📘 Algebra · Chapter MN02🎯 NDA Level : High Priority
Complex numbers extend the real number system to solve equations like x² + 1 = 0 — which have no real solution. For NDA Mathematics, this chapter is consistently tested every year. Questions are concept-heavy but calculation-light, making it ideal for quick high-yield preparation. The three pillars are: powers of iota, modulus/argument/conjugate, and cube roots of unity.
📌 What to expect in NDA (based on 2022–2024 papers): (1) Finding i^n for large n using the cycle of 4; (2) Modulus |z| = √(a²+b²); (3) Conjugate and its properties; (4) Simplifying expressions using 1+ω+ω² = 0 and ω³ = 1; (5) Value of (1+ω)^n or (1+ω²)^n; (6) Real and imaginary parts of a given complex expression; (7) Argument of a complex number; (8) Square root of a complex number (less frequent but tested).
Topics at a Glance
① Iota & Powers
i, i², i³, i⁴ — cycle of 4, finding i^n
② Cartesian Form
z = a + ib, real part, imaginary part
③ Modulus
|z| = √(a²+b²), properties
④ Argument & Conjugate
θ = arg(z), z̄ = a − ib
⑤ Square Root of z
√(a+ib) — method via x+iy
⑥ Cube Roots of Unity
ω, ω², 1+ω+ω²=0, ω³=1
1. Iota (i) — The Imaginary Unit & Its Powers
1.1
Definition & the Cycle of Four
Highest frequency topic — asked almost every NDA paper
The imaginary unit i (iota) is defined as i = √(−1), so i² = −1. This single definition generates an endlessly repeating 4-step cycle when you raise i to successive positive integer powers.
i¹
= i
i²
= −1
i³
= −i
i⁴
= 1
After i⁴ = 1, the cycle restarts: i⁵ = i, i⁶ = −1, and so on. The pattern has a period of 4.
⚡ Finding i^n for Any Positive Integer n
Step 1: Divide n by 4 and find the remainder r = n mod 4
Step 2: i^n = i^r
r = 0 → i^n = 1 (e.g., i^⁴⁰ = 1)
r = 1 → i^n = i (e.g., i^⁴¹ = i)
r = 2 → i^n = −1 (e.g., i^⁴² = −1)
r = 3 → i^n = −i (e.g., i^⁴³ = −i)
Special: i^0 = 1. For negative powers: i^(−1) = −i, i^(−2) = −1, i^(−3) = i, i^(−4) = 1.
Worked Example — Large Power of i
Find i^⁹⁹⁷.
997 ÷ 4 = 249 remainder 1. So i^⁹⁹⁷ = i^¹ = i.
Find i^⁻³.
i^(−3) = 1/i³ = 1/(−i) = −(1/i) = −(i/i²) = −(i/−1) = i. Or: i^(−3) = i^(4−3) = i^1 = i. (Add 4 until positive.)
⚠ Common Traps with Iota:
• i^4k = 1 for any integer k (including k=0). Students sometimes write i^4 = i — wrong.
• √(−a) = i√a only if a > 0. √(−4) = 2i, NOT −2i or √4·√(−1).
• i² = −1 (not +1). This is the most basic and most often misapplied fact.
📌 Sum of consecutive powers: i^n + i^(n+1) + i^(n+2) + i^(n+3) = 0 always (sum of one full cycle = 0). This is sometimes asked directly.
📋 TOPIC-WISE PYQ
Powers of Iota — NDA-Pattern Questions
Q1. The value of i⁵⁷ + i⁵⁸ + i⁵⁹ + i⁶⁰ is:
(a) 0 (b) i (c) −1 (d) 1
Answer: (a) 0
Sum of any 4 consecutive powers of i is always 0. Here remainders of 57,58,59,60 mod 4 are 1,2,3,0 → i + (−1) + (−i) + 1 = 0.
Q2. What is the value of i^(4n+3) where n is any positive integer?
