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

To solve the given problem, we start by evaluating the set \(S = \{ z \in \mathbb{C} : 4z^2 + \bar{z} = 0 \}\). Here, \(z\) is a complex number that can be written as \(z = x + yi\), where \(x\) and \(y\) are real numbers, and \(i\) is the imaginary unit. The conjugate \(\bar{z}\) is \(x - yi\).

Substituting \(z = x + yi\) and \(\bar{z} = x - yi\) into the equation, we have:

\(4(x + yi)^2 + (x - yi) = 0\).

Expanding \(4(x + yi)^2\), we get:

\(4(x^2 + 2xyi - y^2) = 4x^2 - 4y^2 + 8xyi\).

Thus, the equation becomes:

\(4x^2 - 4y^2 + 8xyi + x - yi = 0\).

Separating real and imaginary parts:

• Real part: \(4x^2 - 4y^2 + x = 0\)
• Imaginary part: \(8xy - y = 0\)

For the imaginary part:

\(y(8x - 1) = 0\).

This implies either \(y = 0\) or \(8x = 1\) (which gives \(x = \frac{1}{8}\)).

Let's consider each case:

• Case 1: \(y = 0\): Substituting into the real part:
• Case 2: \(8x = 1\) gives \(x = \frac{1}{8}\), so the real part \(4\left(\frac{1}{8}\right)^2 - 4y^2 + \frac{1}{8} = 0\) simplifies to:

From the solutions:

• Solution 1: \(z = 0\)
• Solution 2: \(z = -\frac{1}{4}\)
• Solution 3: \(z = \frac{1}{8} + i\frac{\sqrt{3}}{8}\)
• Solution 4: \(z = \frac{1}{8} - i\frac{\sqrt{3}}{8}\)

We calculate \(|z|^2\) for each:

• For \(z = 0\), \(|z|^2 = 0\)
• For \(z = -\frac{1}{4}\), \(|z|^2 = \left(-\frac{1}{4}\right)^2 = \frac{1}{16}\)
• For \(z = \frac{1}{8} + i\frac{\sqrt{3}}{8}\) and \(z = \frac{1}{8} - i\frac{\sqrt{3}}{8}\), \(|z|^2 = \left(\frac{1}{8}\right)^2 + \left(\frac{\sqrt{3}}{8}\right)^2 = \dfrac{1}{64} + \dfrac{3}{64} = \dfrac{4}{64} = \dfrac{1}{16}\)

Total sum:

\(0 + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{3}{16}\)

Thus, \(\sum_{z \in S} |z|^2\) is \(\frac{3}{16}\).