NDA Complex Numbers previous year questions with solutions

Answer: (c) 1

Method. x = 4 + i satisfies x² - 8x + 17 = 0; divide the cubic by this quadratic

x − 4 = i, so (x − 4)² = -1, i.e. x² - 8x + 17 = 0.
(x² - 8x + 17)(x + 10) = x³ + 2x² - 63x + 170.
So x³ + 2x² - 63x + 171 = (x² - 8x + 17)(x + 10) + 1 = 0 + 1 = 1.

Answer: (c) 1

Method. the exponents are the squares 1², 2², …, 24²

576 = 24², so the product is i^{12 + 22 + … + 242}.
1² + 2² + … + 24² = 24.25.496 = 4900.
4900 = 4.1225, so i⁴⁹⁰⁰ = 1.

Answer: (a) None

Method. complex numbers with non-zero imaginary part cannot be ordered

The complex numbers carry no order relation compatible with arithmetic, so > and < have no meaning between non-real numbers.
Every number in I, II and III has a non-zero imaginary part.
So none of the three inequalities is valid.

Answer: (b) 0

Method. 1 + i + i² = i, and any four consecutive powers of i add to 0

i^n-1 + iⁿ + iⁿ⁺¹ = i^n-1(1 + i + i²) = i^n-1·i = iⁿ.
So the sum is 20Σn=1 iⁿ = i + i² + … + i²⁰.
These are five complete blocks i + i² + i³ + i⁴ = i − 1 − i + 1 = 0.
Sum = 0.

Answer: (c) 5⁠√2⁠

Method. find ⁠√(12⁠ + 5i) as p + iq; the other root is its conjugate

Let ⁠√(12⁠ + 5i) = p + iq: p² - q² = 12, 2pq = 5, and p² + q² = |12 + 5i| = 13.
So p² = 252, q² = 12; with pq > 0, ⁠√(12⁠ + 5i) = 5 + i⁠√2⁠ (principal root).
Similarly ⁠√(12⁠ −⁠ 5i) = 5 − i⁠√2.⁠
Sum = 10⁠√2⁠ = 5⁠√2.⁠

Answer: (b) 1 − i⁠√2⁠

Method. square each option and compare with -i

-⁠√-1⁠ = -i.
(b): (1 − i⁠√2)²⁠ = 1 − 2i + i²2 = -2i2 = -i. (holds)
(a): (1 + i)² = 2i; (c): (1 + i⁠√2)²⁠ = i; (d): (i⁠√2)²⁠ = -12.
Only 1 − i⁠√2⁠ is a square root of -i.

Answer: (a) -1

Method. powers of i repeat in blocks of four, each block summing to 0

Read the upper limit as N = 8m + 7 for some whole number m ≥ 0 (the page uses the letter n for both the index and the limit).
i + i² + i³ + i⁴ = i − 1 − i + 1 = 0, and every block of four consecutive powers sums to 0.
N = 8m + 7 = 4(2m + 1) + 3: the first 4(2m + 1) terms sum to 0, leaving the last three terms i^N-2 + i^N-1 + i^N, which equal i¹ + i² + i³ because N − 2 ≡ 1 (mod 4).
Sum = i − 1 − i = -1.

The printed upper limit 8n+7 reuses the summation index n; read as any number of the form 8m + 7, the sum is -1 for every m, so (a) is the single correct option under every sensible reading and the question is not treated as defective.

Answer: (a) ⁠√(a⁠ + ib) + ⁠√(a⁠ −⁠ ib)

Method. square the candidate; ⁠√(a⁠ + ib)·⁠√(a⁠ −⁠ ib) = |a + ib|

Let p = ⁠√(a⁠ + ib) and take ⁠√(a⁠ −⁠ ib) as its conjugate p̅ (the principal roots are conjugate whenever b ≠ 0 or a ≥ 0); then p·p̅ = |p|² = |a + ib| = ⁠√(a²⁠ + b²).
(a): (p + p̅)² = p² + p̅² + 2pp̅ = (a + ib) + (a − ib) + 2⁠√(a²⁠ + b²) = 2a + 2⁠√(a²⁠ + b²). (holds)
(b): (p − p̅)² = 2a − 2⁠√(a²⁠ + b²), not the required number.
(c), (d): (2a ± ib)² = 4a² - b² ± 4iab, not 2a + 2⁠√(a²⁠ + b²) in general.
So ⁠√(a⁠ + ib) + ⁠√(a⁠ −⁠ ib) is a square root.

Answer: (b) Two

Method. i^-2n = (i²)^-n = (-1)^-n

i^-2n = (i²)^-n = (-1)^-n = (-1)ⁿ.
For even n this is 1, for odd n it is -1.
So i^-2n takes exactly two values, 1 and -1.