(a) i (b) −i (c) −1 (d) 1
Answer: (b) −i
4n+3 divided by 4 always gives remainder 3. So i^(4n+3) = i³ = −i.
Q3. The value of i^(−999) is:
(a) i (b) −1 (c) −i (d) 1
Answer: (a) i
i^(−999) = i^(−999 + 1000) = i^1 = i. (Add the smallest multiple of 4 that makes the power non-negative: 999 = 4×249+3, so −999 ≡ 1 mod 4 → i^1 = i.)
🔥 TRICKY QUESTIONS
Iota — Classic NDA Traps
🧩 T1. Find the value of i¹ + i² + i³ + ... + i¹⁰⁰.
Solution: 0.
100 is divisible by 4, so there are exactly 25 complete cycles. Each cycle (i¹+i²+i³+i⁴) = i+(−1)+(−i)+1 = 0. Total = 25 × 0 = 0. Trap: Trying to add term by term instead of grouping in cycles of 4.
🧩 T2. Is √(−1) × √(−1) = √(−1 × −1) = √1 = 1?
Solution: NO. The answer is −1, not 1.
√(−1) × √(−1) = i × i = i² = −1. The rule √a × √b = √(ab) only holds when a, b ≥ 0. It CANNOT be applied to negative numbers. This is a conceptual trap that appears in statement-based NDA questions.
2. Cartesian Form, Modulus, Argument & Conjugate
2.1
Cartesian Form — a + ib
The standard representation — everything else builds on this
Every complex number z is written as z = a + ib, where a and b are real numbers.
Real & Imaginary Parts
Re(z) = a — the real part
Im(z) = b — the imaginary part (b is real!)
z is purely real if b = 0 (e.g., z = 5)
z is purely imaginary if a = 0 (e.g., z = 3i)
z = 0 iff a = 0 AND b = 0 simultaneously
Equality of Complex Numbers
If a + ib = c + id, then a = c AND b = d
Two complex numbers are equal iff both real parts and imaginary parts match
No concept of "greater than" or "less than" between complex numbers
Only equality/inequality of moduli can be compared
For NDA: The triangle inequality is frequently tested in statement-based questions. |z|² = z·z̄ is very useful in division problems.
Examples of Modulus
|3 + 4i| = √(9+16) = √25 = 5
|1 + i| = √(1+1) = √2
|−5| = 5 (purely real: b=0)
|7i| = 7 (purely imaginary: a=0)
|0| = 0
On the Argand Plane
|z| = distance from origin O to point P(a,b)
|z₁ − z₂| = distance between points P₁ and P₂
All z with |z| = r lie on a circle radius r centred at O
|z| = 1 → unit circle
2.3
Argument — Angle in the Argand Plane
θ = arg(z) — principal value lies in (−π, π]
The argument of z = a + ib is the angle θ that the line OP makes with the positive real axis, measured anticlockwise. The principal argument is the unique value in (−π, π].
Properties tested in both standalone MCQs and within larger problems
⚡ Conjugate Definition & Properties
If z = a + ib, then z̄ (conjugate) = a − ib
z + z̄ = 2a = 2·Re(z) (purely real)
z − z̄ = 2ib = 2i·Im(z) (purely imaginary)
z · z̄ = a² + b² = |z|² (always real and positive)
(z̄)̄ = z (double conjugate returns original)
z̄₁ + z̄₂ = (z₁+z₂)̄ (conjugate of sum = sum of conjugates)
z̄₁ · z̄₂ = (z₁·z₂)̄ (conjugate of product = product of conjugates)
z is purely real ⟺ z = z̄. z is purely imaginary ⟺ z = −z̄. These two conditions are directly tested.
Argand Plane — Geometric View
Fig 1: Argand Plane — z and its conjugate z̄ as mirror images across the real axis; |z| is the distance OP; θ is the argument.
🧩 T3. If |z₁| = |z₂|, does it follow that z₁ = z₂?
NO. Equal moduli only means both points are the same distance from the origin (lie on the same circle). For example, |3+4i| = |3−4i| = 5, but 3+4i ≠ 3−4i. Trap: Students assume equal modulus means equal complex numbers. This is a statement-verification question in NDA.
