The number of solutions for the equations $|z - 1| = |z - 2| = |z - i|$ is

  • A
    One solution
  • B
    $3$ solutions
  • C
    $2$ solutions
  • D
    No solution

Explore More

Similar Questions

Suppose $z_1, z_2, z_3$ are the vertices of an equilateral triangle inscribed in the circle $|z| = 2$. If $z_1 = 1 + i\sqrt{3}$,then the values of $z_3$ and $z_2$ are respectively:

The number of values of $z \in \mathbb{C}$,satisfying the equations $|z - (4 + 8i)| = \sqrt{10}$ and $|z - (3 + 5i)| + |z - (5 + 11i)| = 4\sqrt{5}$,is:

If $Z$ is a complex number such that $|Z| \leq 3$ and $-\frac{\pi}{2} \leq \operatorname{amp}(Z) \leq \frac{\pi}{2}$,then the area of the region formed by the locus of $Z$ is

Let $A = \{z \in \mathbb{C} : |z - 2 - i| = 3\}$, $B = \{z \in \mathbb{C} : \operatorname{Re}(z - iz) = 2\}$ and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to . . . . . . .

Let $x_1, x_2, x_3, x_4$ be the roots of the equation $4x^4 + 8x^3 - 17x^2 - 12x + 9 = 0$. If $(4+x_1^2)(4+x_2^2)(4+x_3^2)(4+x_4^2) = \frac{125}{16}m$,then the value of $m$ is:

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