Question
Let $${z_1}\,{\text{and }}{z_2}$$ be two roots of the equation $${z^2} + az + b = 0,$$ $$z$$ being complex. Further , assume that the origin, $${z_1}\,{\text{and }}{z_2}$$ form an equilateral triangle. Then
A.
$${a^2} = 4b$$
B.
$${a^2} = b$$
C.
$${a^2} = 2b$$
D.
$${a^2} = 3b$$
Answer :
$${a^2} = 3b$$
Solution :
$$\eqalign{
& {z^2} + az + b = 0\,;\,\,\,{z_1} + {z_2} = - a \, \& \, {z_1}{z_2} = b \cr
& 0,{z_1},{z_2}\,\,{\text{from}}\,{\text{an}}\,{\text{equilateral}}\,\,\Delta \cr
& \therefore {0^2} + {z_1}^2 + {z_2}^2 = 0.{z_1} + {z_1}.{z_2} + {z_2}.0 \cr} $$
$$\left( {{\text{For}}\,{\text{an}}\,{\text{equilateral}}\,{\text{triangle}},\,\,{z_1}^2 + {z_2}^2 + {z_3}^2 = {z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}} \right)$$
$$\eqalign{
& \Rightarrow \,\,{z_1}^2 + {z_2}^2 = {z_1}{z_2} \cr
& \Rightarrow \,\,{\left( {{z_1} + {z_2}} \right)^2} = 3{z_1}{z_2} \cr
& \therefore \,\,{a^2} = 3b \cr} $$