Key Concepts and Formulas

• Complex Number Representation: A complex number $z$ can be represented as $z = x + iy$, where $x$ and $y$ are real numbers, and $i = \sqrt{-1}$. The modulus of $z$ is given by $|z| = \sqrt{x^2 + y^2}$.
• Equating Real and Imaginary Parts: For a complex equation $A + iB = 0$ to hold, both the real part $A$ and the imaginary part $B$ must be equal to zero.
• Discriminant of a Quadratic Equation: For a quadratic equation $Ax^2 + Bx + C = 0$ to have real roots, the discriminant $D = B^2 - 4AC$ must be greater than or equal to zero, i.e., $D \ge 0$. Squaring both sides of an equation can introduce extraneous roots; therefore, verification is necessary.

Step-by-Step Solution

Step 1: Substitute $z = x + iy$ into the given equation

The given equation is $z + \alpha|z - 1| + 2i = 0$. We substitute $z = x + iy$, where $x$ and $y$ are real numbers. This allows us to work with real and imaginary parts separately. $(x + iy) + \alpha|(x + iy) - 1| + 2i = 0$ $(x + iy) + \alpha|(x - 1) + iy| + 2i = 0$ $(x + iy) + \alpha\sqrt{(x - 1)^2 + y^2} + 2i = 0$

Step 2: Separate the equation into real and imaginary parts

We group the real and imaginary terms together to separate the equation into its real and imaginary parts. This is crucial for applying the condition that both parts must be zero. $\left(x + \alpha\sqrt{(x - 1)^2 + y^2}\right) + i(y + 2) = 0$ For this equation to hold, both the real and imaginary parts must be zero.

• Imaginary Part: $y + 2 = 0 \quad \Rightarrow \quad y = -2$ This tells us that the imaginary part of $z$ must be $-2$.
• Real Part: $x + \alpha\sqrt{(x - 1)^2 + y^2} = 0$ Substitute $y = -2$ into the real part equation: $x + \alpha\sqrt{(x - 1)^2 + (-2)^2} = 0$ $x + \alpha\sqrt{(x - 1)^2 + 4} = 0 \quad (*)$

Step 3: Isolate the square root term and square both sides

We isolate the square root term to prepare for squaring both sides. Squaring eliminates the square root, simplifying the equation, but we must remember to check for extraneous solutions later. Rearrange equation $(*)$: $\alpha\sqrt{(x - 1)^2 + 4} = -x$ Square both sides: $\alpha^2((x - 1)^2 + 4) = (-x)^2$ $\alpha^2(x^2 - 2x + 1 + 4) = x^2$ $\alpha^2(x^2 - 2x + 5) = x^2$

Step 4: Rearrange into a quadratic equation in $x$

We rearrange the equation into a standard quadratic form in terms of $x$. This allows us to use the discriminant to determine the conditions for real solutions. Distribute $\alpha^2$ and move $x^2$ to the left side: $\alpha^2 x^2 - 2\alpha^2 x + 5\alpha^2 - x^2 = 0$ Group terms by powers of $x$: $(\alpha^2 - 1)x^2 - 2\alpha^2 x + 5\alpha^2 = 0 \quad (**)$ This is a quadratic equation in $x$. For a solution $z = x + iy$ to exist, $x$ must be a real number.

Step 5: Apply the discriminant condition for real solutions of $x$

For equation $(**)$ to have real solutions for $x$, its discriminant ($D$) must be greater than or equal to zero ($D \ge 0$). The discriminant $D = B^2 - 4AC \ge 0$, where $A = (\alpha^2 - 1)$, $B = -2\alpha^2$, $C = 5\alpha^2$. $(-2\alpha^2)^2 - 4(\alpha^2 - 1)(5\alpha^2) \ge 0$ $4\alpha^4 - 20\alpha^2(\alpha^2 - 1) \ge 0$ $4\alpha^4 - 20\alpha^4 + 20\alpha^2 \ge 0$ $-16\alpha^4 + 20\alpha^2 \ge 0$ Divide by $4$: $-4\alpha^4 + 5\alpha^2 \ge 0$ Factor out $\alpha^2$: $\alpha^2(-4\alpha^2 + 5) \ge 0$ Multiply by $-1$ and reverse the inequality sign: $\alpha^2(4\alpha^2 - 5) \le 0$ Since $\alpha$ is a real number, $\alpha^2 \ge 0$. For the product $\alpha^2(4\alpha^2 - 5)$ to be less than or equal to zero, we must have: $4\alpha^2 - 5 \le 0$ $4\alpha^2 \le 5$ $\alpha^2 \le \frac{5}{4}$ Combining with $\alpha^2 \ge 0$, we get: $0 \le \alpha^2 \le \frac{5}{4}$ Taking the square root of all parts, we find the range for $\alpha$: $-\sqrt{\frac{5}{4}} \le \alpha \le \sqrt{\frac{5}{4}}$ $-\frac{\sqrt{5}}{2} \le \alpha \le \frac{\sqrt{5}}{2}$

Step 6: Identify $p$ and $q$

The least real value of $\alpha$ is $p = -\frac{\sqrt{5}}{2}$. The largest real value of $\alpha$ is $q = \frac{\sqrt{5}}{2}$.

Step 7: Calculate $4(p^2 + q^2)$

$4(p^2 + q^2) = 4\left(\left(-\frac{\sqrt{5}}{2}\right)^2 + \left(\frac{\sqrt{5}}{2}\right)^2\right)$ $= 4\left(\frac{5}{4} + \frac{5}{4}\right)$ $= 4\left(\frac{10}{4}\right)$ $= 10$

Common Mistakes & Tips

• Always separate real and imaginary parts: This is the fundamental first step for solving equations with complex numbers.
• Sign check after squaring: Squaring both sides of an equation to remove a square root can introduce extraneous solutions. The original equation $\sqrt{A} = B$ implies $A = B^2$ AND $B \ge 0$. Failing to check the sign condition ($B \ge 0$) is a common mistake.
• Discriminant for real roots: Remember that $D \ge 0$ is the condition for a quadratic equation to have real solutions.

Summary

This problem demonstrates solving complex number equations by separating them into real and imaginary components. The existence of a solution for the complex variable $z$ depends on the existence of real values for its components $x$ and $y$, which transforms the problem into finding the range of a parameter ($\alpha$) for which a related quadratic equation in $x$ has real roots. We found the least and largest values of $\alpha$, $p$ and $q$, and used them to calculate $4(p^2 + q^2)$.

The final answer is \boxed{10}.