Question

Consider a square on the complex plane. The complex numbers corresponding to its four vertices are the four distinct roots of the equation with integer coefficients $x^4+p{x}^3+q{x}^2+rx+s=0$, then the minimum area of the square is

Answer 1

$x^4+p{x}^3+q{x}^2+rx+s=0$

Let $\alpha ,\beta ,\gamma$ and $\delta$ be the roots of the equations.

Such that,

$\Rightarrow P\left(x\right)=x^4+p{x}^3+q{x}^2+rx+s=(x-\alpha )(x-\beta )(x-\gamma )(x-\delta )$

$\therefore$ The area is given by, $\frac{1}{2}(x-\alpha )(x-\beta )(x-\gamma )(x-\delta )$

$\Rightarrow (x-\alpha )(x-\beta )(x-\gamma )(x-\delta )=2[x^4+p{x}^3+q{x}^2]+rx+s$

When we factorize the RHS,

$x(x^3+p{x}^2+qx+r)$

$x^4+(\alpha +\beta +\gamma +\delta )x^3+.....$

Clearly, $p=(\alpha +\beta +\gamma +\delta )$

$\Rightarrow \frac{1}{2}(p-\alpha )(p-\beta )(p-\gamma )(p-\delta )=1$

$(p-\alpha )(p-\beta )(p-\gamma )(p-\delta )=2.$

Hence, the answer is $2.$