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If |z1|=|z2|=|z3|=1 then value of |z1-z3...

If `|z_1|=|z_2|=|z_3|=1` then value of `|z_1-z_3|^2+|z_3-z_1|^2+|z_1-z_2|^2` cannot exceed

A

`6`

B

`9`

C

`12`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
`(b)` Let `y=|z_(1)-z_(2)|^(2)+|z_(2)-z_(3)|^(2)+|z_(3)-z_(1)|^(2)`
`=(z_(1)-z_(2))(barz_(1)-barz_(2))+(z_(2)-z_(3))(barz_(2)-barz_(3))+(z_(3)-z_(1))(barz_(3)-barz_(1))`
`=6-(z_(1)barz_(2)+z_(2)barz_(1)+z_(2)barz_(3)+barz_(2)z_(3)+z_(3)barz_(1)+z_(1)barz_(3))`...........`(i)`
Now we know
`|z_(1)+z_(2)+z_(3)|^(2) ge 0`
`implies 3+(z_(1)barz_(2)+z_(2)barz_(1)+z_(1)barz_(3)+z_(3)barz_(1)+z_(2)barz_(3)+barz_(2)z_(3)) ge 0`.........`(ii)`
From `(i)` and `(ii)`, `y le 9`
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