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Question

If |z1|=|z2|=|z3|=1 and z1+z2+z3=0 then area of the triangle whose vertices are z1,z2,z3 is

A
334
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B
34
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C
1
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D
2
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Solution

The correct option is A 334
Given:-
|z1|=|z2|=|z3|=1and
z1+z2+z3=0
To find the area of triangle whose vertices arez1,z2,z3
|z1+z2|2+|z1z2|2=|z1|2+|z2|2+2|z1||z2|+|z1|2+|z2|22|z1||z2|
|z3|2+|z1z2|2=2(|z1|2+|z2|2)
1+|z1z2|2=2(1+1)
1+|z1z2|2=4
|z1z2|2=3
|z1z2|=3
Similarly,|z3z2|=|z3z1|=3
Thus, the is an equilateral triangle whose wide length is3
Hence, the area of equilateral triangle is34a2
Where a is the length of the side of
area of triangle=34a2
=34(3)2
=334
Hence, the area of is334

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