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Modulus of a Complex Number

Let the compl...

Question

Let the complex number $z_{1}, z_{_{2}}, z_{3}$ be the vertices of an equilxeral triangle. let z $z_{0}$ be the circumcentre of the triangle , then $z_{1}^{2} + z_{2}^{2} + z_{3}^{2} -$

z_{0}^{2}

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- z_{0}^{2}

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3 z_{0}^{2}

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- 3 z_{0}^{2}

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Solution

The correct option is B $3 z_{0}^{2}$
Circumcentre and centroid of equilateral triangle coincides.

Thus , $z_{0} = \frac{1}{3} (z_{1} + z_{2} + z_{3})$

$(3 z_{0})^{2} = (z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + 2 (z_{1} z_{2} + z_{2} z_{3} + z_{1} z_{3})$

so, the triangle is equilateral ,

then

$z_{1} z_{2} + z_{2} z_{3} + z_{1} z_{3} = z_{1}^{2} + z_{2}^{2} + z_{3}^{2}$

$g z_{0}^{2} = 3 (z_{1}^{2} + z_{2}^{2} + z_{3}^{2})^{2}$

$(z_{1}^{2} + z_{2}^{2} + z_{3}^{2})^{2} = 3 z_{0}^{2}$

hence, C is the correct answer.

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Similar questions

Q. Let the complex numbers

z_{1}, z_{2}

and

z_{3}

be the vertices of an equilateral triangle. Let

z_{0}

be the circumcentre of the triangle, then

z_{1}^{2} + z_{2}^{2} + z_{3}^{2} =

Q. I. If

\frac{z_{1}}{z_{2}} + \frac{z_{2}}{z_{1}} = 1

then origin

z_{1}, z_{2}

form equilateral triangle.
II.

z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 3 z_{0}^{2}

then

z_{0}

is circumcentre of equilateral triangle with the vertices

z_{1}, z_{2}, z_{3}

Let the complex numbers z₁,z₂and z₃ be the vertices of an equilateral triangle. Let z₀ be the circumcentre of the triangle, then $z_{1}^{2} + z_{2}^{2} + z_{3}^{2}$ =

Q. Let the complex number

z_{1}, z_{_{2}}, z_{3}

be the vertices of an equilateral triangle. Let z

z_{0}

be the circumcentre of the triangle , then proove that

{z_{1}}^{2} + {z_{2}}^{2} + {z_{3}}^{2} = 3 {z_{0}}^{2}

Q. Let the complex number

z_{1}, z_{2},

and

z_{3}

be the vertices of an equilateral triangle. Let

z_{0}

be the circumcentre of the triangle.

Then show that

z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 3 z_{0}^{2}

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Let the complex number z1,z2,z3 be the vertices of an equilxeral triangle. let zz0 be the circumcentre of the triangle , then z21+z22+z23−

The correct option is B 3z20Circumcentre and centroid of equilateral triangle coincides.Thus , z0=13(z1+z2+z3)(3z0)2=(z21+z22+z23+2(z1z2+z2z3+z1z3)so, the triangle is equilateral ,then z1z2+z2z3+z1z3=z21+z22+z23gz20=3(z21+z22+z23)2(z21+z22+z23)2=3z20hence, C is the correct answer.

Let the complex number $z_{1}, z_{_{2}}, z_{3}$ be the vertices of an equilxeral triangle. let z $z_{0}$ be the circumcentre of the triangle , then $z_{1}^{2} + z_{2}^{2} + z_{3}^{2} -$