If the complex numbers Z 1- Z 2- z 3 represents vertices of a triangle ABC satisfy the equation z 3-1- If there is a complex number Z 0 satisfying 1- z 0- z 1- 1- z 0- z 2-1- z 0- z 3-0 then Z 0 isA- Not Necessarily at originB- Centroid of A B CC- an Excentre of A B CD- a point on Circum circle of - A B C

The correct option is B Z_1,Z_2,Z_3 are cube roots of unity ∴ ABC is equilateral α_r = 1/Z_0 Z_r,r = 1,2,3 Z_r = Z_0 1/α_r Z^3 = 1 ⇒ Z_0 1/α_r^3 = 1 (Z_0α ...

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Standard XII

Mathematics

Complex Numbers

If the comple...

Question

If the complex numbers $z_{1}, z_{2}, z_{3}$ represents vertices of a triangle ABC satisfy the equation $z^{3}$ = 1. If there is a complex number $z_{0}$ satisfying $\frac{1}{z_{0} - z_{1}} + \frac{1}{z_{0} - z_{2}} + \frac{1}{z_{0} - z_{3}}$ = 0 then $z_{0}$ is

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Solution

The correct option is B

$Z_{1}, Z_{2}, Z_{3}$ are cube roots of unity

$∴ △ A B C$ is equilateral $α_{r}$ = $\frac{1}{Z_{0} - Z_{r}}$ ,r = 1,2,3

$Z_{r}$ = $Z_{0} - \frac{1}{α_{r}}$

$Z^{3}$ = 1 $\Rightarrow ⟮ Z_{0} - \frac{1}{α_{r}} ⟯^{3}$ = 1

$(Z_{0} α_{r} - 1)^{3}$ = $α_{r}^{3}$

$α_{r}^{3} (Z_{0}^{3} - 1) - 3 Z_{0}^{2} α_{r}^{2} + 3 Z_{0} α_{r} - 1$ = 0

$α_{1} + α_{2} + α_{3}$ = $\frac{3 Z_{0}^{2}}{Z_{0}^{3} - 1}$ = $\frac{3 Z_{0}^{2}}{Z_{0}^{3} - 1} \Rightarrow Z_{0}$ = 0

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If the complex numbers z1,z2,z3 represents vertices of a triangle ABC satisfy the equation z3 = 1. If there is a complex number z0 satisfying 1z0−z1+1z0−z2+1z0−z3 = 0 then z0 is

The correct option is B Z1,Z2,Z3 are cube roots of unity ∴△ABC is equilateral αr = 1Z0−Zr,r = 1,2,3 Zr = Z0−1αr Z3 = 1 ⇒⟮Z0−1αr⟯3 = 1 (Z0αr−1)3 = α3r α3r(Z30−1)−3Z20α2r+3Z0αr−1 = 0 α1+α2+α3 = 3Z20Z30−1 = 3Z20Z30−1⇒Z0 = 0

If the complex numbers $z_{1}, z_{2}, z_{3}$ represents vertices of a triangle ABC satisfy the equation $z^{3}$ = 1. If there is a complex number $z_{0}$ satisfying $\frac{1}{z_{0} - z_{1}} + \frac{1}{z_{0} - z_{2}} + \frac{1}{z_{0} - z_{3}}$ = 0 then $z_{0}$ is