The locus of the points z which satisfy the c | vedclass

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

Question

(B) Let $z = x + iy$. Then $\frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy}$.Multiplying the numerator and denominator by the conjugate of the

Vedclass · Accepted Answer

$A$ circle

Vedclass · Answer

(B) Let $z = x + iy$. Then $\frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy}$.Multiplying the numerator and denominator by the conjugate of the denominator,we get $\frac{((x - 1) + iy)((x + 1) - iy)}{(x + 1)^2 + y^2} = \frac{(x^2 + y^2 - 1) + i(2y)}{(x + 1)^2 + y^2}$.Given $	ext{arg} \left( \frac{z - 1}{z + 1} ight) = \frac{\pi}{3}$,we have $	an^{-1} \left( \frac{2y}{x^2 + y^2 - 1} ight) = \frac{\pi}{3}$.This implies $\frac{2y}{x^2 + y^2 - 1} = 	an \frac{\pi}{3} = \sqrt{3}$.Rearranging the terms,we get $x^2 + y^2 - 1 = \frac{2}{\sqrt{3}}y$,or $x^2 + y^2 - \frac{2}{\sqrt{3}}y - 1 = 0$.This is the equation of a circle.

The locus of the points $z$ which satisfy the condition $\text{arg} \left( \frac{z - 1}{z + 1} \right) = \frac{\pi}{3}$ is

Similar Questions

The minimum value of $|z-1|+|z-5|$ is

If $z_1, z_2, z_3$ are points in the Argand plane,then $\left| \begin{array}{ccc} z_1 & \overline{z_1} & 1 \\ z_2 & \overline{z_2} & 1 \\ z_3 & \overline{z_3} & 1 \end{array} \right| = $

If the point $\left(\frac{k-1}{k}, \frac{k-2}{k}\right)$ lies on the locus of $z$ satisfying the inequality $\left|\frac{z+3i}{3z+i}\right| < 1$,then the interval in which $k$ lies is

If $z=x+iy$ and the point $P$ in the Argand plane represents $z$,then the locus of $z$ satisfying the equation $|z-2|+|z-2i|=4$ is

Let $S = \{z \in \mathbb{C} : \left|\frac{z-6i}{z-2i}\right| = 1 \text{ and } \left|\frac{z-8+2i}{z+2i}\right| = \frac{3}{5}\}$. Then $\sum_{z \in S} |z|^2$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module