Key Concepts and Formulas

• Modulus of a complex number: For $z = x + iy$, its modulus is $|z| = \sqrt{x^2 + y^2}$. The square of the modulus is $|z|^2 = x^2 + y^2$.
• Geometric interpretation of $|z - z_0| \leq r$: This represents a closed disk (circle and its interior) centered at $z_0$ with radius $r$.
• Geometric interpretation of $|z - z_1| \leq |z - z_2|$: This represents the closed half-plane containing $z_1$, bounded by the perpendicular bisector of the segment connecting $z_1$ and $z_2$.

Step-by-Step Solution

Let $z = a + ib$, where $a, b \in \mathbb{Z}$.

Step 1: Analyze the inequality $|z - 1| \leq 1$

• What & Why: We substitute $z = a + ib$ into the inequality and simplify to get an algebraic inequality in terms of $a$ and $b$. This will help us define a region in the complex plane.
• Substitution: $| (a + ib) - 1 | \leq 1$ $| (a - 1) + ib | \leq 1$
• Applying the definition of modulus: $\sqrt{(a - 1)^2 + b^2} \leq 1$
• Squaring both sides: $(a - 1)^2 + b^2 \leq 1$
• Reasoning: This represents all points $(a, b)$ inside or on the circle centered at $(1, 0)$ with radius $1$.

Step 2: Analyze the inequality $|z - 5| \leq |z - 5i|$

• What & Why: We substitute $z = a + ib$ into the inequality and simplify. This gives us another algebraic inequality that restricts the possible values of $a$ and $b$.
• Substitution: $| (a + ib) - 5 | \leq | (a + ib) - 5i |$ $| (a - 5) + ib | \leq | a + i(b - 5) |$
• Applying the definition of modulus: $\sqrt{(a - 5)^2 + b^2} \leq \sqrt{a^2 + (b - 5)^2}$
• Squaring both sides: $(a - 5)^2 + b^2 \leq a^2 + (b - 5)^2$
• Expanding: $a^2 - 10a + 25 + b^2 \leq a^2 + b^2 - 10b + 25$
• Simplifying: $-10a \leq -10b$
• Dividing by $-10$ and flipping the inequality: $a \geq b$
• Reasoning: This inequality represents all points $(a, b)$ that lie on or below the line $a = b$.

Step 3: Find integer points satisfying both conditions

• What & Why: We need to find all pairs of integers $(a, b)$ that satisfy both inequalities derived in Steps 1 and 2. We'll do this by considering possible integer values for $a$ and $b$.

• We need to find integer pairs $(a, b)$ such that:

  1. $(a - 1)^2 + b^2 \leq 1$
  2. $a \geq b$

• Case 1: $b = 0$ $(a - 1)^2 \leq 1 \implies -1 \leq a - 1 \leq 1 \implies 0 \leq a \leq 2$. Possible integer values for $a$ are $0, 1, 2$. The points are $(0, 0), (1, 0), (2, 0)$. Checking $a \geq b$:

  • $(0, 0)$: $0 \geq 0$ (True)
  • $(1, 0)$: $1 \geq 0$ (True)
  • $(2, 0)$: $2 \geq 0$ (True)

• Case 2: $b = 1$ $(a - 1)^2 + 1 \leq 1 \implies (a - 1)^2 \leq 0 \implies a - 1 = 0 \implies a = 1$. The point is $(1, 1)$. Checking $a \geq b$:

  • $(1, 1)$: $1 \geq 1$ (True)

• Case 3: $b = -1$ $(a - 1)^2 + 1 \leq 1 \implies (a - 1)^2 \leq 0 \implies a - 1 = 0 \implies a = 1$. The point is $(1, -1)$. Checking $a \geq b$:

  • $(1, -1)$: $1 \geq -1$ (True)

• Case 4: $|b| \geq 2$ If $|b| \geq 2$, then $b^2 \geq 4$. Since $(a - 1)^2 \geq 0$, we have $(a - 1)^2 + b^2 \geq 4 > 1$, which violates the first inequality. Therefore, no solutions exist for $|b| \geq 2$.

The set of complex numbers $z$ is $\{0, 1, 2, 1 + i, 1 - i\}$.

Step 4: Calculate the sum of the square of the modulus

• What & Why: We calculate $|z|^2$ for each of the complex numbers we found, and then sum them up as requested by the problem.
• For $z = 0$: $|0|^2 = 0^2 + 0^2 = 0$
• For $z = 1$: $|1|^2 = 1^2 + 0^2 = 1$
• For $z = 2$: $|2|^2 = 2^2 + 0^2 = 4$
• For $z = 1 + i$: $|1 + i|^2 = 1^2 + 1^2 = 2$
• For $z = 1 - i$: $|1 - i|^2 = 1^2 + (-1)^2 = 2$

Sum $= 0 + 1 + 4 + 2 + 2 = 9$

Common Mistakes & Tips

• Forgetting Integer Constraints: The condition $a, b \in \mathbb{Z}$ is crucial. Without it, there would be infinitely many solutions.
• Sign Errors: Pay careful attention to signs when expanding and simplifying inequalities.
• Geometric Intuition: Sketching the regions defined by the inequalities can help visualize the problem and check your work.

Summary

We found all complex numbers $z = a + ib$ that satisfy the given conditions by analyzing the geometric meaning of the inequalities, converting them to algebraic form, and finding integer solutions for $a$ and $b$. We then calculated the square of the modulus for each of these complex numbers and summed them up. The sum of the square of the modulus of the elements in the set is $9$.

The final answer is $\boxed{9}$.