The points in the set {z in mathbb{C} : arg l | vedclass

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

Question

(A) Given that,$\arg \left(\frac{z-2}{z-6i}\right) = \frac{\pi}{2}$.Let $z = x + iy$.The expression represents the locus of points $z$ such that the a

Vedclass · Accepted Answer

circle

Vedclass · Answer

(A) Given that,$\arg \left(\frac{z-2}{z-6i}ight) = \frac{\pi}{2}$.Let $z = x + iy$.The expression represents the locus of points $z$ such that the angle subtended by the segment joining $A(2, 0)$ and $B(0, 6)$ at $z$ is $\frac{\pi}{2}$.Using the property $\arg(z_1) - \arg(z_2) = \arg\left(\frac{z_1}{z_2}ight)$,we have:$\arg(z-2) - \arg(z-6i) = \frac{\pi}{2}$.Substituting $z = x + iy$:$	an^{-1}\left(\frac{y}{x-2}ight) - 	an^{-1}\left(\frac{y-6}{x}ight) = \frac{\pi}{2}$.Using the identity $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1}\left(\frac{A-B}{1+AB}ight) = \frac{\pi}{2}$,the denominator $1+AB$ must be $0$ as $A-B$ is finite.$1 + \left(\frac{y}{x-2}ight)\left(\frac{y-6}{x}ight) = 0$.$x(x-2) + y(y-6) = 0$.$x^2 - 2x + y^2 - 6y = 0$.This is the equation of a circle with center $(1, 3)$ and radius $\sqrt{1^2 + 3^2} = \sqrt{10}$.

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

Similar Questions

If ${z_1}, {z_2}, {z_3}, {z_4}$ are the affixes of four points in the Argand plane and $z$ is the affix of a point such that $|z - z_1| = |z - z_2| = |z - z_3| = |z - z_4|$,then ${z_1}, {z_2}, {z_3}, {z_4}$ are

Let $z$ be a complex number such that $|z + 2| = |z - 2|$ and $\arg\left(\frac{z + 3}{z - i}\right) = \frac{\pi}{4}$. Then $|z|^2$ is equal to:

The area (in sq units) of the triangle whose vertices are the points represented by the complex numbers $0, z$,and $z e^{i \alpha}$ $(0 < \alpha < \pi)$ is:

Let $z_{1}$ and $z_{2}$ be two complex numbers such that $\arg(z_{1}-z_{2})=\frac{\pi}{4}$ and $z_{1}, z_{2}$ satisfy the equation $|z-3|=\operatorname{Re}(z)$. Then the imaginary part of $z_{1}+z_{2}$ is equal to ..... .

Let $a = \text{Im}\left( \frac{1 + z^2}{2iz} \right)$,where $z$ is any non-zero complex number. The set $A = \{ a : |z| = 1 \text{ and } z \ne \pm 1 \}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module