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MathMedMain 2026 • 4th Apr - Shift 2

Sum of Squared Modulus of Complex Roots

Let S={z∈C:z2+4z+16=0}S = \{z \in \mathbb{C} : z^2 + 4z + 16 = 0\}. Then ∑z∈S∣z+3i∣2\sum_{z \in S} |z + \sqrt{3}i|^2 is equal to:

Options

A

42

B

23

C

27

D

38

Correct

Topics & Concepts

Complex Numbers and Quadratic EquationsComplex Numbers

Step-by-Step Solution

To find the value of ∑z∈S∣z+3i∣2\sum_{z \in S} |z + \sqrt{3}i|^2, we first determine the elements of the set SS, which are the roots of the quadratic equation: z2+4z+16=0z^2 + 4z + 16 = 0

Using the quadratic formula, we have: z=−4±42−4(1)(16)2(1)z = \frac{-4 \pm \sqrt{4^2 - 4(1)(16)}}{2(1)} z=−4±16−642=−4±−482=−4±43i2=−2±23iz = \frac{-4 \pm \sqrt{16 - 64}}{2} = \frac{-4 \pm \sqrt{-48}}{2} = \frac{-4 \pm 4\sqrt{3}i}{2} = -2 \pm 2\sqrt{3}i

Thus, the set SS consists of two complex roots: z1=−2+23iandz2=−2−23iz_1 = -2 + 2\sqrt{3}i \quad \text{and} \quad z_2 = -2 - 2\sqrt{3}i

Now, we calculate ∣z+3i∣2|z + \sqrt{3}i|^2 for each root:

  1. For z1=−2+23iz_1 = -2 + 2\sqrt{3}i: z1+3i=−2+23i+3i=−2+33iz_1 + \sqrt{3}i = -2 + 2\sqrt{3}i + \sqrt{3}i = -2 + 3\sqrt{3}i ∣z1+3i∣2=(−2)2+(33)2=4+27=31|z_1 + \sqrt{3}i|^2 = (-2)^2 + (3\sqrt{3})^2 = 4 + 27 = 31

  2. For z2=−2−23iz_2 = -2 - 2\sqrt{3}i: z2+3i=−2−23i+3i=−2−3iz_2 + \sqrt{3}i = -2 - 2\sqrt{3}i + \sqrt{3}i = -2 - \sqrt{3}i ∣z2+3i∣2=(−2)2+(−3)2=4+3=7|z_2 + \sqrt{3}i|^2 = (-2)^2 + (-\sqrt{3})^2 = 4 + 3 = 7

Summing the two squared moduli, we get: ∑z∈S∣z+3i∣2=∣z1+3i∣2+∣z2+3i∣2=31+7=38\sum_{z \in S} |z + \sqrt{3}i|^2 = |z_1 + \sqrt{3}i|^2 + |z_2 + \sqrt{3}i|^2 = 31 + 7 = 38

Therefore, the correct option is D.

Sum of Squared Modulus of Complex Roots | Mathematics PYQ Solution - JEE Challenger