Question:

The sum of the square of the modulus of the elements in the set \[ \{z = a + ib : a, b \in \mathbb{Z}, z \in \mathbb{C}, |z-1| \leq 1, |z-5| \leq |z-5i|\} \] is ________.

Updated On: Mar 20, 2025
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Correct Answer: 9

Solution and Explanation

Step 1: Analyze the conditions

The first condition is:

\( |z - 1| \leq 1 \).

Substitute \( z = x + iy \), where \( x, y \in \mathbb{R} \):

\( |z - 1| = \sqrt{(x - 1)^2 + y^2} \leq 1 \).

Squaring both sides:

\( (x - 1)^2 + y^2 \leq 1. \) (1)

The second condition is:

\( |z - 5| \leq |z - 5i|. \)

Substitute \( z = x + iy \):

\( \sqrt{(x - 5)^2 + y^2} \leq \sqrt{x^2 + (y - 5)^2}. \)

Squaring both sides:

\( (x - 5)^2 + y^2 \leq x^2 + (y - 5)^2. \)

Simplify:

\( -10x - 10y \leq 0. \)

\( x + y \geq 0. \) (2)

Step 2: Solve the inequalities

From condition (1), \( (x - 1)^2 + y^2 \leq 1 \), the points lie within or on a circle centered at \( (1, 0) \) with radius 1.

From condition (2), \( x + y \geq 0 \), the points lie above or on the line \( y = -x \).

Step 3: Discretize \( x, y \) as integers

Since \( x, y \in \mathbb{Z} \), we identify the integer points satisfying both conditions. These points are:

\( (0, 0), (1, 0), (2, 0), (1, 1), (1, -1). \)

Step 4: Compute the square of the modulus

For each point \( z_k = x_k + iy_k \), the modulus squared is \( |z_k|^2 = x_k^2 + y_k^2 \). Calculate for each point:

  • \( |z_1|^2 = |0 + 0i|^2 = 0^2 + 0^2 = 0, \)
  • \( |z_2|^2 = |1 + 0i|^2 = 1^2 + 0^2 = 1, \)
  • \( |z_3|^2 = |2 + 0i|^2 = 2^2 + 0^2 = 4, \)
  • \( |z_4|^2 = |1 + i|^2 = 1^2 + 1^2 = 2, \)
  • \( |z_5|^2 = |1 - i|^2 = 1^2 + (-1)^2 = 2. \)

Step 5: Sum the squares of the modulus

\( \sum_{k=1}^{5} |z_k|^2 = 0 + 1 + 4 + 2 + 2 = 9. \)

Final Answer is \( 9 \).

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