Answer: (b) 1

Method. modulus of a product = product of moduli; |i| = |-i| = 1

|i²ⁿ⁺¹(-i)^2n-1| = |i|²ⁿ⁺¹·|-i|^2n-1 = 1²ⁿ⁺¹·1^2n-1 = 1.
(Directly: (-i)^2n-1 = (-1)^2n-1i^2n-1 = -i^2n-1, so the number is -i⁴ⁿ = -1, whose modulus is 1.)
A modulus is never negative, so (a) is impossible; the answer is 1.

Answer: (c) -6i − 3

Method. x² = 2i, then build the higher even powers from it

x² = (1 + i)² = 1 + 2i − 1 = 2i
x⁴ = (2i)² = -4
x⁶ = x⁴ · x² = -4 · 2i = -8i
x⁶ + x⁴ + x² + 1 = -8i − 4 + 2i + 1 = -6i − 3.

No correct option. This question is defective as printed in UPSC's paper.

Method. (intended reading) expand the 3 × 3 determinant and equate real and imaginary parts

Reading the array as a determinant:
Δ = x[(1)(-i) − (i)(2i)] - (-3i)[(y)(-i) − (i)(0)] + 1[(y)(2i) − (1)(0)]
= x(-i + 2) + 3i(-iy) + 2iy = x(2 − i) + 3y + 2iy
= (2x + 3y) + i(2y − x).
Equating with 6 + 11i: 2x + 3y = 6 and 2y − x = 11.
From the second, x = 2y − 11; then 4y − 22 + 3y = 6, so y = 4 and x = -3.

Defective as printed: the 3 × 3 array is printed in square brackets, so it is a matrix, and a matrix cannot equal the number 6 + 11i. Read as a determinant, x = −⁠3, y = 4 give 6 + 11i, option (a); the other pairs give 18 + 5i, −⁠6 − 11i and −⁠18 − 5i.

Answer: (a) 0

Method. four consecutive powers of i sum to 0

i¹⁰⁰⁰ = (i⁴)²⁵⁰ = 1.
i¹⁰⁰⁰ + i¹⁰⁰¹ + i¹⁰⁰² + i¹⁰⁰³ = 1(1 + i + i² + i³) = 1 + i − 1 − i = 0.

Answer: (c) -2i

Method. split the sum; any four consecutive powers of i sum to 0

11Σn=2 (iⁿ + iⁿ⁺¹) = 11Σn=2 iⁿ + 12Σn=3 iⁿ.
11Σn=2 iⁿ: the eight terms n = 2 to 9 sum to 0; left with i¹⁰ + i¹¹ = -1 − i.
12Σn=3 iⁿ: the eight terms n = 3 to 10 sum to 0; left with i¹¹ + i¹² = -i + 1.
Total = (-1 − i) + (1 − i) = -2i.

Shortcut. iⁿ + iⁿ⁺¹ = iⁿ(1 + i); 11Σn=2 iⁿ = -1 − i, so the sum is (1 + i)(-1 − i) = -(1 + i)² = -2i.

Answer: (a) 0

Method. powers of i repeat in fours; four consecutive powers sum to 0

i²ⁿ + i^{2n + 1} + i^{2n + 2} + i^{2n + 3} = i²ⁿ(1 + i + i² + i³)
1 + i + i² + i³ = 1 + i − 1 − i = 0
So the value is i²ⁿ × 0 = 0.

Answer: (c) 1

Method. solve the quadratic: x = cos θ ± i sin θ

x = 2cos θ ± ⁠√(4cos²θ⁠ −⁠ 4)2 = cos θ ± ⁠√(-sin²θ)⁠ = cos θ ± i sin θ.
|x| = ⁠√(cos²θ⁠ + sin²θ) = 1.

Shortcut. Product of the roots is 1 and they are conjugates (or both ±1), so |x|² = 1.

Answer: (d) 1

Method. Z₁/⁠Z₂ purely imaginary puts it on the imaginary axis, equidistant from 1 and -1

3Z₁4Z₂ purely imaginary means Z₁Z₂ = ik with k real, k ≠ 0.
|Z₁ + Z₂Z₁ - Z₂| = |ik + 1||ik − 1| = ⁠√(k²⁠ + 1)⁠√(k²⁠ + 1) = 1.

Answer: (b) ⁠√2⁠

Method. expand the 3 × 3 determinant along the first row, then take the modulus

Δ = i(3i·3 − 2.1) − 2i(2i·3 − 2.3) + 1(2i·1 − 3i·3)
= i(9i − 2) − 2i(6i − 6) + (2i − 9i)
= (-9 − 2i) + (12 + 12i) − 7i = 3 + 3i.
Z = ⁠13⁠(3 + 3i) = 1 + i, so |Z| = ⁠√(1⁠ + 1) = ⁠√2.⁠

Answer: (c) 1

Method. |z₁ + z₂| = |z₁ - z₂| makes z₁/⁠z₂ purely imaginary

Put w = z₁z₂. The condition becomes |w + 1| = |w − 1|.
So w is equidistant from 1 and -1: it lies on the perpendicular bisector, the imaginary axis.
(Squaring: (a + 1)² + b² = (a − 1)² + b² gives 4a = 0, with w = a + ib.)
Hence Re(w) = 0 and Re(z₁z₂) + 1 = 1.

