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the roots of the equation z3+az2+bz+c=0 (where a,b,c are complex numbers )are the vertices of an equilateral triangle in the Argand plane.if ab=9(c+1) find the area of the triangle
Let the roots be p,q,r.
We know that if p,q,r are the vertices of an equilateral triangle then the following relation holds.
p2 + q2+r2 = pq+qr+rp. In terms of the coefficients we get a2=3b. We are further given that ab=9(c+1).
Substituting for b,c we can write the equation as z3+az2+a2z/3 + (a3/27 - 1) = 0 or (z+a/3)3 = 1
Hence, we see that by translating the origin to z=-a/3, p,q,r form an equilateral triangle inscribed in a circle of radius 1.
Such a triangle has an area 3√3/4
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