Let z ∈ C be such that (z^2+3i)/(z-2+i) = 2+3i. Then the sum of all possible values of z^2 is

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q4 – JEE-MAIN 2025

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Home Maths Complex & QuadraticQuestion #88

Let

z \in C

be such that

\frac{z^{2} + 3 i}{z - 2 + i} = 2 + 3 i

. Then the sum of all possible values of

z^{2}

Clue 1 of 3: Algebraic Manipulation

Begin by clearing the denominator to transform the given equation into a standard polynomial form.

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Step 1: Rearrange the equation into a quadratic form✦ Active

The given equation is $\frac{z^{2} + 3 i}{z - 2 + i} = 2 + 3 i$ . Multiply both sides by $(z - 2 + i)$ to clear the denominator:

z^{2} + 3 i = (2 + 3 i) (z - 2 + i)

z^{2} + 3 i = 2 z - 4 + 2 i + 3 i z - 6 i + 3 i^{2}

z^{2} + 3 i = 2 z - 4 + 2 i + 3 i z - 6 i - 3

z^{2} + 3 i = (2 + 3 i) z - 7 - 4 i

Rearrange the terms to form a quadratic equation in $z$ :

z^{2} - (2 + 3 i) z + (7 + 7 i) = 0

This is a quadratic equation $a z^{2} + b z + c = 0$ with $a = 1$ , $b = - (2 + 3 i)$ , and $c = (7 + 7 i)$ . Note that $z \neq 2 - i$ for the original expression to be defined. We can verify that $z = 2 - i$ is not a root of the quadratic equation, so both roots will be valid.

Step 2: Apply Vieta's formulas for roots

z_{1}, z_{2}

○ Expand

Let $z_{1}$ and $z_{2}$ be the roots of the quadratic equation $z^{2} - (2 + 3 i) z + (7 + 7 i) = 0$ . According to Vieta's formulas:

z_{1} + z_{2} = - \frac{b}{a} = - \frac{- (2 + 3 i)}{1} = 2 + 3 i

z_{1} z_{2} = \frac{c}{a} = \frac{7 + 7 i}{1} = 7 + 7 i

Step 3: Calculate the sum of squares of the roots○ Expand

The sum of all possible values of $z^{2}$ is $z_{1}^{2} + z_{2}^{2}$ . We use the algebraic identity $z_{1}^{2} + z_{2}^{2} = (z_{1} + z_{2})^{2} - 2 z_{1} z_{2}$ :

z_{1}^{2} + z_{2}^{2} = (2 + 3 i)^{2} - 2 (7 + 7 i)

z_{1}^{2} + z_{2}^{2} = (2^{2} + 2 (2) (3 i) + (3 i)^{2}) - (14 + 14 i)

z_{1}^{2} + z_{2}^{2} = (4 + 12 i + 9 i^{2}) - (14 + 14 i)

z_{1}^{2} + z_{2}^{2} = (4 + 12 i - 9) - (14 + 14 i)

z_{1}^{2} + z_{2}^{2} = (- 5 + 12 i) - (14 + 14 i)

z_{1}^{2} + z_{2}^{2} = - 5 - 14 + 12 i - 14 i

z_{1}^{2} + z_{2}^{2} = - 19 - 2 i

💡 Teacher's Secret Hint

Ensure careful expansion of complex number squares and distribution of multiplication.

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Maths Question 4 – JEE-MAIN 2025

Clue 1 of 3: Algebraic Manipulation

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