Let $$z = x + iy$$   be a complex number where $$x$$ and $$y$$ are integers. Then the area of the rectangle whose vertices are the roots of the equation: $$z{\overline z ^3} + \overline z {z^3} = 350$$    is

Solution :
Given $$z = x + iy$$   where $$x$$ and $$y$$ are integer
Also $$z{\overline z ^3} + \overline z {z^3} = 350$$
$$\eqalign{ & \Rightarrow \,\,{\left| z \right|^2}\left( {{{\overline z }^2} + {z^2}} \right) = 350 \cr & \Rightarrow \,\,\left( {{x^2} + {y^2}} \right)\left( {{x^2} - {y^2}} \right) = 175 \cr & \Rightarrow \,\,\left( {{x^2} + {y^2}} \right)\left( {{x^2} - {y^2}} \right) = 25 \times 7\,\,\,\,.....\left( {\text{i}} \right) \cr & {\text{or}}\,\,\,\left( {{x^2} + {y^2}} \right)\left( {{x^2} - {y^2}} \right) = 35 \times 5\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr & \because \,\,\,x\,{\text{and}}\,y\,{\text{are}}\,\operatorname{integers} \cr & \therefore \,\,\,{x^2} + {y^2} = 25\,\,\,{\text{and}}\,{x^2} - {y^2} = 7\,\,\,\,\,\,\,\,\,\left[ {{\text{From}}\,{\text{eq}}\,\left( {\text{i}} \right)} \right] \cr & \Rightarrow \,\,{x^2} = 16\,{\text{and}}\,{y^2} = 9 \cr & \Rightarrow \,\,x = \pm 4\,{\text{and}}\,y = \pm 3 \cr & \therefore \,\,{\text{Vertices}}\,{\text{of}}\,{\text{rectangle}}\,{\text{are}} \cr & \,\,\left( {4,3} \right),\left( {4, - 3} \right),\left( { - 4, - 3} \right),\left( { - 4,3} \right). \cr & {\text{So, area of rectangle}} = 8 \times 6 = 48\,{\text{sq}}{\text{. units}} \cr & {\text{Now from eq}}{\text{.}}\left( {{\text{ii}}} \right) \cr & {\text{or }}{x^2} + {y^2} = 35\,\,{\text{and }}{x^2} - {y^2} = 5 \cr} $$
$$ \Rightarrow \,\,{x^2} = 20,$$   which is not possible for any integral value of $$x$$