Answer: (c) 1

Method. set the real part of z − 1z + 1 to 0

Let z = x + iy. z − 1z + 1 = (x − 1 + iy)(x + 1 − iy)(x + 1)² + y² = x² + y² - 1 + 2iy(x + 1)² + y².
Purely imaginary ⟹ real part 0 ⟹ x² + y² - 1 = 0.
So |z|² = 1, i.e. |z| = 1.

Answer: (a) 1

Method. |z₁z₂| = |z₁||z₂|; conjugates have equal moduli

1 − i sin θ is the conjugate of 1 + i sin θ (sin θ is real), so both have modulus ⁠√(1⁠ + sin² θ).
|z| = ⁠√(1⁠ + sin² θ)⁠√(1⁠ + sin² θ) = 1.

Answer: (c) nπ

Method. rationalise z and set its imaginary part to 0

Multiply numerator and denominator by 1 + i sin θ:
z = (1 + i sin θ)²1 + sin² θ = 1 − sin² θ1 + sin² θ + i2 sin θ1 + sin² θ.
z is purely real ⟺ 2 sin θ = 0 ⟺ sin θ = 0 ⟺ θ = nπ.
(d) misses the odd multiples of π, e.g. θ = π gives z = 1; (a) includes θ = π2, where z = i.

Answer: (b) (2n + 1)π2

Method. rationalise z and set its real part to 0

From z = 1 − sin² θ1 + sin² θ + i2 sin θ1 + sin² θ:
z is purely imaginary ⟺ 1 − sin² θ = 0 (and then sin θ = ±1, so the imaginary part ±1 is non-zero).
cos² θ = 0 ⟺ θ = (2n + 1)π2.
Check: θ = π2 gives z = 1 + i1 − i = i.

Answer: (c) 1

Method. |zⁿ| = |z|ⁿ

⁠√-32⁠ − 12 = -12 + ⁠√32i.⁠
Its modulus is ⁠√(14⁠ + ⁠34⁠) = 1.
So the modulus of its 200th power is 1²⁰⁰ = 1.

Answer: (b) 2

Method. rationalise 2/⁠Z and add

2Z = 21 + i = 2(1 − i)(1 + i)(1 − i) = 2(1 − i)2 = 1 − i.
Z + 2Z = (1 + i) + (1 − i) = 2.
|Z + 2Z| = |2| = 2.

Answer: (a) 1 only

Method. conjugate of a quotient; Z·Z^-1 = 1

Statement 1: Z·Z^-1 = 1, so taking conjugates Z̅·(Z̅^-1̅) = 1̅ = 1, i.e. (Z̅^-1̅) = (Z̅)^-1. Correct.
Statement 2: for any Z ≠ 0, ZZ^-1 = 1, whereas |Z|² can be any positive number; e.g. Z = 2 gives ZZ^-1 = 1 but |Z|² = 4. So statement 2 is not true in general. Incorrect.
(The true identity is ZZ̅ = |Z|², with the conjugate, not the inverse.)
Only statement 1 is correct.

Settled as printed: statement 2 reads ZZ^-1 = |Z|², which fails for every Z with |Z| ≠ 1, so it is a valid false statement and (a) is the one correct option; had ZZ̅ = |Z|² been meant, the answer would be (c), but nothing on the page forces that reading.

No correct option. This question is defective as printed in UPSC's paper.

Method. Z ± Z̅ in terms of x and y

Let Z = x + iy, so Z̅ = x − iy.
1. Z − Z̅ = 2iy: its real part is 0, so it is a (purely) imaginary number. Correct.
2. Z + Z̅ = 2x: its imaginary part is 0, so it is real. Correct.
Both statements are correct.

Defective as printed: for a real Z the difference Z − Z̅ is 0, and whether 0 counts as an imaginary number depends on the convention, which the question does not fix. Counting 0 as (purely) imaginary, both statements hold, option (c); taking an imaginary number to have a non-zero imaginary part, statement 1 fails for real Z, option (b). Statement 2 holds in every case.

Answer: (b) 1

Method. |cos θ + i sin θ| = |cos θ − i sin θ| = 1

|cos θ + i sin θ| = ⁠√(cos²⁠ θ + sin² θ) = 1, and likewise |cos θ − i sin θ| = 1.
Modulus of a quotient = quotient of moduli = 11 = 1.

Answer: (c) 1

Method. simplify the denominator: (1 − i)² = -2i

(1 − i)² = 1 − 2i + i² = -2i
1 − (1 − i)² = 1 + 2i
z = 1 + 2i1 + 2i = 1, so |z| = 1.

