Question

Find the real values of the parameter $a$ for which at least one complex number $z=x+iy$ satisfies both the equality $|z+\sqrt{2}|=a^2-3a+2$ and the inequality $|z+i\sqrt{2}|<a^2$.

• A. $a\in (2,\infty )$
• B. $a\in (-\infty ,2]$
• C. $a\in [2,\infty )$
• D. $a\in (-\infty ,2)$

Answer 1

The correct option is A $a\in (2,\infty )$

Let $z=x+iy$

$\Rightarrow |z+\sqrt{2}|=\sqrt{x^2+2\sqrt{2}x+2+y^2}=a^2-3a+2$ .................(1)

Also,

$|z+i\sqrt{2}|=\sqrt{y^2+2\sqrt{2}y+2+x^2}<a^2$..........................(2)

Since, modulus of a complex number is greater than zero,

Hence, $a^2-3a+2>0$ $\left[\text{from (1)}\right]$

$\Rightarrow (a-2)(a-1)>0$

Solving the inequality, we get,

$\Rightarrow a\in (-\infty ,1)\cup (2,\infty )$ ... (3)

and because of (2), $a$ cannot be zero

So, $a\ne 0$ .... (4)

The distance between $-\sqrt{2}$ and $-i\sqrt{2}$ is $2$. So, for the first circle to overlap with the interior of the second circle the following two inequalities need to hold:

$r_1+r_2>2$

$\Rightarrow 2a^2-3a+2>2$

$\Rightarrow {(a-\frac{3}{4})}^2>\frac{9}{16}$

$\Rightarrow a\in (-\infty ,0)\cup (\frac{3}{2},\infty )$ ... (5)

and $r_1<r_2+2$

$a^2-3a+2<a^2+2$

$a>0$ ... (6)

From (3), (4), (5), (6),

We can say that

$a>2$