + 4 = 0$.This is a circle with center $(-\frac{5}{2}, 0)$ and radius $r = \sqrt{(\frac{5}{2})^2 - 4} = \sqrt{\frac{25}{4} - 4} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.The condition $z_1 - z_2 = z_4 - z_3$ implies $z_1 + z_3 = z_2 + z_4$,which means the diagonals $PR$ and $QS$ bisect each other. Thus,$PQRS$ is a parallelogram.$A$ parallelogram inscribed in a circle must be a rectangle.The area of a rectangle inscribed in a circle is maximized when it is a square.The diameter of the circle is $2r = 2(\frac{3}{2}) = 3$.For a square inscribed in a circle of radius $r$,the diagonal is the diameter $d = 2r = 3$.The area of a square with diagonal $d$ is $\frac{1}{2}d^2 = \frac{1}{2}(3)^2 = \frac{9}{2}$.

If $A = \{z : |\frac{z - 2}{z + 2}| = 3, z \in C\}$ and $z_1, z_2, z_3, z_4 \in A$ are $4$ complex numbers representing points $P, Q, R, S$ respectively on the complex plane such that $z_1 - z_2 = z_4 - z_3$,then the maximum value of the area of quadrilateral $PQRS$ is:

Similar Questions

If $a = \cos \alpha + i\sin \alpha$,$b = \cos \beta + i\sin \beta$,$c = \cos \gamma + i\sin \gamma$ and $\frac{b}{c} + \frac{c}{a} + \frac{a}{b} = 1$,then $\cos (\beta - \gamma ) + \cos (\gamma - \alpha ) + \cos (\alpha - \beta )$ is equal to

The points $z_1, z_2, z_3, z_4$ in the complex plane are the vertices of a parallelogram taken in order,if and only if

If a complex number $z$ satisfies $|z^2-1|=|z|^2+1$,then $z$ lies on

The set of all $\alpha \in R$,for which $w = \frac{1 + (1 - 8\alpha)z}{1 - z}$ is a purely imaginary number,for all $z \in C$ satisfying $|z| = 1$ and $\text{Re}(z) \neq 1$,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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module