Previous year questions: Complex Numbers and Quadratic Equations

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Six items are real JEE Main 2026 questions taken from the official National Testing Agency papers, and four are from the NCERT Exemplar for Class 11 Mathematics. Each heading shows the paper and the marks. The stems were checked against the official PDF pages and the answers against the official key. Try each question before reading the solution.

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

Real numbers x and y satisfy 50 times (2x divided by (1 + 3i) minus y divided by (1 minus 2i)) equal to 31 + 17i. Find 10(x minus 3y).

Answer: Rationalise each fraction first. The first fraction becomes 2x(1 minus 3i) divided by 10, that is x(1 minus 3i) by 5. The second becomes y(1 plus 2i) by 5. Combine over the single denominator 5: the bracket is (x minus y) minus (3x plus 2y)i, all divided by 5. Multiplying by 50 gives 10(x minus y) minus 10(3x plus 2y)i equal to 31 plus 17i. Compare real and imaginary parts: x minus y is 31 by 10, and 3x plus 2y is minus 17 by 10. Solving the pair gives x equal to 0.9 and y equal to minus 2.2. Then x minus 3y is 0.9 plus 6.6, that is 7.5, and 10 times that is 75. The official key is 75.

Q2 (JEE Main 2026, 2 Apr Shift 2, 4 marks)

In the complex plane, C1 is the circle with centre at the origin and radius r, and C2 is the circle of radius 5 with centre at 3 + 4i. The circle C2 lies inside C1. If z1 moves on C1, z2 moves on C2, and the minimum value of |z1 minus z2| is 2, find the maximum value of |z1 minus z2|.

Answer: The modulus is distance, so both quantities are extreme distances between the two circles. The centre of C2 has modulus 5, because the square root of 9 plus 16 is 5, and its radius is 5. For C2 to lie inside C1, the radius r of C1 must be at least the distance between centres plus the radius of C2, that is at least 10.

The closest approach is the gap between the circles along the line of centres, which is r minus 10. Given that this equals 2, r must be 12. The maximum distance is measured in the opposite direction along the same line and equals r plus 10, that is 22. So the required maximum is 22, matching the official key.

Q3 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

Let S be the set of complex z satisfying z squared plus 4z plus 16 equal to 0. Find the sum over z in S of the square of the modulus of (z plus root 3 i).

Answer: Solve the quadratic first. The discriminant is 16 minus 64, that is minus 48, so the square root is 4 root 3 i. The roots are (minus 4 plus or minus 4 root 3 i) by 2, that is minus 2 plus 2 root 3 i and minus 2 minus 2 root 3 i.

Take the first root. Adding root 3 i gives minus 2 plus 3 root 3 i, whose modulus squared is 4 plus 27, that is 31. Take the second root. Adding root 3 i gives minus 2 minus root 3 i, whose modulus squared is 4 plus 3, that is 7. The sum over the two elements of S is 31 plus 7 equal to 38, which is the official answer.

Q4 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

The complex numbers 1 and 1 plus i root 2 are two roots of the cubic x cubed plus ax squared plus bx plus c equal to 0, with real coefficients. Find the integral from minus 1 to 1 of the cubic.

Answer: A cubic with real coefficients has the third root equal to the conjugate of the complex root, so the third root is 1 minus i root 2. The three roots are 1, 1 plus i root 2 and 1 minus i root 2. Their sum is 3, so a equals minus 3. The sum of pairwise products is b: 1 times (1 plus i root 2) plus 1 times (1 minus i root 2) plus the product of the conjugate pair, which is 3. That gives b equal to 2 plus 3, that is 5. The product of the roots is minus c; the product is 1 times 3 equal to 3, so c is minus 3.

Now integrate term by term. The integral of x cubed, bx and ax squared are all odd powers and contribute nothing between symmetric limits. Only c contributes: its integral is c times (1 minus minus 1), that is 2c, which is minus 6. Wait, a times x squared is not odd, so include it: the integral of ax squared gives a times 2 by 3, that is minus 2. Adding, the value is minus 2 minus 6, that is minus 8, matching the official key.

Q5 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

The roots of x squared minus x plus p equal to 0 are alpha and beta, and the roots of x squared minus 4x plus q equal to 0 are gamma and delta, with p and q integers. If alpha, beta, gamma and delta are in geometric progression in that order, find the absolute value of p plus q.

Answer: Let the four terms be a, ar, ar squared and ar cubed. The first pair sums to 1 and the second pair sums to 4. So a(1 plus r) equals 1 and ar squared (1 plus r) equals 4. Dividing the second by the first gives r squared equal to 4, so r is 2 or minus 2.

