Question

Let \(S = \{z \in \mathbb{C} : z^2 + 4z + 16 = 0\}\). Then \(\sum_{z \in S} |z + \sqrt{3}i|^2\) is equal to:

• A. 42
• B. 23
• C. 27
• D. 38

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We must find the complex roots of the quadratic equation \(z^2 + 4z + 16 = 0\), representing the elements of set \(S\). We then substitute these roots into the magnitude expression \(|z + \sqrt{3}i|^2\) and compute their sum.
Step 2: Key Formula or Approach:
The roots of a quadratic equation \(az^2 + bz + c = 0\) are given by:
\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] The squared magnitude of a complex number \(w = x + iy\) is \(|w|^2 = x^2 + y^2\).
Step 3: Detailed Explanation:
Solve \(z^2 + 4z + 16 = 0\):
\[ z = \frac{-4 \pm \sqrt{16 - 64}}{2} = \frac{-4 \pm \sqrt{-48}}{2} \] \[ z = \frac{-4 \pm 4\sqrt{3}i}{2} = -2 \pm 2\sqrt{3}i \] Set \(S\) has two elements: \(z_1 = -2 + 2\sqrt{3}i\) and \(z_2 = -2 - 2\sqrt{3}i\).
Evaluate \(|z + \sqrt{3}i|^2\) for \(z_1\):
\[ z_1 + \sqrt{3}i = -2 + 3\sqrt{3}i \] \[ |z_1 + \sqrt{3}i|^2 = (-2)^2 + (3\sqrt{3})^2 = 4 + 27 = 31 \] Evaluate \(|z + \sqrt{3}i|^2\) for \(z_2\):
\[ z_2 + \sqrt{3}i = -2 - \sqrt{3}i \] \[ |z_2 + \sqrt{3}i|^2 = (-2)^2 + (-\sqrt{3})^2 = 4 + 3 = 7 \] Compute the final sum:
\[ \sum_{z \in S} |z + \sqrt{3}i|^2 = 31 + 7 = 38 \] Step 4: Final Answer:
The sum is 38.