Key Concepts and Formulas

• Modulus of a Complex Number: For $z = a + bi$, $|z| = \sqrt{a^2 + b^2}$ represents the distance from the origin in the complex plane. $|z - z_0|$ is the distance between the complex numbers $z$ and $z_0$.
• Geometric Interpretation of Complex Inequalities: The inequality $r_1 < |z - z_0| < r_2$ represents the annulus (ring-shaped region) centered at $z_0$ with inner radius $r_1$ and outer radius $r_2$.
• Integer Lattice Points: We are looking for points with integer coordinates ($a, b \in \mathbb{Z}$) within the annulus.

Step-by-Step Solution

Step 1: Substitute z = a + bi into the inequality

We are given the inequality $1 < |z - 3 + 2i| < 4$ and $z = a + bi$, where $a, b \in \mathbb{Z}$. Substituting $z$ into the inequality, we get: $1 < |(a + bi) - 3 + 2i| < 4$

Step 2: Group the real and imaginary parts

Rearranging the terms inside the modulus: $1 < |(a - 3) + (b + 2)i| < 4$

Step 3: Express the modulus in terms of a and b

Using the definition of the modulus of a complex number: $1 < \sqrt{(a - 3)^2 + (b + 2)^2} < 4$

Step 4: Square the inequality

Squaring all parts of the inequality to eliminate the square root (since all parts are positive, the inequality signs remain the same): $1^2 < (a - 3)^2 + (b + 2)^2 < 4^2$ $1 < (a - 3)^2 + (b + 2)^2 < 16$

Step 5: Introduce new variables X and Y

Let $X = a - 3$ and $Y = b + 2$. Since $a$ and $b$ are integers, $X$ and $Y$ are also integers. The inequality becomes: $1 < X^2 + Y^2 < 16$

Step 6: Count integer pairs (X, Y) satisfying the inequality

We need to find the number of integer pairs $(X, Y)$ such that $1 < X^2 + Y^2 < 16$. We will examine possible integer values of $Y$ and find corresponding integer values of $X$. Remember that $X^2$ and $Y^2$ must be non-negative.

• Case 1: Y = 0 $1 < X^2 < 16$. $X^2$ can be $4$ or $9$. Thus, $X = \pm 2$ or $X = \pm 3$. This gives 4 pairs: $(\pm 2, 0), (\pm 3, 0)$.

• Case 2: Y = ±1 $1 < X^2 + 1 < 16$, which means $0 < X^2 < 15$. $X^2$ can be $1, 4, 9$. Thus, $X = \pm 1, \pm 2, \pm 3$. This gives 6 pairs for $Y = 1$ and 6 pairs for $Y = -1$, totaling 12 pairs: $(\pm 1, \pm 1), (\pm 2, \pm 1), (\pm 3, \pm 1)$.

• Case 3: Y = ±2 $1 < X^2 + 4 < 16$, which means $-3 < X^2 < 12$. $X^2$ can be $0, 1, 4, 9$. Thus, $X = 0, \pm 1, \pm 2, \pm 3$. This gives 7 pairs for $Y = 2$ and 7 pairs for $Y = -2$, totaling 14 pairs: $(0, \pm 2), (\pm 1, \pm 2), (\pm 2, \pm 2), (\pm 3, \pm 2)$.

• Case 4: Y = ±3 $1 < X^2 + 9 < 16$, which means $-8 < X^2 < 7$. $X^2$ can be $0, 1, 4$. Thus, $X = 0, \pm 1, \pm 2$. This gives 5 pairs for $Y = 3$ and 5 pairs for $Y = -3$, totaling 10 pairs: $(0, \pm 3), (\pm 1, \pm 3), (\pm 2, \pm 3)$.

• Case 5: Y = ±4 $1 < X^2 + 16 < 16$, which means $-15 < X^2 < 0$. There are no possible values for $X$. This gives 0 pairs.

Step 7: Calculate the total number of integer pairs

Adding the number of pairs from each case: $4 + 12 + 14 + 10 + 0 = 40$

Common Mistakes & Tips

• Forgetting the Integer Constraint: This constraint significantly reduces the number of possible solutions.
• Incorrectly Handling Inequalities: Ensure you are squaring all parts of the inequality and maintaining the correct direction.
• Missing Points: Be systematic in your counting to avoid missing any valid pairs.

Summary

We transformed the given complex number inequality into an equivalent inequality involving integer coordinates $(X, Y)$. By systematically enumerating all possible integer pairs $(X, Y)$ within the defined annulus, we found that there are a total of 40 such pairs.

Final Answer

The final answer is \boxed{40}.