Answer: (a) 0

Method. z simplifies to 1, a positive real number

As in the previous item, 1 − (1 − i)² = 1 − (-2i) = 1 + 2i, so z = 1 + 2i1 + 2i = 1.
z = 1 lies on the positive real axis, so its principal argument is 0.

Answer: (a) Zero (No solution)

Method. |1 − 2i| = ⁠√5,⁠ so the equation is 5^x/⁠2 = 5^x

|1 − 2i| = ⁠√(1⁠ + 4) = ⁠√5.⁠
The equation becomes (⁠√5)^x⁠ = 5^x, i.e. 5^x/⁠2 = 5^x.
So x2 = x, giving x = 0 only.
There is no non-zero integral solution.

Answer: (c) 1

Method. use αα = 1 to write 1 − αβ = α(α − β)

|α| = 1 gives αα = 1.
1 − αβ = αα − αβ = α(α − β) = α(α − β).
|1 − αβ| = |α| |α − β| = |α − β|, which is non-zero since α ≠ β.
So |α − β1 − αβ| = 1.

Answer: (a) 0

Method. a conjugate is the mirror image in the real axis, so its argument is the negative

Let z = r(cos θ + i sin θ), with θ the principal value in (−⁠π, π]. Then z̅ = r(cos θ − i sin θ) = r(cos(-θ) + i sin(-θ)).
So amp(z̅) = -amp(z), and amp(z) + amp(z̅) = 0.
(Only for a negative real z do the principal values give π + π = 2π, which is the same angle as 0.)

Answer: (b) 2π3

Method. polar form: z = 2e^iπ/⁠32e^-iπ/⁠3

1 + i⁠√3⁠ = 2(cos π3 + i sin π3) and 1 − i⁠√3⁠ = 2(cos π3 − i sin π3).
arg z = arg(1 + i⁠√3)⁠ − arg(1 − i⁠√3)⁠ = π3 − (-π3) = 2π3.
Check: z = (1 + i⁠√3)²4⁠ = -2 + 2i⁠√34⁠ = -12 + ⁠√32i,⁠ which lies in the second quadrant with argument 2π3.

Answer: (b) -π4

Method. rationalise, then read off the argument from the quadrant

11 + i = 1 − i(1 + i)(1 − i) = 12 − 12i.
Real part positive, imaginary part negative: fourth quadrant, with tan θ = -1.
Principal argument (in (-π, π]) = -π4.

Shortcut. arg(11 + i) = -arg(1 + i) = -π4.

Answer: (a) 240°

Method. rationalise (or divide polar forms); the quotient is -1/⁠2 − i⁠√3/⁠2,⁠ at 240°

Multiply numerator and denominator by 1 − i⁠√3:⁠
(1 − i⁠√3)²1⁠ + 3 = 1 − 2i⁠√3⁠ − 34 = -12 − i⁠√32.⁠
Both parts negative: third quadrant, with reference angle 60°.
Argument = 180° + 60° = 240° (equivalently -120° as the principal value); of the options, 240°.

Shortcut. 1 − i⁠√3⁠ has argument -60° and 1 + i⁠√3⁠ has +60°, so the quotient has argument -120°, i.e. 240°.

Answer: (c) -3π/⁠4

Method. -1 − i is in the third quadrant: arg = -(π − π/⁠4)

-1 − i has real part -1 < 0 and imaginary part -1 < 0: third quadrant.
Reference angle: tan^-1(11) = π4.
Principal argument (in (-π, π]) = -(π − π4) = -3π4.

Answer: (b) 2(cos π6 + i sin π6)

Method. modulus ⁠√(3⁠ + 1) = 2; argument tan^-1(1/⁠√3)⁠ = π/⁠6

r = |⁠√3⁠ + i| = ⁠√(3⁠ + 1) = 2.
cos θ = ⁠√32,⁠ sin θ = 12, so θ = π6.
⁠√3⁠ + i = 2(cos π6 + i sin π6).

Answer: (a) 1, 0

Method. simplify the denominator: (1 − i)² = -2i

(1 − i)² = 1 − 2i + i² = -2i
1 − (1 − i)² = 1 + 2i
So the number is 1 + 2i1 + 2i = 1.
1 lies on the positive real axis: modulus 1, principal argument 0.

Answer: (c) 8

Method. compare moduli, then check the argument

|1 + i|ⁿ = (⁠√2)ⁿ⁠ = 16 = (⁠√2)⁸⁠ forces n = 8.
Check: (1 + i)² = 2i, so (1 + i)⁸ = (2i)⁴ = 16. (holds)
The least (indeed the only) value is n = 8.

Answer: (b) 2

Method. 1 − i1 + i = -i and 1 + i1 − i = i

1 + i1 − i = (1 + i)²2 = 2i2 = i, and 1 − i1 + i = 1i = -i.
The equation becomes (-i)^2m i²ⁿ = (-1)^m (-1)ⁿ = (-1)^{m + n} = 1.
So m + n is even, which means m − n is even.
The smallest positive value of m − n is 2.

