If a z1 b z2 c z3 are the vertices of an equilateral triangle abc then value of arg z2 z3 2z1 z3 z2
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then value of arg z2 z3 2z1 z3 z2 is tan-1(√3)=π/3
Given,
=A(Z₁),B(Z₂),C(Z₃) are vertices fo an equlilateral ∆ABC
value of arg (z₂+z₂-z₃/z₃-z₂)
let the vertices of triangle
=z₁(0,a√3)
=z₂(-a,o)
=z₃(a,o)
=z₁=0+ia√3=ia√3
=z₂=-a+io=-a
=z₃=+a+io=a
now
=z₂+z₃-2z₁=a+a-=2(ia√3)=i2a√3,z₃+z₂=a-(-a)=2a
so,
= z₂+z₃+z₁/z₃-z₂=+i2a√3/+2a=i√3
=arg(z₂+z₃-2z₁/z₃-z₂)=tan-¹(√3)=π/3
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