• # question_answer 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

If $z=x+iy$ is a complex number satisfying the given conditions, then ${{a}^{2}}-3a+2=|z+\sqrt{2}|\,=\,|z+i\sqrt{2}+\sqrt{2}-i\sqrt{2}|$ $\le |z+i\sqrt{2}|+\sqrt{2}|1-i|$ $<{{a}^{2}}+2$ Þ $-3a<0\,\,\Rightarrow a>0$ .....(i) Since $|z+\sqrt{2}|={{a}^{2}}-3a+2$ represents a circle with centre at $A(-\sqrt{2},0)$ and radius $\sqrt{{{a}^{2}}-3a+2}$, and $|z+\sqrt{2}i|$$<{{a}^{2}}$ represents the interior of the circle with centre at $B(0,-\sqrt{2})$ and radius $a$, therefore there will be a complex number satisfying the given condition and the given inequality if the distance $AB$ is less than the sum or difference of the radii of the two circles, i.e., if  $\sqrt{{{(-\sqrt{2}-0)}^{2}}+{{(0+\sqrt{2})}^{2}}}<\sqrt{{{a}^{2}}-3a+2}\pm a$ Þ $2\pm a<\sqrt{{{a}^{2}}-3a+2}$ Þ $4+{{a}^{2}}\pm 4a<{{a}^{2}}-3a+2$ Þ$-a<-2$or $7a<-2$Þ $a>2$or $a<-\frac{7}{2}$ But $a>0$ from (i), therefore$a>2$.