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 3√34Given:-|z1|=|z2|=|z3|=1andz1+z2+z3=0 To find the area of triangle whose vertices arez1,z2,z3|z1+z2|2+|z1−z2|2=|z1|2+|z2|2+2|z1||z2|+|z1|2+|z2|2−2|z1||z2||−z3|2+|z1−z2|2=2(|z1|2+|z2|2)1+|z1−z2|2=2(1+1)1+|z1−z2|2=4|z1−z2|2=3|z1−z2|=√3 Similarly,|z3−z2|=|z3−z1|=√3 Thus, the△ is an equilateral triangle whose wide length is√3 Hence, the area of equilateral triangle is√34a2 Where a is the length of the side of △ area of triangle=√34a2=√34(√3)2=3√34 Hence, the area of△ is3√34

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