🧩 T4. z is purely imaginary. Then z + z̄ = ?
0. z + z̄ = 2·Re(z) = 2a. If z is purely imaginary, a = 0. So z + z̄ = 0. Corollary: z is purely real ⟺ z = z̄. z is purely imaginary ⟺ z = −z̄. Both conditions appear as direct MCQs.
🧩 T5. |z + z̄| = ? vs |z| + |z̄|
z + z̄ = 2a (purely real), so |z + z̄| = 2|a|. |z| + |z̄| = 2|z| (since |z| = |z̄| = √(a²+b²)).
These are equal only when b = 0 (z is real). In general, 2|a| ≤ 2√(a²+b²). This tests Triangle Inequality + conjugate properties together.
3. Square Root of a Complex Number
3.1
Method: Let √(a + ib) = x + iy
Set up two equations — solve systematically
To find √(a + ib), assume it equals x + iy where x, y are real. Squaring both sides and comparing real and imaginary parts gives a solvable system.
⚡ Square Root Method — Step by Step
Let √(a + ib) = x + iy
Square both sides: a + ib = (x² − y²) + i(2xy)
Comparing parts:
x² − y² = a ...(1)
2xy = b ...(2)
Also use: x² + y² = √(a² + b²) = |z| ...(3)
From (1) and (3):
x² = (|z| + a)/2 → x = ±√[(|z| + a)/2]
y² = (|z| − a)/2 → y = ±√[(|z| − a)/2]
Sign of y is determined by equation (2): 2xy = b
If b > 0: x and y have same sign
If b < 0: x and y have opposite signs
There are always two square roots: +( x + iy) and −(x + iy). Both are valid.
Worked Example — Square Root
Find √(3 + 4i).
Here a = 3, b = 4. |z| = √(9+16) = 5.
x² = (5+3)/2 = 4 → x = ±2.
y² = (5−3)/2 = 1 → y = ±1.
Since b = 4 > 0, x and y same sign → x = 2, y = 1 or x = −2, y = −1. √(3+4i) = 2 + i or −(2+i) = −2−i.
⚠ Exam Note: The square root of a complex number always gives two values (just like square roots of real numbers). If the question asks for "the square root", both ±(x+iy) are answers. NDA occasionally gives this as a 4-option MCQ — choose the correct pair.
📋 TOPIC-WISE PYQ
Square Root of Complex Number — NDA-Pattern Questions
Key: ω and ω² are complex conjugates of each other: ω̄ = ω².
📌 Memory Aid: ω = "half of (−1 + i√3)" — remember the √3 is under i, not a². The three cube roots of unity are 1, ω, ω². They are equally spaced at 120° apart on the unit circle.
4.2
Properties of ω — The Most Tested Results
Memorise ALL properties — they are used across many NDA questions
⚡ Core Properties of Cube Roots of Unity
P1. ω³ = 1 (and by extension ω⁶ = 1, ω⁹ = 1, ...)
P2. 1 + ω + ω² = 0 ← most important
P3. ω² = 1/ω (equivalently: ω · ω² = ω³ = 1)
P4. ω̄ = ω² (they are conjugates)
P5. |ω| = |ω²| = 1 (both lie on unit circle)
P6. (ω)^(3k) = 1 for any integer k
P7. ω^n repeats with period 3: check n mod 3
P2 is the single most tested property. Every NDA set on ω eventually uses 1+ω+ω²=0 somewhere in the solution.
