Complex numbers for NDA

Complex numbers look abstract at first, but NDA questions on them are among the most mechanical in the paper. A few rules, applied cleanly, answer almost everything: the cycle of powers of i, the conjugate trick for division, the quadrant rule for the argument and two identities for ω. Candidates lose marks here through small slips, such as a wrong quadrant or a sign in i², rather than through hard ideas.

Typical question types are:

• simplifying powers and sums of powers of i;
• modulus, conjugate and argument of a given number;
• dividing or simplifying a fraction such as (1 + i)/(1 − i);
• the square root of a complex number;
• expressions in ω, the cube root of unity;
• the locus of z from a condition on its modulus.

The basics

Write a complex number as z = a + ib, where a = Re(z) and b = Im(z) are real and i² = −1.

Powers of i repeat in a cycle of four: i¹ = i, i² = −1, i³ = −i, i⁴ = 1. To find iⁿ, divide n by 4 and use the remainder (remainder 0 means iⁿ = 1). A useful consequence: any four consecutive powers of i add up to 0, since i + (−1) + (−i) + 1 = 0.

Conjugate and modulus. For z = a + ib:

• conjugate z̄ = a − ib;
• modulus, written |z|, equals √(a² + b²), the distance of z from the origin;
• z × z̄ = a² + b², the square of the modulus;
• z + z̄ = 2a, and z − z̄ = 2ib.

Division. Multiply the numerator and the denominator by the conjugate of the denominator; the denominator becomes real.

Rules for the modulus. The modulus of a product is the product of the moduli, and the modulus of a quotient is the quotient of the moduli. For a sum there is only an inequality: the modulus of (z₁ + z₂) is at most the sum of the two moduli (the triangle inequality).

Argument and polar form

The argument θ is the angle z makes with the positive real axis. Take the basic angle α = tan⁻¹(b/a) using positive values, then place it in the right quadrant. The principal argument lies in (−π, π].

The polar form is z = r(cos θ + i sin θ), where r is the modulus. When two numbers are multiplied, the moduli multiply and the arguments add (adjusting by 2π to stay in the principal range).

De Moivre's theorem: for any integer n, (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. It makes high powers easy once a number is in polar form.

Square roots of a complex number

To find √(a + ib), set it equal to x + iy and square:

• x² − y² = a and 2xy = b;
• also x² + y² equals the modulus of a + ib.

Adding and subtracting the first and third equations gives x² and y². The sign of b tells you whether x and y have the same sign (b > 0) or opposite signs (b < 0).

Cube roots of unity

Solving z³ = 1 means (z − 1)(z² + z + 1) = 0, so the roots are 1 and the two roots of z² + z + 1 = 0:

• ω = (−1 + i√3)/2 and ω² = (−1 − i√3)/2.

Two identities follow and solve nearly every question:

• ω³ = 1, so powers of ω repeat in a cycle of three;
• 1 + ω + ω² = 0, because the sum of the roots of z² + z + 1 = 0 is −1.

Useful rearrangements: 1 + ω = −ω², 1 + ω² = −ω and ω + ω² = −1. Also, ω² is the conjugate of ω, and both have modulus 1.

Loci in brief

If z = x + iy, a condition on z becomes an equation in x and y:

• the modulus of z equals r: a circle of radius r centred at the origin;
• the modulus of (z − z₁) equals the modulus of (z − z₂): the perpendicular bisector of the segment joining z₁ and z₂.

Worked NDA-style MCQs

Q1. The value of i + i² + i³ + … + i¹⁰² is:
(a) 0 (b) i (c) −1 + i (d) 1 + i

The first 100 terms form 25 groups of four consecutive powers, each adding to 0. What remains is i¹⁰¹ + i¹⁰² = i + (−1). Answer: (c).

Q2. The smallest positive integer n for which ((1 + i)/(1 − i))ⁿ = 1 is:
(a) 1 (b) 2 (c) 4 (d) 8

(1 + i)/(1 − i) = (1 + i)² ÷ (1 − i)(1 + i) = (1 + 2i − 1) ÷ 2 = i. So we need iⁿ = 1, and the smallest such n is 4. Answer: (c).

