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Sum of |z+√3 i|^2 for Roots of z^2+4z+16: 38 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
Let S={z∈C:z2+4z+16=0}S=\{z\in\mathbb{C}:z^2+4z+16=0\}. Then ∑z∈S∣z+3 i∣2\displaystyle\sum_{z\in S}\left|z+\sqrt3\,i\right|^2 is equal to
A4242
B2323
C2727
D3838correct
Solution
Step 1: Solve z2+4z+16=0z^2+4z+16=0:
(z+2)2=−12 ⇒ z=−2±23 i.(z+2)^2=-12\ \Rightarrow\ z=-2\pm2\sqrt3\,i.
 So S={−2+23 i, −2−23 i}S=\{-2+2\sqrt3\,i,\ -2-2\sqrt3\,i\}. Step 2: Compute each ∣z+3 i∣2\left|z+\sqrt3\,i\right|^2:
∣−2+23 i+3 i∣2=∣−2+33 i∣2=4+27=31,\left|-2+2\sqrt3\,i+\sqrt3\,i\right|^2=\left|-2+3\sqrt3\,i\right|^2=4+27=31,
∣−2−23 i+3 i∣2=∣−2−3 i∣2=4+3=7.\left|-2-2\sqrt3\,i+\sqrt3\,i\right|^2=\left|-2-\sqrt3\,i\right|^2=4+3=7.
 Step 3: Sum:
31+7=38.31+7=38.
 Correct answer: (4)
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