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Maths Question 2 – JEE-MAIN 2026

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Let S={z∈C|z2+4z+16=0}. Then ∑z∈S|z+3i|2 is equal to:

First, find the roots of the given quadratic equation z2+4z+16=0 in the complex plane.

● Video Walkthrough
Step 1: Find the roots of the quadratic equation✦ Active

The given quadratic equation is z2+4z+16=0. We use the quadratic formula z=−b±b2−4ac2a to find its roots.

z=−4±42−4(1)(16)2(1)=−4±16−642=−4±−482 z=−4±43i2=−2±23i

So, the roots are z1=−2+23i and z2=−2−23i.

Step 2: Calculate |z+3i|2 for each root○ Expand

For z1=−2+23i:

z1+3i=(−2+23i)+3i=−2+33i |z1+3i|2=(−2)2+(33)2=4+(9×3)=4+27=31

For z2=−2−23i:

z2+3i=(−2−23i)+3i=−2−3i |z2+3i|2=(−2)2+(−3)2=4+3=7
💡 Teacher's Secret Hint

Remember that for a complex number a+bi, its modulus squared is a2+b2.

Step 3: Sum the calculated values○ Expand

The required sum is the sum of the modulus squared values calculated in the previous step.

∑z∈S|z+3i|2=|z1+3i|2+|z2+3i|2=31+7=38
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