Let S = {z : 3 le |2z - 3(1 + i)| le 7} be a | vedclass

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

Question

(B) Given the set $S = \{z : 3 \le |2z - 3(1 + i)| \le 7\}$.Dividing by $2$, we get: $\frac{3}{2} \le |z - \frac{3}{2}(1 + i)| \le \frac{7}{2}$.This r

Vedclass · Accepted Answer

$ \frac{3}{2} $

Vedclass · Answer

(B) Given the set $S = \{z : 3 \le |2z - 3(1 + i)| \le 7\}$.Dividing by $2$, we get: $\frac{3}{2} \le |z - \frac{3}{2}(1 + i)| \le \frac{7}{2}$.This represents an annulus centered at $C(\frac{3}{2}, \frac{3}{2})$ with inner radius $r_1 = \frac{3}{2}$ and outer radius $r_2 = \frac{7}{2}$.We want to find the minimum distance from a point $P(-\frac{5}{2}, -\frac{3}{2})$ to the set $S$.The distance $PC = \sqrt{(\frac{3}{2} - (-\frac{5}{2}))^2 + (\frac{3}{2} - (-\frac{3}{2}))^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5$.The minimum distance from $P$ to the annulus is $PC - r_2 = 5 - \frac{7}{2} = \frac{3}{2}$.

Let $S = \{z : 3 \le |2z - 3(1 + i)| \le 7\}$ be a set of complex numbers. Then $\min_{z \in S} |z + \frac{1}{2}(5 + 3i)|$ is equal to:

Similar Questions

The equation $\text{Re}(z^2) = 1$ represents which of the following?

If the amplitude of $z-2-3i$ is $\frac{\pi}{4}$,then the locus of $z=x+iy$ is:

$ABCD$ is a rhombus. Its diagonals $AC$ and $BD$ intersect at the point $M$ and satisfy $BD = 2AC$. If the points $D$ and $M$ represent the complex numbers $1 + i$ and $2 - i$ respectively,then $A$ represents the complex number

The largest value of $r$ for which the region represented by the set $\{ \omega \in \mathbb{C} : |\omega - 4 - i| \le r \}$ is contained in the region represented by the set $\{ z \in \mathbb{C} : |z - 1| \le |z + i| \}$ is equal to

The locus of a point $z$ satisfying $|z|^2 = \operatorname{Re}(z)$ is a circle with centre

Vedclass Test Series

Exam Paper Generator

Online Exam Module