Complex Numbers JEE 2025: Determinant and Modulus Solution

How is the Complex Numbers JEE 2025 determinant and modulus question solved?

The answer is 0 ordered pairs. In this Complex Numbers JEE 2025 question, the determinant allows three roots, but neither admissible ordered pair satisfies the modulus condition with any of them.

Source caution: the bank labels this numerical-answer question 2025 but records 2023-01-24, Morning, meaning 24 January 2023. Its exact year and shift require verification against the official paper at jeemain.nta.nic.in. No answer options are supplied.

Choose the two integers independently from the two-element set, excluding a zero sum:

a, b \in {- 3, 3}, a + b \neq 0 .

Count the ordered pairs for which at least one complex number satisfies both equations:

| \frac{z - a}{z + b} | = 1,

| \begin{matrix} z + 1 & ω & ω^{2} \\ ω & z + ω^{2} & 1 \\ ω^{2} & 1 & z + ω \end{matrix} | = 1, z \in C .

The two non-real cube roots of unity are the roots of:

x^{2} + x + 1 = 0, x \in {ω, ω^{2}} .

A pair counts once if any root satisfies both conditions. Finding determinant roots alone does not establish that the pair works.

How does the first-row expansion reduce the determinant to a cube?

The determinant equals the cube of the unknown, so setting it equal to one leaves exactly the three cube roots of unity. The first-row expansion shows how every lower-power term cancels.

Write the determinant as:

D = | \begin{matrix} z + 1 & ω & ω^{2} \\ ω & z + ω^{2} & 1 \\ ω^{2} & 1 & z + ω \end{matrix} | .

Use these identities throughout:

1 + ω + ω^{2} = 0, ω^{3} = 1, ω^{4} = ω .

The supplied solution expands along the first row, with alternating cofactor signs:

D = & (z + 1) [(z + ω^{2}) (z + ω) - 1] & - ω [ω (z + ω) - ω^{2}] & + ω^{2} [ω - (z + ω^{2}) ω^{2}] .

In the first bracket, the constant terms cancel and the cube-root sum gives the linear term:

(z + ω^{2}) (z + ω) - 1 & = z^{2} + z (ω + ω^{2}) + ω^{3} - 1 & = z^{2} - z & = z (z - 1) .

The second bracket loses its two identical constant terms:

ω (z + ω) - ω^{2} = ω z + ω^{2} - ω^{2} = ω z .

The third bracket simplifies using the fourth-power identity:

ω - (z + ω^{2}) ω^{2} = ω - ω^{2} z - ω^{4} = - ω^{2} z .

Retain the outside cofactor multipliers while collecting terms:

D & = (z + 1) z (z - 1) - ω (ω z) + ω^{2} (- ω^{2} z) & = z (z^{2} - 1) - ω^{2} z - ω^{4} z & = z (z^{2} - 1) - z (ω + ω^{2}) & = z (z^{2} - 1) + z & = z^{3} .

The question sets the determinant equal to one, not zero. Therefore:

D = 1 ⟹ z^{3} = 1 ⟹ z \in {1, ω, ω^{2}} .

These are candidates from the determinant equation only. Each still needs to pass the modulus condition for an admissible pair.

How does the modulus equation give a condition on the difference?

The modulus equation requires the difference between the two integers to equal the sum of a determinant root and its square. Check the denominator before multiplying, and establish the conjugate identity before substituting it.

The possible poles are: $z = - b \in {- 3, 3} .$

Neither has cube equal to one:

3^{3} = 27, (- 3)^{3} = - 27 .

Thus the quotient is defined for every determinant candidate, and:

| \frac{z - a}{z + b} | = 1 ⟺ | z - a | = | z + b | .

Both sides are nonnegative, so squaring preserves equivalence. Since the two integers are real:

(z - a) (\bar{z} - a) = (z + b) (\bar{z} + b) .

Expand both products:

z \bar{z} - a (z + \bar{z}) + a^{2} = z \bar{z} + b (z + \bar{z}) + b^{2} .

Cancel the common product and rearrange:

a^{2} - b^{2} - (a + b) (z + \bar{z}) = 0 .

Here the determinant restriction justifies replacing the conjugate:

z^{3} = 1 ⟹ | z |^{3} = 1 ⟹ | z | = 1,

z \bar{z} = 1 ⟹ \bar{z} = \frac{1}{z} = z^{2} .

This is not a general complex-number identity. It works here because the unknown is a cube root of unity.

Factoring now gives:

(a + b) [a - b - (z + z^{2})] = 0 .

Use the given nonzero-sum condition before dividing:

a + b \neq 0 ⟹ a - b = z + z^{2} .

