Find the real values of a which satisfies both the equality $|z+\sqrt2|=a^2-3a+2$ and the inequality $|z+i\sqrt 2|
What are the real values of parameter 'a' for which at least one
  complex number z satisfies both the equality $|z+\sqrt {2}|=a^2-3a+2$ and
  the inequality $|z+i\sqrt 2|<a^2$ ?
Ok I tried to consider the equations as circles centred at $-\sqrt{2}$ and $-i\sqrt{2}$ respectively.Can't think of what to do after that.Suggestions please!
 A: The equation
$$|z+\sqrt{2}|=a^2-3a+2 \tag{1}$$
defines a circle of radius $r_1=a^2-3a+2$ centered at $-\sqrt{2}$.
The inequality 
$$|z+i\sqrt 2|<a^2 \tag{2}$$ 
defines the interior of a circle of radius $r_2=a^2$ centered at $-i\sqrt{2}$.
As others have pointed out, we must have
$$\begin{align}
a^2-3a+2>0 &\implies \\[2ex]
(a-1)(a-2)>0 &\implies \\[2ex]
a<1\text{ or }a>2
\tag{3}
\end{align}$$
and
$$a\ne0 \tag{4}$$
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:
$$\begin{align}
r_1+r_2>2 &\implies \\[2ex]
2a^2-3a+2>2 &\implies \\[2ex]
(a-\tfrac{3}{4})^2 > \tfrac{9}{16} &\implies \\[2ex]
a\in(-\infty,0)\cup(\tfrac{3}{2},\infty) \tag{5}
\end{align}$$
and 
$$\begin{align}
r_1<r_2+2 &\implies \\[2ex]
a^2-3a+2<a^2+2 &\implies \\[2ex]
a>0 \tag{6}
\end{align}$$
Putting the restrictions from (3),(4),(5),(6) together, solutions for $z$ exist for all $a$ such that $a>2,\quad a\in\mathbb{R}$.
