Ever wondered what happens when you take the square root of a negative number? Real numbers refuse to answer, so mathematicians invented a brand-new number called iota (i). A complex number packs a real part and an imaginary part into one tidy object z = a + ib. This chapter is short, formula-driven and a near-guaranteed source of 2–3 NDA marks every year.
Why Complex Numbers Exist
In the real number system, the equation x2 + 1 = 0 has no solution, because no real number squared gives −1. To fix this gap, we define a new quantity called iota, written as i, such that i2 = −1.
Once i exists, every quadratic equation suddenly has a solution. That single idea unlocks a whole number system used across the NDA syllabus — in quadratic equations, trigonometry and even rotation geometry.
i = √(−1), so i2 = −1. A complex number is written z = a + ib, where a is the real part Re(z) and b is the imaginary part Im(z). Note: b itself is real.
If b = 0, the number is purely real (like 5). If a = 0 and b ≠ 0, it is purely imaginary (like 3i). Every real number is therefore also a complex number with zero imaginary part. This is why we say the real numbers are a small part of the much larger family of complex numbers.
The whole set of complex numbers is denoted by the letter C. Unlike real numbers, complex numbers cannot be arranged on a single line from smallest to largest, so phrases like greater than or less than have no meaning for them. We can only compare their moduli, which you will meet shortly. For the NDA exam, the most important habit is to always keep the real and imaginary parts neatly separated while you work.
Powers of Iota
The powers of i repeat in a cycle of four. This is one of the most frequently tested micro-facts in NDA.
- i1 = i
- i2 = −1
- i3 = −i
- i4 = 1
After that the pattern restarts. To find any high power, divide the exponent by 4 and keep only the remainder.
To evaluate in, compute n mod 4. If the remainder is 0→1, 1→i, 2→−1, 3→−i. Example: i50 has remainder 2, so i50 = −1.
√(−a) × √(−b) is not √(ab). The rule √x × √y = √(xy) fails for negatives. Always convert to i first: √(−4) × √(−9) = 2i × 3i = 6i2 = −6, not +6.
Adding, Multiplying and Dividing
Treat i like an ordinary algebra symbol, then replace i2 with −1 at the end.
Addition and subtraction
Add real parts to real parts and imaginary parts to imaginary parts:
(a + ib) + (c + id) = (a + c) + i(b + d).
Multiplication
(a + ib)(c + id) = (ac − bd) + i(ad + bc).
Division
To divide, multiply numerator and denominator by the conjugate of the denominator. This clears i from the bottom.
The conjugate of z = a + ib is z̅ = a − ib. A number times its conjugate is always real: z · z̅ = a2 + b2.
The reason this trick works is simple. When we multiply by the conjugate, the cross terms involving i cancel out, leaving only a real denominator. That makes the division clean and lets us read off the real and imaginary parts directly. Master this single step and division of complex numbers will never trouble you in the exam hall.
Conjugate and Its Properties
The conjugate flips the sign of the imaginary part only. Geometrically, it is the mirror image of z across the real (x) axis.
- Conjugate of a sum = sum of conjugates
- Conjugate of a product = product of conjugates
- z + z̅ = 2 Re(z) — always real
- z − z̅ = 2i Im(z) — always imaginary
A complex number is purely real when z = z̅, and purely imaginary when z = −z̅. NDA loves to test these two quick checks.
Geometrically, taking a conjugate is the same as reflecting the point across the horizontal axis, so the distance from the origin stays exactly the same. That is why a number and its conjugate always share the same modulus. Keeping this picture in mind helps you answer many short questions almost instantly, without grinding through algebra.
Modulus and the Argand Plane
We can plot z = a + ib as the point (a, b) on a graph called the Argand plane, where the x-axis is real and the y-axis is imaginary.
The modulus |z| is the distance of this point from the origin.
|z| = √(a2 + b2). Also |z|2 = z · z̅. The modulus is always ≥ 0 and is zero only when z = 0.
Useful modulus rules: |z1z2| = |z1||z2| and |z1/z2| = |z1| ÷ |z2|. The modulus of a number and its conjugate are equal: |z| = |z̅|.
The Argand plane turns abstract algebra into easy pictures. Adding two complex numbers becomes the same as adding two arrows, and the modulus becomes a simple length you could measure with a ruler. Many tricky-looking NDA questions become obvious once you sketch the points. Because of this geometric view, a complex number is often thought of as a point or as a directed arrow from the origin, and both ideas are useful depending on the problem.