Answer: (a) -1

Method. polar form: ⁠√3⁠ ± i = 2(cos π6 ± i sin π6)

⁠√3⁠ + i = 2e^iπ/⁠6 and ⁠√3⁠ − i = 2e^-iπ/⁠6.
⁠√3⁠ + i⁠√3⁠ − i = e^iπ/⁠3.
Cubing: e^iπ = cos π + i sin π = -1.

Shortcut. Numerator and denominator are conjugates, so the ratio has modulus 1 and argument 2·π6.

Answer: (d) -8

Method. (1 ± i)² = ±2i

(1 + i)² = 1 + 2i − 1 = 2i, so (1 + i)⁴ = (2i)² = -4.
(1 − i)² = -2i, so (1 − i)⁴ = (-2i)² = -4.
Sum = -4 − 4 = -8.

Answer: (a) 2

Method. simplify the base to a power of i, then use i⁴ = 1

1 − i1 + i = (1 − i)²(1 + i)(1 − i) = 1 − 2i + i²2 = -2i2 = -i.
So we need (-i)ⁿ² = 1; the powers of -i repeat with period 4, so n² must be a multiple of 4.
n² is a multiple of 4 exactly when n is even; the smallest positive even n is 2.
Check: (-i)⁴ = 1. Answer: n = 2.

Answer: (c) 2i

Method. polar form: (⁠√3⁠ + i)/⁠2 = cos π/⁠6 + i sin π/⁠6, (i - ⁠√3)/⁠2⁠ = cos 5π/⁠6 + i sin 5π/⁠6; De Moivre

i + ⁠√32⁠ = cos π6 + i sin π6; i - ⁠√32⁠ = cos 5π6 + i sin 5π6.
By De Moivre, the first power is cos 2019π6 + i sin 2019π6; 2019π6 = 336π + π2, so it equals cos π2 + i sin π2 = i.
The second is cos 5 × 2019π6 + i sin 5 × 2019π6; 10095π6 = 1682π + π2, so it also equals i.
Sum = i + i = 2i.

Shortcut. (i - ⁠√3)/⁠2⁠ = -((i̅ + ⁠√3̅)/⁠2̅);⁠ with an odd power, the second term is -i̅ = i, so the sum is 2i.

Answer: (b) 4

Method. (1 + i)/⁠(1 − i) = i, and iⁿ = 1 first at n = 4

1 + i1 − i = (1 + i)²(1 − i)(1 + i) = 1 + 2i − 12 = i
iⁿ = 1 needs n a multiple of 4.
Smallest positive n = 4.

Answer: (b) γβ

Method. the cube roots of -8 are α, αω, αω²

The three roots are -2, -2ω, -2ω²; whatever the labelling, β = αt and γ = αt², where t = βα is ω or ω² (t³ = 1).
Numerator = α²(p² + t²q² + t⁴r²) = α²(p² + t²q² + t r²).
Denominator = α²(t²p² + t⁴q² + r²) = α²(t²p² + t q² + r²).
t × (denominator) = α²(t³p² + t²q² + t r²) = numerator, so the ratio is t.
t = βα = αt²αt = γβ.

Answer: (c) 0

Method. reduce the powers of ω mod 3, then use 1 + ω + ω² = 0

α = ω. Since 19 = 3.6 + 1, 35 = 3.11 + 2, 25 = 3.8 + 1, 38 = 3.12 + 2:
1 + α¹⁹ - α³⁵ = 1 + ω − ω² = -ω² - ω² = -2ω².
1 − 3α²⁵ + α³⁸ = 1 − 3ω + ω² = -ω − 3ω = -4ω.
(-2ω²)¹⁰⁰ = 2¹⁰⁰ ω²⁰⁰ = 2¹⁰⁰ ω²; (-4ω)⁵⁰ = 2¹⁰⁰ ω⁵⁰ = 2¹⁰⁰ ω².
Difference = 0.

Answer: (a) 6ab

Method. expand the squares and use 1 + ω + ω² = 0, ω³ = 1

x² = a² + 2ab + b²
y² = a²ω² + 2abω³ + b²ω⁴ = a²ω² + 2ab + b²ω
z² = a²ω⁴ + 2abω³ + b²ω² = a²ω + 2ab + b²ω²
Adding: a²(1 + ω² + ω) + b²(1 + ω + ω²) + 6ab = 6ab.

Answer: (c) 87

Method. divide the equation by x to get x + 1/⁠x = 1

x ≠ 0, so dividing x² - x + 1 = 0 by x gives x + 1x = 1.
(x − 1x)² = (x + 1x)² - 4 = 1 − 4 = -3.
Then (x − 1x)⁴ = 9 and (x − 1x)⁸ = 81.
Sum = -3 + 9 + 81 = 87.