Derived Results
ω + ω² = −1
ω · ω² = 1 (product)
(1+ω) = −ω²
(1+ω²) = −ω
ω² + 1/ω = 0
Powers of ω (cycle 3)
ω⁰ = 1, ω¹ = ω, ω² = ω²
ω³ = 1, ω⁴ = ω, ω⁵ = ω²
ω^n → find n mod 3
mod 3 = 0 → 1
mod 3 = 1 → ω
mod 3 = 2 → ω²
Key Expressions
(1+ω)³ = (−ω²)³ = −ω⁶ = −1
(1+ω²)³ = (−ω)³ = −ω³ = −1
ω² + ω + 1 = 0 (same as P2)
a+b+c=0 ⟹ a³+b³+c³=3abc
1³+ω³+(ω²)³ = 3·1·ω·ω²= 3
Argand Diagram — Three Cube Roots of Unity
Fig 2: Argand Plane — Three cube roots of unity (1, ω, ω²) on the unit circle, each 120° apart. The centroid of the triangle they form is the origin.
📋 TOPIC-WISE PYQ
Cube Roots of Unity — NDA-Pattern Questions
Q9. If ω is a complex cube root of unity, then 1 + ω + ω² = ?
(a) 1 (b) −1 (c) 0 (d) i
Answer: (c) 0
This is the fundamental property. 1+ω+ω² = 0 always, when ω is a non-real cube root of unity.
Q10. The value of (1 + ω)³ is:
(a) 1 (b) ω (c) −ω² (d) −1
Answer: (d) −1
Since 1+ω+ω²=0, we get 1+ω = −ω². So (1+ω)³ = (−ω²)³ = −(ω²)³ = −(ω³)² = −(1)² = −1.
Q11. If ω is a cube root of unity, then ω^100 + ω^200 + ω^300 = ?
(a) −1 (b) 0 (c) 3 (d) 1
Answer: (c) 3
100 = 3×33+1 → ω^100 = ω¹ = ω. 200 = 3×66+2 → ω^200 = ω². 300 = 3×100 → ω^300 = ω^0 = 1.
Wait — re-check: 300 ÷ 3 = 100 exactly, remainder 0 → ω^300 = (ω³)^100 = 1^100 = 1. Hmm — but also ω^100 = ω¹ and ω^200 = ω². So sum = ω + ω² + 1 = 0. (Re-check option: (b) 0 is the answer.) Corrected Answer: (b) 0. Sum = ω + ω² + 1 = 0 by the fundamental property.
Q12. If 1, ω, ω² are cube roots of unity, the value of (2+3ω+2ω²)² is:
(a) 9ω (b) 9ω² (c) ω (d) ω²
Answer: (b) 9ω²
Use 1+ω+ω²=0 → ω+ω²=−1. Rewrite: 2+3ω+2ω² = 2(1+ω²)+3ω = 2(−ω)+3ω [since 1+ω²=−ω] = −2ω+3ω = ω. Hmm — actually: 2+3ω+2ω² = 2+2ω²+3ω = 2(1+ω²)+3ω. Since 1+ω² = −ω → 2(−ω)+3ω = ω. So (ω)² = ω². Answer: (d) ω². Strategy: Always group terms using 1+ω+ω²=0 to simplify the bracket before squaring.
🔥 TRICKY QUESTIONS
Cube Roots of Unity — High-Value Traps
🧩 T6. Show that 1³ + ω³ + (ω²)³ = 3. But what is 1 + ω³ + ω⁶?
1 + ω³ + ω⁶ = 1 + 1 + 1 = 3.
ω³ = 1 and ω⁶ = (ω³)² = 1² = 1. So the sum = 3. Trap: Students write ω³ = ω and ω⁶ = ω². Wrong! ω³ = 1, not ω. The cube root cycle is 3, so any multiple of 3 in the exponent gives 1.
🧩 T7. If ω is a cube root of unity, evaluate (1 − ω + ω²)⁴ + (1 + ω − ω²)⁴.
🧩 T8. If α, β are complex cube roots of unity, what is αβ?
αβ = ω · ω² = ω³ = 1.
The two non-real cube roots of unity are ω and ω². Their product is always 1 (since ω³ = 1).
Their sum ω + ω² = −1. These are the Vieta's formulas for x² + x + 1 = 0: sum of roots = −1, product of roots = 1.
📋 Master Formula Sheet — MN02 Complex Numbers
All high-yield formulae in one place for rapid pre-exam revision.