Assertion $(A)$: If $z$ is a complex number such that $|z| \geq 3$,then the least value of $|z + \frac{3}{z}|$ is $1$.
Reason $(R)$: $|z_1 - z_2| \leq |z_1| + |z_2|$,for any two complex numbers $z_1, z_2$.
The correct option among the following is:

  • A
    $A$ is true,$R$ is true and $R$ is the correct explanation for $A$.
  • B
    $A$ is true,$R$ is true but $R$ is not the correct explanation for $A$.
  • C
    $A$ is true but $R$ is false.
  • D
    $A$ is false but $R$ is true.

Explore More

Similar Questions

If ${z_r} = \cos \frac{{r\alpha }}{{{n^2}}} + i\sin \frac{{r\alpha }}{{{n^2}}}$,where $r = 1, 2, 3, \dots, n$,then $\mathop {\lim }\limits_{n \to \infty } {z_1}{z_2}{z_3} \dots {z_n}$ is equal to

Difficult
View Solution

If $(3 + i)z = (3 - i)\bar{z}$,then the complex number $z$ is

$z = \frac{3 + 2i \sin \theta}{1 - 2i \sin \theta}, \quad (i = \sqrt{-1})$ will be purely imaginary if $\theta =$

If $Z_1, Z_2, Z_3$ are three complex numbers with unit modulus such that $|Z_1-Z_2|^2+|Z_1-Z_3|^2=4$,then $Z_1 \overline{Z_2}+\overline{Z_1} Z_2+Z_1 \overline{Z_3}+\overline{Z_1} Z_3=$

Let $z_1$ and $z_2$ be any two non-zero complex numbers such that $3|z_1| = 4|z_2|$. If $z = \frac{3z_1}{2z_2} + \frac{2z_2}{3z_1}$,then:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo