Interpret the equation carefully

The given equation is $z + a|z-1| + 2i = 0,$ where $z \in \mathbb C$ and $a \in \mathbb R$.

We need the least and greatest real values of $a$ for which this equation has at least one solution $z$.

Write $z$ in terms of real and imaginary parts

Let $z = x+iy,$ where $x,y \in \mathbb R$.

Then the equation becomes $(x+iy) + a|x+iy-1| + 2i = 0.$

Since $|x+iy-1| = \sqrt{(x-1)^2+y^2},$ and this is a real number, we can separate real and imaginary parts.

Equate real and imaginary parts

The equation is $x + a\sqrt{(x-1)^2+y^2} + i(y+2)=0.$

Therefore, $y+2=0 \implies y=-2,$ and $x + a\sqrt{(x-1)^2+4}=0.$

So we must find all real $a$ such that there exists real $x$ satisfying $x + a\sqrt{(x-1)^2+4}=0.$

Express $a$ as a function of $x$

Rearranging, $a = -\frac{x}{\sqrt{(x-1)^2+4}}.$

Thus the possible values of $a$ are exactly the range of $f(x) = -\frac{x}{\sqrt{(x-1)^2+4}}, \quad x\in\mathbb R.$

Find the range of $f(x)$

Let $t=f(x)=-\frac{x}{\sqrt{(x-1)^2+4}}.$

Since the denominator is always positive, we square to find the extreme possible values: $t^2 = \frac{x^2}{(x-1)^2+4} = \frac{x^2}{x^2-2x+5}.$

For fixed real $x$, this is well-defined. To find the maximum of $t^2$, write $\frac{x^2}{x^2-2x+5} \le 1?$ But that is not sharp enough. Instead, let us solve directly using the condition that for a given $a$, the equation in $x$ has a real solution.

Use a quadratic condition

From $x = -a\sqrt{(x-1)^2+4},$ squaring both sides gives $x^2 = a^2\big((x-1)^2+4\big)=a^2(x^2-2x+5).$

Hence $(1-a^2)x^2 + 2a^2x - 5a^2 = 0.$

For real $x$ to exist, this quadratic must have a real root. So its discriminant must satisfy $\Delta \ge 0.$

Compute:

So $-\frac{\sqrt5}{2} \le a \le \frac{\sqrt5}{2}.$

Check endpoint validity

At $a=\pm \frac{\sqrt5}{2},$ the discriminant is $0$, so real $x$ exists. Hence the endpoints are included.

Therefore, $p=-\frac{\sqrt5}{2}, \qquad q=\frac{\sqrt5}{2}.$

Compute the required value

$p^2=q^2=\frac54.$ Therefore, $4(p^2+q^2)=4\left(\frac54+\frac54\right)=4\cdot \frac52=10.$

Final answer

$\boxed{10}$