If z 1- z 2- - z n lie on the circle - z -2 then the value of - z 1- z 2- z n-4-1- z 1-1- z 2-1- z n -A- nB- -nC- 1D- 0

As z_1, z_2, ⋯ z_n lie on the circle |z| = 2, |z_i| = 2 ⇒ |z_i|^2 = 4 ⇒ z_iz_i = 4 for i = 1, 2, 3, ⋯ , n Thus 1z_i = 14z_i for i = 1, 2, ⋯, n. So, |z_1 + z_2 + ...

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Byju's Answer

Standard XII

Mathematics

Properties of Conjugate of a Complex Number

If z 1, z 2, ...

Question

If $z_{1}, z_{2}, \dots z_{n}$ lie on the circle $| z | = 2$ then the value of $| z_{1} + z_{2} + \dots + z_{n} | - 4 ∣ ∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \dots + \frac{1}{z_{n}} ∣ ∣ ∣ =$

0

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Solution

The correct option is A $0$
As $z_{1}, z_{2}, \dots z_{n}$ lie on the circle $| z | = 2,$
$| z_{i} | = 2 \Rightarrow | z_{i} |^{2} = 4$
$\Rightarrow z_{i} {¯ ¯ ¯ z}_{i} = 4$ for $i = 1, 2, 3, \dots, n$
Thus $\frac{1}{z_{i}} = \frac{1}{4} {¯ ¯ ¯ z}_{i}$ for $i = 1, 2, \dots, n .$
So, $| z_{1} + z_{2} + \dots + z_{n} | - 4 ∣ ∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \dots + \frac{1}{z_{n}} ∣ ∣ ∣$
$\Rightarrow | z_{1} + z_{2} + \dots + z_{n} | - 4 ∣ ∣ ∣ \frac{1}{4} {¯ ¯ ¯ z}_{1} + \frac{1}{4} {¯ ¯ ¯ z}_{1} + \dots + \frac{1}{4} {¯ ¯ ¯ z}_{n} ∣ ∣ ∣$
$= | z_{1} + z_{2} + \dots + z_{n} | - ∣ ∣ {¯ ¯ ¯ z}_{1} + {¯ ¯ ¯ z}_{1} + \dots + {¯ ¯ ¯ z}_{n} ∣ ∣$
$= | z_{1} + z_{2} + \dots + z_{n} | - ∣ ∣ ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ z_{1} + z_{1} + \dots + z_{n} ∣ ∣$
$= 0$ $[∵ | z | = | ¯ ¯ ¯ z |]$

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If z1,z2,⋯zn lie on the circle |z|=2 then the value of |z1+z2+⋯+zn|−4∣∣∣1z1+1z2+⋯+1zn∣∣∣=

If $z_{1}, z_{2}, \dots z_{n}$ lie on the circle $| z | = 2$ then the value of $| z_{1} + z_{2} + \dots + z_{n} | - 4 ∣ ∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \dots + \frac{1}{z_{n}} ∣ ∣ ∣ =$