Which ordered pairs survive both conditions?

Neither admissible pair survives. Both have zero difference, whereas every determinant root requires a difference of either two or minus one.

The original four ordered pairs are:

(- 3, - 3), (- 3, 3), (3, - 3), (3, 3) .

Exclude the opposite-sign pairs because their sums vanish:

(- 3, 3), (3, - 3) .

Both remaining pairs have: $a - b = 0 .$

Check the required difference for every root:

For the real root:

z = 1 : z + z^{2} = 2 .

For the first non-real root:

z = ω : ω + ω^{2} = - 1 .

For the second non-real root:

z = ω^{2} : ω^{2} + ω^{4} = ω^{2} + ω = - 1 .

The actual difference matches neither required value: $0 \notin {2, - 1} .$

Neither pair admits any valid root. The question counts ordered pairs, not roots, so enter:

0

Why is multiplying two pairs by three roots invalid?

Multiplication treats candidate choices as independently valid choices without checking the modulus equation. A hypothetical incorrect method lists two admissible pairs and three determinant roots, then writes: $2 \times 3 = 6 .$

Both equations must hold for the same root and the same pair. For each admissible pair, compatibility would require:

0 = 2 or 0 = - 1 .

Both statements are false, so all six candidate pair-root combinations fail. There is also a difference between counting triples and ordered pairs:

(a, b, z) versus (a, b) .

Even if several roots worked for one pair, that pair would count only once. Here, 6 is an illustrative wrong numerical result, not an official option, a documented student answer, or a proven explanation of the old stored value.

Can you solve two related questions using the same method?

The same modulus expansion gives two complex numbers in the first exercise and two ordered pairs in the second. These are original related practice, not additional verified PYQs.

Which cube roots are equally distant from one and minus two?

The two non-real cube roots satisfy the condition; the real root does not. Original related practice 1: find all solutions and state their principal arguments:

z^{3} = 1, | z - 1 | = | z + 2 | .

Squaring and expanding gives:

z \bar{z} - (z + \bar{z}) + 1 = z \bar{z} + 2 (z + \bar{z}) + 4,

$z + \bar{z} = - 1 .$ Among the three cube roots, the solutions and their principal arguments are:

z = - \frac{1}{2} + i \frac{\sqrt{3}}{2}, Arg z = \frac{2 π}{3},

z = - \frac{1}{2} - i \frac{\sqrt{3}}{2}, Arg z = - \frac{2 π}{3} .

How many pairs work when the set is minus one, zero and one?

Exactly two ordered pairs work. Original related practice 2: retain the original determinant and modulus equations, but use:

a, b \in {- 1, 0, 1}, a + b \neq 0 .

Reuse the established conditions:

D = z^{3} = 1, a - b \in {2, - 1} .

Difference two gives only: $(a, b) = (1, - 1),$ which is excluded by its zero sum. Difference minus one gives:

(a, b) = (- 1, 0), (0, 1) .

Both work with either non-real cube root. Their respective denominators are:

z, z + 1 .

Neither denominator vanishes at either root: $z \in {ω, ω^{2}} .$

Therefore:

2 ordered pairs, not 4 pair-root combinations .

On your next attempt, write the compatibility condition before listing answers. Then count each surviving pair exactly once.

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Frequently asked questions

What is the answer to the Complex Numbers JEE 2025 determinant question?

The answer is 0 ordered pairs. The nonzero-sum condition leaves only (-3, -3) and (3, 3), both with a - b = 0. The determinant roots require a - b to be either 2 or -1, so neither pair works.

How does the determinant simplify to z^3?

Expand along the first row with alternating cofactor signs. Using 1 + ω + ω² = 0, ω³ = 1 and ω⁴ = ω, the expansion becomes z(z² - 1) - z(ω + ω²) = z³. Since the determinant equals 1, the candidates are 1, ω and ω².

How do you simplify the modulus condition in this question?

First check that z + b is nonzero for every determinant root, then rewrite the condition as |z - a| = |z + b|. Squaring and expanding gives (a + b)[a - b - (z + conjugate(z))] = 0. Since a + b is nonzero and z³ = 1 implies conjugate(z) = z², the condition becomes a - b = z + z².

Why can't I multiply two ordered pairs by three roots?

The two pairs and three roots are only candidates; each pair-root combination must satisfy the modulus equation as well. All six combinations fail that check here. The question counts ordered pairs, so even if several roots worked for one pair, that pair would count only once.

Is this question from JEE Main 2025 or 2023?

The question bank labels it 2025 but records 24 January 2023, Morning. Its exact year and shift remain unverified and should be checked against the official paper at jeemain.nta.nic.in.