If z_1, z_2, z_3 in mathbb{C} are the vertice | vedclass

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

Question

(A) For an equilateral triangle with vertices $z_1, z_2, z_3$ and centroid $z_0$,we have the property $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) For an equilateral triangle with vertices $z_1, z_2, z_3$ and centroid $z_0$,we have the property $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$. Also,the centroid is given by $z_0 = \frac{z_1 + z_2 + z_3}{3}$,which implies $z_1 + z_2 + z_3 = 3z_0$. We want to evaluate $\sum_{k=1}^3 (z_k - z_0)^2 = (z_1 - z_0)^2 + (z_2 - z_0)^2 + (z_3 - z_0)^2$. Expanding this,we get $(z_1^2 + z_2^2 + z_3^2) - 2z_0(z_1 + z_2 + z_3) + 3z_0^2$. Substituting $z_1 + z_2 + z_3 = 3z_0$,the expression becomes $(z_1^2 + z_2^2 + z_3^2) - 2z_0(3z_0) + 3z_0^2 = (z_1^2 + z_2^2 + z_3^2) - 3z_0^2$. Since $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$,and using the identity $(z_1 + z_2 + z_3)^2 = z_1^2 + z_2^2 + z_3^2 + 2(z_1 z_2 + z_2 z_3 + z_3 z_1)$,we have $(3z_0)^2 = 3(z_1^2 + z_2^2 + z_3^2)$,so $z_1^2 + z_2^2 + z_3^2 = 3z_0^2$. Thus,the expression is $3z_0^2 - 3z_0^2 = 0$.

If $z_1, z_2, z_3 \in \mathbb{C}$ are the vertices of an equilateral triangle,whose centroid is $z_0$,then $\sum_{k=1}^3 (z_k - z_0)^2$ is equal to

Similar Questions

The equation $|z+1-i|=|z-1+i|$ represents a (where $z$ is a complex number)

The equation of the locus of $z$ such that $\left|\frac{z-i}{z+i}\right|=2$,where $z=x+iy$ is a complex number,is

If the amplitude of $z-2-3i$ is $\frac{\pi}{4}$,then the locus of $z=x+iy$ is:

In the Argand plane,the vector $z = 4 - 3i$ is turned in the clockwise sense through $180^o$ and stretched three times. The complex number represented by the new vector is

If $|z + 4| \le 3$,then the maximum value of $|z + 1|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module