The locus of the point z satisfying the equat | vedclass

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

Question

(A) Given equation: $|iz - 1| + |z - i| = 2$ $|i(z - 1/i)| + |z - i| = 2$ $|i(z + i)| + |z - i| = 2$ $|i| |z + i| + |z - i| = 2$ Since $|i| = 1$,we ha

Vedclass · Accepted Answer

$A$ straight line

Vedclass · Answer

(A) Given equation: $|iz - 1| + |z - i| = 2$ $|i(z - 1/i)| + |z - i| = 2$ $|i(z + i)| + |z - i| = 2$ $|i| |z + i| + |z - i| = 2$ Since $|i| = 1$,we have $|z - (-i)| + |z - i| = 2$ This is of the form $|z - z_1| + |z - z_2| = 2a$,where $z_1 = -i$ and $z_2 = i$. The distance between $z_1$ and $z_2$ is $|z_2 - z_1| = |i - (-i)| = |2i| = 2$. Since the sum of the distances from $z$ to two fixed points $z_1$ and $z_2$ is equal to the distance between the points themselves $(|z - z_1| + |z - z_2| = |z_1 - z_2|)$,the locus of $z$ is the line segment joining $z_1$ and $z_2$. Thus,the locus is a straight line.

The locus of the point $z$ satisfying the equation $|iz - 1| + |z - i| = 2$ is

Similar Questions

When $\frac{z + i}{z + 2}$ is purely imaginary,the locus described by the point $z$ in the Argand diagram is a

Let $S = \{z \in \mathbb{C} : 4z^2 + \overline{z} = 0\}$. Then $\sum_{z \in S} |z|^2$ is equal to:

Let $z=x+iy$ be a complex number,where $x$ and $y$ are integers and $i=\sqrt{-1}$. Then the area of the rectangle whose vertices are the roots of the equation $z\bar{z}^3+\bar{z}z^3=350$ is

If $|z_1|=1, |z_2|=2, |z_3|=3$ and $|9z_1z_2 + 4z_1z_3 + z_2z_3| = 12$,then the value of $|z_1 + z_2 + z_3|$ is equal to :-

The inequality $|z - 4| < |z - 2|$ represents the region given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module