Complex Numbers JEE 2024: Why Both Roots Are Valid

What is the complex numbers JEE 2024 ordered-pair question?

The supplied complex numbers JEE 2024 solution selects the positive answer, but the printed conditions allow both signs. A complex number satisfies the following conditions; find the real component of its conjugate after adding two.

The chapter is Complex Numbers and Quadratic Equations. Hard is the question bank’s difficulty tag, not a measured student success rate.

The bank labels this JEE Main 2024 but lists 2023-01-24, Morning. That date does not verify a 2024 sitting. The supplied record does not enumerate the options, so we cannot reconstruct them. Its “Correct answer: 1” identifies option [1], not the numerical value one.

How do you rationalise the quotient and extract its imaginary part?

Rationalisation makes the quotient’s imaginary part equal to the imaginary coordinate of the original complex number: the denominator is one by the modulus condition. The cross terms leave one imaginary term, while the product of the imaginary factors contributes a positive square to the real part.

Set the real coordinates:

Squaring the modulus condition gives:

The given modulus is one, so the original denominator cannot be zero. The quotient is therefore defined.

Multiply the numerator and denominator by the conjugate of the denominator:

Expand the numerator:

The imaginary terms combine:

Hence:

Use the modulus equation to replace the entire denominator by one:

Equate the coefficient of the imaginary unit to the given imaginary component:

Do not divide the imaginary components of the original numerator and denominator. Extract the imaginary part after rationalising.

What is the keyed answer, and are both roots valid?

The supplied solution selects the positive value and identifies option [1], but both roots satisfy the printed conditions. Neither condition removes the negative root, so the bank’s keyed answer must be distinguished from the full set of allowed values.

Substitute the imaginary coordinate into the modulus equation:

Simplify the radical while retaining both branches:

Conjugation changes the imaginary sign, not the real component:

Supplied keyed answer, option [1]: $\frac{2\sqrt{6}}{5}.$

Qualification: the printed conditions alone also allow the negative value. Check both candidates:

For either sign:

An additional restriction would select the positive branch, for example: $Re(z)>-2.$

That restriction is not stated in the question. Taking the principal square root of the right-hand side does not justify discarding the negative solution of the squared equation.

What algebra error does the supplied note associate with option [3]?

The supplied solution’s note associates the previously keyed option [3] with two mistakes: dropping the factor two when simplifying the radical, then adding one. The record does not supply that option’s full wording or evidence about how often students selected it.

The incorrect simplification fails because:

The correct positive square root is:

Extracting a perfect-square factor requires retaining its square root as a multiplier. A quick squaring check catches the lost factor:

Adding one changes the target without justification. The requested quantity has already been isolated:

What changes when the modulus is two?

The rationalised denominator becomes four, not one or two. This original practice question, not another verified PYQ, tests whether you use the squared modulus rather than copy the denominator from the previous solution.

Find every possible value of the requested real part:

Set real coordinates and square the modulus:

$(a+3{)}^2+b^2=4.$ Rationalising gives:

Therefore:

Return to the modulus equation: $(a+3{)}^2=4-1=3,$

Both signs survive because no further restriction is given. Use the squared modulus as the rationalised denominator: it is four here.

How does an explicit condition choose the correct branch?

A restriction on the real component can remove one root. In this original practice question, the real component lies below negative one, so the requested shifted real component must be negative. Apply that restriction after finding both algebraic candidates.

The conditions and target are:

Set real coordinates:

Rationalise the quotient:

Its imaginary component gives:

Hence:

Now apply the explicit restriction:

Thus:

Carry both roots until an explicit condition selects one.

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