Maths Q2 – JEE-MAIN 2026

Let

z_{1}, z_{2} \in C

be the distinct solutions of the equation

z^{2} + 4 z - (1 + 12 i) = 0

. Then

| z_{1} |^{2} + | z_{2} |^{2}

is equal to :

Recall Vieta's formulas for the sum and product of roots of a quadratic equation, and properties of complex modulus.

Step 1: Apply Vieta's Formulas✦ Active

For the given quadratic equation $z^{2} + 4 z - (1 + 12 i) = 0$ , let $z_{1}$ and $z_{2}$ be its roots. According to Vieta's formulas:

z_{1} + z_{2} = - 4

z_{1} z_{2} = - (1 + 12 i)

From the sum of roots, we can find the square of its modulus:

| z_{1} + z_{2} |^{2} = | - 4 |^{2} = 16

Step 2: Calculate

| z_{1} - z_{2} |^{2}

○ Expand

First, we find $(z_{1} - z_{2})^{2}$ using the identity $(z_{1} - z_{2})^{2} = (z_{1} + z_{2})^{2} - 4 z_{1} z_{2}$ :

(z_{1} - z_{2})^{2} = (- 4)^{2} - 4 (- (1 + 12 i))

(z_{1} - z_{2})^{2} = 16 + 4 + 48 i = 20 + 48 i

Now, we calculate the modulus of this complex number, which is equal to $| z_{1} - z_{2} |^{2}$ :

| z_{1} - z_{2} |^{2} = | 20 + 48 i | = \sqrt{20^{2} + 48^{2}}

| z_{1} - z_{2} |^{2} = \sqrt{400 + 2304} = \sqrt{2704} = 52

💡 Teacher's Secret Hint

Remember that $| w^{2} | = | w |^{2}$ . So, $| (z_{1} - z_{2})^{2} | = | z_{1} - z_{2} |^{2}$ .

Step 3: Apply the Parallelogram Law○ Expand

The parallelogram law for complex numbers states:

| z_{1} + z_{2} |^{2} + | z_{1} - z_{2} |^{2} = 2 (| z_{1} |^{2} + | z_{2} |^{2})

Substitute the values calculated in the previous steps:

16 + 52 = 2 (| z_{1} |^{2} + | z_{2} |^{2})

68 = 2 (| z_{1} |^{2} + | z_{2} |^{2})

Solving for $| z_{1} |^{2} + | z_{2} |^{2}$ :

| z_{1} |^{2} + | z_{2} |^{2} = \frac{68}{2} = 34

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Maths Question 2 – JEE-MAIN 2026