Complex NumbersmediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Complex Numbers: Let Distinct Solutions 12i Equal (JEE Main 2026)

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

Question
Let z1,z2∈Cz_1, z_2 \in \mathbb{C} be the distinct solutions of z2+4z−(1+12i)=0z^2 + 4z - (1+12i) = 0. Then ∣z1∣2+∣z2∣2|z_1|^2 + |z_2|^2 is equal to:
A18
B22
C29
D34correct
Solution
Step 1: Apply the quadratic formula
z=−4±16+4(1+12i)2=−2±5+12iz = \frac{-4 \pm \sqrt{16 + 4(1+12i)}}{2} = -2 \pm \sqrt{5+12i}
 Step 2: Evaluate 5+12i\sqrt{5+12i} Set 5+12i=a+bi\sqrt{5+12i} = a+bi with a,b∈Ra, b \in \mathbb{R}. Then a2−b2=5a^2 - b^2 = 5 and 2ab=122ab = 12.
(a2+b2)2=(a2−b2)2+(2ab)2=25+144=169  ⟹  a2+b2=13(a^2+b^2)^2 = (a^2-b^2)^2 + (2ab)^2 = 25 + 144 = 169 \implies a^2+b^2 = 13
 Solving: a2=9a^2 = 9, b2=4b^2 = 4, giving 5+12i=3+2i\sqrt{5+12i} = 3+2i. Step 3: Determine z1z_1 and z2z_2
z1=−2+(3+2i)=1+2i,z2=−2−(3+2i)=−5−2iz_1 = -2+(3+2i) = 1+2i, \qquad z_2 = -2-(3+2i) = -5-2i
 Step 4: Compute the required sum
∣z1∣2=12+22=5,∣z2∣2=(−5)2+(−2)2=29|z_1|^2 = 1^2+2^2 = 5, \qquad |z_2|^2 = (-5)^2+(-2)^2 = 29
∣z1∣2+∣z2∣2=5+29=34|z_1|^2+|z_2|^2 = 5+29 = 34
 Answer: (4)
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