If $\log _{\frac{1}{\sqrt{3}}}\left\{\frac{|z|^2-|z|+1}{2+|z|}\right\}>-2$,then $z$ lies inside

  • A
    a triangle
  • B
    an ellipse
  • C
    a circle
  • D
    a square

Explore More

Similar Questions

The locus of a point on the Argand plane represented by the complex number $z$, when $z$ satisfies the condition $\left|\frac{z-1+i}{z+1-i}\right|=\left|\operatorname{Re}\left(\frac{z-1+i}{z+1-i}\right)\right|$ is

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

Difficult
View Solution

Let $C$ be the set of all complex numbers. Let $S_{1}=\{z \in C:|z-2| \leq 1\}$ and $S_{2}=\{z \in C: z(1+i)+\overline{z}(1-i) \geq 4\}$. Then,the maximum value of $\left|z-\frac{5}{2}\right|^{2}$ for $z \in S_{1} \cap S_{2}$ is equal to:

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

The locus of $z$ satisfying $\left|\frac{z-i}{z-2i}\right|=2$ is a

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