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

If $|z|=1$ and $z \neq \pm 1$,then all the values of $\frac{z}{1-z^2}$ lie on

If $|z-2|=|z-1|$,where $z$ is a complex number,then the locus of $z$ is a straight line:

If ${z_1} = 1 + i$,${z_2} = -2 + 3i$,and ${z_3} = \frac{ai}{3}$,where ${i^2} = -1$,are collinear,then the value of $a$ is:

If the point $P$ represents the complex number $z=x+iy$ in the Argand plane and if $\frac{z+i}{z-1}$ is a purely imaginary number, then the locus of $P$ is:

If the roots of the equation $Z^3+i Z^2+2 i=0$ are the vertices of a triangle $ABC$,then that triangle $ABC$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module