Question

If at least one value of complex number $z=x+iy$ satisfies the condition $|z+\sqrt{2}|=\sqrt{a^2-3a+2}$ 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 $a>2$

Let $z=x+iy$ be a complex number satisfying the given condition.

then $a^2-3a+2=|z+\sqrt{2}|=|z+i\sqrt{2}-i\sqrt{2}+\sqrt{2}|$

$\Rightarrow a^2-3a+2\le |z+i\sqrt{2}|+\sqrt{2}|1-i|<a^2+2\Rightarrow a>0$

Since $a^2-3a+2=|z+\sqrt{2}|$ represents a circle with center at A $(-\sqrt{2},0)$ and radius $\sqrt{a^2-3a+2}$

and $|z+i\sqrt{2}|<a^2$ represents the interior of the circle with center at B $(0,\sqrt{2})$ and radius $a$.

Thus 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 two radii of 2 circles. i.e.

$\Rightarrow \sqrt{{(-\sqrt{2}-0)}^2+{(0+\sqrt{2})}^2}<\sqrt{a^2-3a+2}\pm a\Rightarrow 2\pm a<\sqrt{a^2-3a+2}$

Square both sides to get, $-a<-2or7a<-2\Rightarrow a>2,a<\frac{-2}{7}$

But $a>0$ thus $a>2$