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 = 12
• C. a < 2
• D. None of these

Answer 1

The correct option is A a > 2

If z = x + iy is a complex number satisfying the given conditions, then
$a^2-3a+2=|z+\sqrt{2}|=|z+i\sqrt{2}+\sqrt{2}-i\sqrt{2}|\le |z+i\sqrt{2}|+\sqrt{2}|1-i|<a^2+2$
$\Rightarrow -3a<0\Rightarrow a>0$ .....(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
$\sqrt{(-\sqrt{2}-0{)}^2+(0+\sqrt{2}{)}^2}<\sqrt{a^2-3a+2}\pm a$
$\Rightarrow 2\pm a<\sqrt{a^2-3a+2}$
$\Rightarrow 4+a^2\pm 4a<a^2-3a+2$
$\Rightarrow$ -a < -2 or 7a < -2 $\Rightarrow$ a > 2 or a < $\frac{7}{2}$
But a > 0 from (i), therefore a > 2.
Area of the triangle = $\frac{1}{2}|z{|}^2$