Q3. The modulus and principal argument of −1 + i√3 are:
(a) 2, π/3 (b) 2, 2π/3 (c) 4, 2π/3 (d) 2, −π/3

Modulus = √(1 + 3) = 2. The basic angle is tan⁻¹(√3) = π/3, and the point is in the second quadrant, so the argument is π − π/3 = 2π/3. Answer: (b).

Q4. A square root of −8 − 6i is:
(a) 1 + 3i (b) 3 − i (c) 1 − 3i (d) 3 + i

Let (x + iy)² = −8 − 6i. Then x² − y² = −8, and x² + y² = √(64 + 36) = 10. Adding, 2x² = 2, so x² = 1 and y² = 9. Since 2xy = −6 is negative, x and y have opposite signs: ±(1 − 3i). Check: (1 − 3i)² = 1 − 6i − 9 = −8 − 6i. Answer: (c).

Q5. If ω is a non-real cube root of unity, (1 + ω − ω²)(1 − ω + ω²) equals:
(a) 0 (b) 1 (c) 4 (d) −4

1 + ω = −ω², so the first bracket is −2ω². 1 + ω² = −ω, so the second is −2ω. The product is 4ω³ = 4. Answer: (c).

Q6. The locus of z satisfying |z − 1| = |z − i| is:
(a) the line y = x (b) the line y = −x (c) a circle (d) the real axis

With z = x + iy: (x − 1)² + y² = x² + (y − 1)². Expanding and cancelling gives −2x = −2y, so y = x. It is the perpendicular bisector of the segment joining 1 and i. Answer: (a).

Common mistakes

• Writing i² = 1 in the middle of a long product. Replace i² by −1 at once.
• Ignoring the quadrant when finding the argument.
• Treating the modulus of a sum as the sum of the moduli.
• Reducing powers of ω by 4 out of habit from powers of i. Powers of ω cycle every 3.

Practice set

1. i⁵⁷ + 1/i¹²⁵ equals: (a) 0 (b) 2i (c) −2i (d) 2
2. The modulus of (3 + 4i)/(5 − 12i) is: (a) 5/13 (b) 13/5 (c) 1 (d) 7/17
3. The conjugate of 1/(2 + i) is: (a) (2 − i)/5 (b) (2 + i)/5 (c) 2 − i (d) (2 + i)/3
4. The principal argument of −√3 + i is: (a) π/6 (b) −π/6 (c) 5π/6 (d) 2π/3
5. ω¹⁰⁰ + ω²⁰⁰ + 1 equals: (a) 0 (b) 1 (c) 3 (d) −1
6. (1 + ω)³ equals: (a) 1 (b) −1 (c) ω (d) 0
7. (1 + i)⁸ equals: (a) 8 (b) 16i (c) 16 (d) −16
8. (cos 15° + i sin 15°)⁴ equals: (a) 1/2 + i√3/2 (b) √3/2 + i/2 (c) i (d) −1/2 + i√3/2

Answers:

1. (a) 0. 57 leaves remainder 1, so i⁵⁷ = i. 125 leaves remainder 1, so 1/i¹²⁵ = 1/i = −i. The sum is 0.
2. (a) 5/13. The modulus of a quotient is 5 ÷ 13.
3. (b). 1/(2 + i) = (2 − i)/5, whose conjugate is (2 + i)/5.
4. (c) 5π/6. Basic angle π/6, second quadrant.
5. (a) 0. ω¹⁰⁰ = ω (100 = 99 + 1) and ω²⁰⁰ = ω² (200 = 198 + 2), so the sum is 1 + ω + ω².
6. (b) −1. (−ω²)³ = −ω⁶ = −1.
7. (c) 16. (1 + i)² = 2i, and (2i)⁴ = 16i⁴ = 16.
8. (a). By De Moivre, cos 60° + i sin 60°.

What to do next

• Memorise the quadrant table and the three ω rearrangements.
• Solve 25 questions from old NDA papers on this chapter, timed at 60 seconds each.
• Complex roots appear again in quadratic equations; De Moivre ties in with trigonometric identities.