Let z_1 and z_2 be two complex numbers such t | vedclass

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

Question

(C) Given $\frac{z_1}{z_2} + \frac{z_2}{z_1} = 1$.Multiplying by $z_1 z_2$,we get $z_1^2 + z_2^2 = z_1 z_2$.This can be rewritten as $z_1^2 + z_2^2 -

Vedclass · Accepted Answer

$z_1, z_2$ and the origin form an equilateral triangle

Vedclass · Answer

(C) Given $\frac{z_1}{z_2} + \frac{z_2}{z_1} = 1$.Multiplying by $z_1 z_2$,we get $z_1^2 + z_2^2 = z_1 z_2$.This can be rewritten as $z_1^2 + z_2^2 - z_1 z_2 = 0$.Let $z_3 = 0$ be the origin.The condition for three points $z_1, z_2, z_3$ to form an equilateral triangle is $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.Substituting $z_3 = 0$,we get $z_1^2 + z_2^2 = z_1 z_2$,which matches our given equation.Therefore,$z_1, z_2$ and the origin form an equilateral triangle.

Let $z_1$ and $z_2$ be two complex numbers such that $\frac{z_1}{z_2} + \frac{z_2}{z_1} = 1$. Then

Similar Questions

If $z$ is a complex number such that $|z| \ge 2$,then the minimum value of $|z + \frac{1}{2}|$ is:

The equation $z\overline{z} + (2 - 3i)z + (2 + 3i)\overline{z} + 4 = 0$ represents a circle of radius

The point $P$ denotes the complex number $z=x+iy$ in the Argand plane. If $\frac{2z-i}{z-2}$ is a purely real number,then the equation of the locus of $P$ is

The complex number $z = x + iy$ which satisfies the equation $\left| \frac{z - 5i}{z + 5i} \right| = 1$ lies on

Let $z$ and $w$ be two non-zero complex numbers such that $|z| = |w|$ and $arg(z) + arg(w) = \pi$. Then $z$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module