If r is 2, then a is 1 by 3, and p is the product alpha beta, which is a squared r, that is 2 by 9, not an integer. That case is rejected because p and q are integers. If r is minus 2, then a is minus 1, p is a squared r equal to minus 2, and q is the product gamma delta, equal to a squared times r to the fifth, which is minus 32. Then p plus q is minus 34 and its absolute value is 34, matching the official key.

Q6 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

Find the number of complex numbers z satisfying both the modulus of (z minus (4 plus 8i)) equal to root 10 and the sum of the moduli of (z minus (3 plus 5i)) and (z minus (5 plus 11i)) equal to 4 root 5.

Answer: Connect the numbers to geometry. Let A be the point (4, 8), B the point (3, 5) and C the point (5, 11). The first equation says z lies on the circle centred at A of radius root 10. The second says the sum of distances from z to B and to C is 4 root 5, which is an ellipse. Note that B and C have midpoint (4, 8), that is A, and the distance BC is 2 root 10. The distance between the foci of the ellipse is 2 root 10, so the semi major axis is 2 root 5 and the semi minor axis is the square root of 20 minus 10, that is root 10. The ellipse is (x minus 4) squared by 20 plus (y minus 8) squared by 10 equal to 1.

Now intersect with the circle of radius root 10 about A. Write X and Y for the offsets from A. The circle is X squared plus Y squared equal to 10 and the ellipse is X squared plus 2Y squared equal to 20. Subtracting gives Y squared equal to 10 and X equal to 0, so there are two solutions, one above and one below A. The official answer is 2.

Q7 (NCERT Exemplar, 3 marks)

If ((1 plus i) divided by (1 minus i)) to the power 100 equals a plus ib, find the ordered pair (a, b).

Answer: Simplify the base first. Multiply numerator and denominator by 1 plus i: the numerator becomes (1 plus i) squared equal to 2i, and the denominator becomes 1 plus 1 equal to 2. So the base is i. The 100th power of i is 1, because 100 is a multiple of 4. Comparing with a plus ib gives a equal to 1 and b equal to 0, so the pair is (1, 0). Source: NCERT Exemplar Class 11 Mathematics, Chapter 5, Exercise 5.3, Short Answer Type, Q5.

Q8 (NCERT Exemplar, 3 marks)

If (1 plus i) squared divided by (2 minus i) equals x plus iy, find x plus y.

Answer: Square the numerator: (1 plus i) squared is 2i. Now divide 2i by 2 minus i by multiplying numerator and denominator by 2 plus i. The denominator becomes 4 plus 1, that is 5. The numerator becomes 2i times (2 plus i), which is 4i plus 2 i squared, that is minus 2 plus 4i. So x is minus 2 by 5 and y is 4 by 5, and their sum is 2 by 5. Source: NCERT Exemplar Class 11 Mathematics, Chapter 5, Exercise 5.3, Short Answer Type, Q4.

Q9 (NCERT Exemplar, 3 marks)

Show that the modulus of (z minus 2) divided by (z minus 3) equal to 2 represents a circle, and find its centre and radius.

Answer: Square both sides to get |z minus 2| squared equal to 4 times |z minus 3| squared. Write z as x plus iy and expand: (x minus 2) squared plus y squared equals 4 times ((x minus 3) squared plus y squared). Expanding and collecting gives 3x squared plus 3y squared minus 20x plus 32 equal to 0. Divide by 3 to get x squared plus y squared minus 20x by 3 plus 32 by 3 equal to 0. Completing the square in x gives centre (10 by 3, 0) and radius squared equal to 100 by 9 minus 32 by 3, which is 4 by 9. So the radius is 2 by 3. Source: NCERT Exemplar Class 11 Mathematics, Chapter 5, Exercise 5.3, Long Answer Type, Q14.

Q10 (NCERT Exemplar, 3 marks)

If (z plus 1) divided by (z minus 1) is purely imaginary and z is not equal to 1, find the modulus of z.

Answer: A purely imaginary number equals the negative of its conjugate. Let w be the fraction. The condition says w plus its conjugate is zero. The conjugate of the fraction is (conjugate of z plus 1) divided by (conjugate of z minus 1). Combine the two fractions over a common denominator and set the numerator to zero. The numerator becomes (z plus 1)(conjugate z minus 1) plus (conjugate z plus 1)(z minus 1). Expanding and cancelling gives 2 times (z times conjugate z minus 1) equal to 0. Therefore z times its conjugate equals 1, and the modulus of z is 1. Source: NCERT Exemplar Class 11 Mathematics, Chapter 5, Exercise 5.3, Long Answer Type, Q15.