Answer: (d) -2¹⁰⁰

Method. use 1 + ω = -ω² and 1 + ω² = -ω

1 + ω − ω² = -ω² - ω² = -2ω²; 1 − ω + ω² = -ω − ω = -2ω.
(-2ω²)¹⁰⁰ = 2¹⁰⁰ ω²⁰⁰ = 2¹⁰⁰ ω², since 200 = 3.66 + 2.
(-2ω)¹⁰⁰ = 2¹⁰⁰ ω¹⁰⁰ = 2¹⁰⁰ ω, since 100 = 3.33 + 1.
Sum = 2¹⁰⁰(ω² + ω) = 2¹⁰⁰(-1) = -2¹⁰⁰.

Answer: (a) 0

Method. two proportional rows make the determinant zero

Row 1: 1 + ω = -ω², 1 + ω² = -ω, ω + ω² = -1, so R₁ = (-ω², -ω, -1).
Row 3: 1ω = ω², 1ω² = ω, so R₃ = (ω², ω, 1).
R₁ = -R₃, so the determinant is 0.

Answer: (a) 1

Method. Z₁/⁠Z₂ is a non-real cube root of unity

Divide by Z₂² (Z₂ ≠ 0, else Z₁ = 0 too): t² + t + 1 = 0 with t = Z₁Z₂.
So t = ω or ω², the non-real cube roots of unity.
|t| = |ω| = 1, so |Z₁Z₂| = 1.

Answer: (b) 0

Method. Z₁/⁠Z₂ = ω or ω², each with real part -1/⁠2

From the stem, t = Z₁Z₂ satisfies t² + t + 1 = 0, so t = -12 ± i⁠√32.⁠
Re(t) = -12.
12 + Re(Z₁Z₂) = 12 − 12 = 0.

Answer: (a) 0

Method. sum of the cube roots of unity, 1 + ω + ω² = 0

Take x = 1, y = ω, z = ω².
xy + yz + zx = ω + ω³ + ω² = ω + 1 + ω² = 0.

Shortcut. xy + yz + zx is the coefficient of t in t³ - 1, which is 0.

Answer: (b) 10(10 − ω), 10(10 − ω²), 90

Method. cube roots of -1000 are -10, -10ω, -10ω²

(z − 100)³ = -1000 = (-10)³.
So z − 100 = -10, -10ω or -10ω².
z = 90, 100 − 10ω = 10(10 − ω), 100 − 10ω² = 10(10 − ω²).

Answer: (b) -1, ω, ω²

Method. factor out (z + 1)

z = -1 is a root: -1 + 2 − 2 + 1 = 0.
z³ + 2z² + 2z + 1 = (z + 1)(z² + z + 1).
z² + z + 1 = 0 gives z = ω, ω².
Roots: -1, ω, ω².

Answer: (d) ω

Method. reduce exponents mod 3 using ω³ = 1

1985 = 3.661 + 2 and 100 = 3.33 + 1.
z = ω: ω¹⁹⁸⁵ + ω¹⁰⁰ + 1 = ω² + ω + 1 = 0. (holds)
z = -1: -1 + 1 + 1 = 1 ≠ 0.
z = -ω: -ω² + ω + 1 = -2ω² ≠ 0; z = -ω²: (-ω²)¹⁹⁸⁵ = -ω³⁹⁷⁰ = -ω, (-ω²)¹⁰⁰ = ω²⁰⁰ = ω², sum -ω + ω² + 1 = -2ω ≠ 0.
So ω is a root of equation-II.

Answer: (c) 2

Method. test the roots -1, ω, ω² of equation-I in equation-II

Equation-I has roots -1, ω, ω².
z = -1: (-1)¹⁹⁸⁵ + (-1)¹⁰⁰ + 1 = -1 + 1 + 1 = 1 ≠ 0, not common.
z = ω: ω² + ω + 1 = 0, common.
z = ω²: ω³⁹⁷⁰ + ω²⁰⁰ + 1 = ω + ω² + 1 = 0 (3970 ≡ 1, 200 ≡ 2 mod 3), common.
Number of common roots = 2.

Answer: (a) ⁠√3⁠

Method. ω + ω² = -1 and |1 − ω| = ⁠√3⁠

Since 1 + ω + ω² = 0, ω + ω² = -1, so |ω + ω²| = 1.
With ω = -12 + ⁠√32i:⁠ 1 − ω = 32 − ⁠√32i,⁠ |1 − ω| = ⁠√(94⁠ + ⁠34⁠) = ⁠√3.⁠
|1 − ωω + ω²| = ⁠√31⁠ = ⁠√3.⁠
(The same holds for ω² in place of ω, by conjugation.)

