Question
The roots of the equation $$1 + z + {z^3} + {z^4} = 0$$ are represented by the vertices of
A.
a square
B.
an equilateral triangle
C.
a rhombus
D.
none of these
Answer :
an equilateral triangle
Solution :
$$\eqalign{ & {\text{Here, }}\left( {1 + z} \right)\left( {1 + {z^3}} \right) = 0 \cr & {\text{or }}{\left( {1 + z} \right)^2}\left( {{z^2} - z + 1} \right) = 0 \cr & \Rightarrow \,\,z = - 1, - 1,\frac{{1 + \sqrt {3}i }}{2}. \cr} $$
The distinct points corresponding to these are
$$\eqalign{ & z = - 1,\frac{{1 + \sqrt {3}i }}{2},\frac{{1 - \sqrt {3}i }}{2} = - 1, - {\omega ^2}, - \omega \cr & AB = \left| {1 - {\omega ^2}} \right| = \left| {{\omega ^2}\left( {\omega - 1} \right)} \right| \cr & AB = \left| {{\omega ^2}} \right|\left| {\omega - 1} \right| = \left| {\omega - 1} \right| \cr & BC = \left| {{\omega ^2} - \omega } \right| = \left| \omega \right|\left| {\omega - 1} \right| = \left| {\omega - 1} \right| \cr & CA = \left| {\omega - 1} \right| \cr & \therefore \,\,AB = BC = CA. \cr} $$
$$\eqalign{ & {\text{Here, }}\left( {1 + z} \right)\left( {1 + {z^3}} \right) = 0 \cr & {\text{or }}{\left( {1 + z} \right)^2}\left( {{z^2} - z + 1} \right) = 0 \cr & \Rightarrow \,\,z = - 1, - 1,\frac{{1 + \sqrt {3}i }}{2}. \cr} $$
The distinct points corresponding to these are
$$\eqalign{ & z = - 1,\frac{{1 + \sqrt {3}i }}{2},\frac{{1 - \sqrt {3}i }}{2} = - 1, - {\omega ^2}, - \omega \cr & AB = \left| {1 - {\omega ^2}} \right| = \left| {{\omega ^2}\left( {\omega - 1} \right)} \right| \cr & AB = \left| {{\omega ^2}} \right|\left| {\omega - 1} \right| = \left| {\omega - 1} \right| \cr & BC = \left| {{\omega ^2} - \omega } \right| = \left| \omega \right|\left| {\omega - 1} \right| = \left| {\omega - 1} \right| \cr & CA = \left| {\omega - 1} \right| \cr & \therefore \,\,AB = BC = CA. \cr} $$