Question

If at least one value of the complex number $z=x+iy$ satisfy the condition $|z+\sqrt{2}|=a^2-3a+2$ and the inequality $|z+i\sqrt{2}|<a^2$ ,then

• A. a>2
• B. a=2
• C. a<2
• D. None of these

Answer 1

The correct option is A a>2

$Ifz=x+iyisacomplexnumbersatisfyingthegivenconditions,then{a}^2-3a+2=|z+\sqrt{2}|=|z+i\sqrt{2}+\sqrt{2}-i\sqrt{2}|\le |z+\sqrt{2}|+\sqrt{2}|1-i|<a^2+2\Rightarrow -3a<0\Rightarrow a>0.....\left(i\right)$
$Since|z+\sqrt{2}|=a^2-3a+2representsacirclewithcentreatA(-\sqrt{2},0)andradius\sqrt{a^2-3a+2},and|z+\sqrt{2i}|<a^{2}representstheinteriorofthecirclewithcentreatB(0,-\sqrt{2})andradiusa,thereforetherewillbeacomplexnumbersatisfyingthegivenconditionandthegiveninequalityifthedistanceislessthanthesumordifferenceoftheradiiofthetwocircles,i.e.,if$
$\sqrt{(-\sqrt{2}-0{)}^2+(0+\sqrt{2}{)}^2}<\sqrt{a^2-3a+2\pm a}\Rightarrow 2\pm a<\sqrt{a^2-3a+2}\Rightarrow 4+a^2\pm 4a<a^2-3a+2\Rightarrow -a<-2or7a<-2\Rightarrow a>2ora<-\frac{7}{2}Buta>0from\left(i\right),thereforea>2.$