Answer: (d) 1⁠√3⁠

Method. roots of x² - x + 1 = 0 are -ω and -ω²

Replacing x by -x turns x² - x + 1 into x² + x + 1, so the roots are α = -ω, β = -ω².
α¹⁰⁰ = ω¹⁰⁰ = ω⁹⁹·ω = ω; β¹⁰⁰ = ω²⁰⁰ = ω¹⁹⁸·ω² = ω².
α¹⁰⁰ + β¹⁰⁰ = ω + ω² = -1; α¹⁰⁰ - β¹⁰⁰ = ω − ω² = i⁠√3.⁠
|-1i⁠√3|⁠ = 1⁠√3.⁠

Answer: (b) 0

Method. factor out x¹⁹⁹

x¹⁹⁹ + x²⁰⁰ + x²⁰¹ = x¹⁹⁹(1 + x + x²).
Given x² + x + 1 = 0, the bracket is 0.
So the value is x¹⁹⁹·0 = 0.

Answer: (a) 1 only

Method. roots of x³ = 8 are 2, 2ω, 2ω²

x³ - 8 = 0 ⟹ x = 2, 2ω, 2ω², where ω = -12 + i⁠√32.⁠
Each root has modulus 2, so the roots lie on the circle |z| = 2, not on a circle of unit radius: statement 2 is wrong.
The three points 2, 2ω, 2ω² are the vertices of an equilateral triangle (arguments 0, 2π3, 4π3), so they are not collinear: statement 1 is correct.
Only statement 1 is correct.

Answer: (c) k²

Method. roots of x² + x + 1 = 0 are ω and ω²

x(x + 1) + 1 = 0 ⟺ x² + x + 1 = 0, whose roots are the non-real cube roots of unity ω and ω².
If k = ω, the other root is ω² = k².
If k = ω², the other root is ω = ω⁴ = (ω²)² = k².
Either way the other root is k².

Shortcut. Product of roots = 1, so other root = 1k = k³k = k² (since k³ = 1).

Answer: (d) 2

Method. α, β = ω, ω²; evaluate ω^j + ω^2j for j = 0 to 3

x² + x + 1 = 0 has roots α = ω, β = ω² (ω³ = 1, 1 + ω + ω² = 0).
j = 0: 1 + 1 = 2
j = 1: ω + ω² = -1
j = 2: ω² + ω⁴ = ω² + ω = -1
j = 3: ω³ + ω⁶ = 1 + 1 = 2
Sum = 2 − 1 − 1 + 2 = 2.

Answer: (b) {-1 + ⁠√3⁠ i2, -1 - ⁠√3⁠ i2}

Method. the roots of x² + x + 1 = 0 are ω and ω², both also roots of x³ − 1 = 0

x³ - 1 = (x − 1)(x² + x + 1), so A = {1, -1 + ⁠√3⁠ i2, -1 - ⁠√3⁠ i2}.
x² + x + 1 = 0 gives x = -1 ± ⁠√(1⁠ −⁠ 4)2 = -1 ± ⁠√3⁠ i2, so B = {-1 + ⁠√3⁠ i2, -1 - ⁠√3⁠ i2}.
B ⊂ A, so A ∩ B = B = {-1 + ⁠√3⁠ i2, -1 - ⁠√3⁠ i2}.

Answer: (d) ω, ω²

Method. factor the cubic as (z + 1)(z² + z + 1) and test each root in the second equation

z³ + 2z² + 2z + 1 = (z³ + 1) + 2z(z + 1) = (z + 1)(z² - z + 1 + 2z) = (z + 1)(z² + z + 1).
Roots: -1, ω, ω².
z = -1: (-1)²⁰¹⁷ + (-1)²⁰¹⁸ + 1 = -1 + 1 + 1 = 1 ≠ 0, not a root.
z = ω: 2017 = 3(672) + 1, 2018 = 3(672) + 2, so ω²⁰¹⁷ + ω²⁰¹⁸ + 1 = ω + ω² + 1 = 0.
z = ω²: ω⁴⁰³⁴ + ω⁴⁰³⁶ + 1 = ω² + ω + 1 = 0 (4034 = 3(1344) + 2, 4036 = 3(1345) + 1).
Common roots: ω, ω².

Answer: (b) 2

Method. the numbers are ω and ω², and ω³ = 1

-1 + i⁠√32⁠ = ω, -1 − i⁠√32⁠ = ω², with ω³ = 1.
ω³ⁿ + ω⁶ⁿ = (ω³)ⁿ + (ω³)²ⁿ = 1 + 1 = 2.

