Question

Let \[ S = \{ z \in \mathbb{C} : 4z^2 + \bar{z} = 0 \}. \] Then \[ \sum_{z \in S} |z|^2 \] is equal to

• A. \( \dfrac{1}{16} \)
• B. \( \dfrac{3}{16} \)
• C. \( \dfrac{5}{64} \)
• D. \( \dfrac{7}{64} \)

Hint:

For equations involving \( z \) and \( \bar{z} \), always write \( z = x + iy \) and equate real and imaginary parts separately.

Answer 1

The Correct Option is B

Let \[ z = x + iy, \quad \bar{z} = x - iy, \] where \( x, y \in \mathbb{R} \).

Step 1: Substitute in the given equation.
\[ 4(x + iy)^2 + (x - iy) = 0. \] \[ 4(x^2 - y^2 + 2ixy) + x - iy = 0. \] \[ (4x^2 - 4y^2 + x) + i(8xy - y) = 0. \]
Step 2: Equate real and imaginary parts.
Real part: \[ 4x^2 - 4y^2 + x = 0 \quad (1) \] Imaginary part: \[ 8xy - y = 0 \Rightarrow y(8x - 1) = 0. \quad (2) \]
Step 3: Solve the cases.
{Case 1:} \( y = 0 \)
From (1): \[ 4x^2 + x = 0 \Rightarrow x(4x + 1) = 0. \] \[ x = 0,\; -\frac{1}{4}. \] Thus, \[ z = 0,\; -\frac{1}{4}. \] {Case 2:} \( 8x - 1 = 0 \Rightarrow x = \frac{1}{8} \)
Substitute in (1): \[ 4\left(\frac{1}{64}\right) - 4y^2 + \frac{1}{8} = 0 \Rightarrow \frac{1}{16} - 4y^2 + \frac{1}{8} = 0. \] \[ 4y^2 = \frac{3}{16} \Rightarrow y^2 = \frac{3}{64}. \] \[ y = \pm \frac{\sqrt{3}}{8}. \] Thus, \[ z = \frac{1}{8} \pm i\frac{\sqrt{3}}{8}. \]
Step 4: Compute \( |z|^2 \) for each solution.
\[ |0|^2 = 0. \] \[ \left|-\frac{1}{4}\right|^2 = \frac{1}{16}. \] \[ \left|\frac{1}{8} \pm i\frac{\sqrt{3}}{8}\right|^2 = \frac{1}{64} + \frac{3}{64} = \frac{4}{64} = \frac{1}{16}. \]
Step 5: Find the required sum.
\[ \sum_{z \in S} |z|^2 = 0 + \frac{1}{16} + \frac{1}{16} = \frac{3}{16}. \]
Final Answer: \[ \boxed{\dfrac{3}{16}} \]