Answer: (c) They form an equilateral triangle

Method. 1, ω, ω² are equally spaced on the unit circle

The cube roots of unity are 1, ω = cos 2π3 + i sin 2π3, ω² = cos 4π3 + i sin 4π3.
All have modulus 1: they lie on the circle of radius 1 (not ⁠√3),⁠ so they are not collinear.
Their arguments 0, 2π3, 4π3 differ by 120°, so they are vertices of an equilateral triangle (side ⁠√3).⁠

Answer: (b) -1

Method. the two numbers are ω and ω²; ωⁿ + ω²ⁿ = -1 when 3 ∤ n

-1 + i⁠√32⁠ = ω and -1 − i⁠√32⁠ = ω², with ω³ = 1 and 1 + ω + ω² = 0.
The expression is ωⁿ + ω²ⁿ.
If n = 3k + 1: ωⁿ + ω²ⁿ = ω + ω² = -1.
If n = 3k + 2: ωⁿ + ω²ⁿ = ω² + ω⁴ = ω² + ω = -1.
So the value is -1.

Answer: (c) 0

Method. 1 + ω + ω² = 0 makes the last factor zero

Since 1 + ω + ω² = 0, the fourth factor is 0.
The other factors are finite: (1 + ω)(1 + ω²) = (-ω²)(-ω) = ω³ = 1 and 1 + ω³ = 2.
So the product is 1 × 2 × 0 = 0.

Answer: (d) Neither I nor II

Method. |z − λi| = |z + λi| is the perpendicular bisector of λi and -λi

The condition is |z + λi| = |z − λi|: z is equidistant from -λi and λi.
With z = x + iy: x² + (y + λ)² = x² + (y − λ)² gives 4λy = 0, so y = 0 (λ > 0).
So z lies on the real axis, not on y = x: I is false.
z is real, so its amplitude is 0 or π, never π4: II is false.

Answer: (b) 4

Method. factorise the cubic by spotting the root z = i

Try z = i: i(i³) + i² - i + i = i(-i) − 1 = 1 − 1 = 0, so z − i is a factor.
Divide: iz³ + z² - z + i = (z − i)(iz² - 1).
So z = i or z² = 1i = -i, which gives |z|² = |-i| = 1.
Every root has |z| = 1, so (|z| + 1)² = 2² = 4.

Answer: (d) circles

Method. write zz̅ = x² + y² and z + z̅ = 2x

The condition becomes x² + y² = |2x| = 2|x|.
For x ≥ 0: x² - 2x + y² = 0, i.e. (x − 1)² + y² = 1.
For x ≤ 0: x² + 2x + y² = 0, i.e. (x + 1)² + y² = 1.
The locus is a pair of circles of radius 1 centred at (1, 0) and (-1, 0), touching at the origin.

Answer: (c) Circle

Method. write zz̅ = |z|² = x² + y² and z + z̅ = 2x

zz̅ + |z|² = 2(x² + y²), and 4(z + z̅) = 4.2x = 8x.
The equation becomes 2x² + 2y² + 8x − 48 = 0, i.e. x² + y² + 4x − 24 = 0.
Completing the square: (x + 2)² + y² = 28.
This is a circle with centre (-2, 0) and radius 2⁠√7.⁠

Answer: (b) 1 + ⁠√5⁠

Method. triangle inequality |z| − 4/⁠|z| ≤ |z − 4/⁠z| gives a quadratic bound on r = |z|

Let r = |z|. |z| − 4|z| ≤ |z − 4z| = 2, so r − 4r ≤ 2.
r² - 2r − 4 ≤ 0, whose roots are r = 1 ± ⁠√5;⁠ so r ≤ 1 + ⁠√5.⁠
It is attained: z = 1 + ⁠√5⁠ (real) satisfies z² - 2z − 4 = 0, i.e. z − 4z = 2.
Maximum |z| = 1 + ⁠√5.⁠

Answer: (c) rectangular hyperbola

Method. write z = x + iy and take the real part

z = x + iy: z² = x² - y² + 2ixy.
z² - i = (x² - y²) + i(2xy − 1), so Re(z² - i) = x² - y².
x² - y² = 2 is a hyperbola with equal semi-axes (a = b = ⁠√2):⁠ a rectangular hyperbola.

Answer: (c) 6

Method. |z + 1| is the distance from -1; triangle inequality over the disc |z + 4| ≤ 3

|z + 4| ≤ 3: z lies in the disc with centre -4 and radius 3.
|z + 1| = |(z + 4) − 3| ≤ |z + 4| + 3 ≤ 3 + 3 = 6.
Equality at z = -7 (on the circle, farthest from -1): |-7 + 1| = 6.
So the maximum is 6.

Shortcut. Max distance from a point to a disc = distance to the centre + radius = |-1 − (-4)| + 3 = 6.

Answer: (c) 4

Method. take moduli: |z|² = 2|z|, then z³ = 2zz̅

Taking moduli: |z|² = 2|z|, so |z| = 0 or |z| = 2.
|z| = 0 gives z = 0, which is a root.
|z| = 2: multiply z² = 2z̅ by z: z³ = 2zz̅ = 2|z|² = 8.
z³ = 8 gives z = 2, 2ω, 2ω²; each has |z| = 2 and satisfies z² = 8z = 8z̅|z|² = 2z̅.
Roots: 0, 2, 2ω, 2ω², four in all.