Argument and Polar Form
The argument (or amplitude) is the angle θ that the line joining the origin to z makes with the positive real axis.
tanθ = b/a. The polar form is z = r(cosθ + i sinθ), where r = |z| and θ = arg(z). The principal argument lies in (−π, π].
You must check the quadrant of (a, b) before finalising θ, because tanθ alone cannot tell which quadrant you are in.
Students compute θ = tan−1(b/a) and stop. For z = −1 + i, the point is in the second quadrant, so the argument is 3π/4, not −π/4. Always sketch the point first.
The compact Euler form writes this as z = reiθ, since eiθ = cosθ + i sinθ.
De Moivre's Theorem
This theorem makes raising a complex number to a power simple once it is in polar form.
(cosθ + i sinθ)n = cos(nθ) + i sin(nθ), for any integer n. In Euler form: (reiθ)n = rneinθ.
So you multiply the angle by n and raise the modulus to the power n. It is far faster than multiplying out brackets. Imagine being asked to raise a complex number to the tenth power by hand. Expanding the bracket would take a full page and invite errors, but De Moivre's theorem finishes it in two short lines. This is exactly why examiners reward students who recognise when polar form is the smart choice.
To find the nth roots of a complex number, use modulus r1/n and angles (θ + 2kπ)/n for k = 0, 1, …, n−1. This always gives exactly n distinct roots.
Cube Roots of Unity
The solutions of x3 = 1 are called the cube roots of unity. They appear again and again in NDA papers, so memorise their properties.
The three cube roots are 1, ω and ω2, where ω = (−1 + i√3)/2. Their key properties:
- 1 + ω + ω2 = 0
- ω3 = 1 (so ω4 = ω, and so on)
- ω and ω2 are conjugates of each other
Whenever you see ω raised to a high power, reduce the exponent using ω3 = 1, exactly like the powers-of-iota trick. For example, ω100 becomes ω1 because 100 leaves remainder 1 when divided by 3. This single shortcut turns frightening-looking expressions into one-step answers, which is exactly the kind of speed the NDA exam rewards under time pressure.
The three cube roots of unity sit at the corners of an equilateral triangle inscribed in a unit circle on the Argand plane.
Worked Example
Let us put modulus, conjugate and division together in one typical problem.
Express z = (3 + 2i) / (1 − i) in the form a + ib, and find |z|.
So Re(z) = 1/2, Im(z) = 5/2 and |z| = √26 ÷ 2. Notice the rhythm of the solution: spot the conjugate, clear the denominator, simplify the numerator, then read off the parts. If you practise five problems with this same four-step rhythm, division will feel completely automatic in the real exam.
Mistakes That Cost Marks
Complex numbers are easy, but careless slips lose marks fast. Watch for these.
Forgetting that i2 = −1 when simplifying. The sign flip from i2 is the most common arithmetic error in division problems.
- Writing √(−a) × √(−b) = √(ab) — wrong, as shown earlier.
- Ignoring the quadrant while finding the argument.
- Treating |z| like a real square root that can be negative — modulus is never negative.
- Mixing up z + z̅ (real) with z − z̅ (imaginary).
Previous-Year Style Question
Here is a question in the exact style NDA uses for this chapter.
Q. If ω is a complex cube root of unity, then the value of (1 + ω)(1 + ω2)(1 + ω4)(1 + ω8) is:
(a) 0 (b) 1 (c) −1 (d) 4
Answer: (b) 1. Since 1 + ω + ω2 = 0, we have 1 + ω = −ω2 and 1 + ω2 = −ω. Also ω4 = ω and ω8 = ω2, so 1 + ω4 = −ω2 and 1 + ω8 = −ω. The product is (−ω2)(−ω)(−ω2)(−ω) = ω6 = (ω3)2 = 1.
Quick Revision
Run through this list the night before the exam.
- i2 = −1; powers of i repeat every 4.
- z = a + ib; conjugate z̅ = a − ib.
- |z| = √(a2 + b2) and z·z̅ = |z|2.
- Polar form: z = r(cosθ + i sinθ) = reiθ.
- De Moivre: (cosθ + i sinθ)n = cosnθ + i sinnθ.
- Cube roots of unity: 1 + ω + ω2 = 0 and ω3 = 1.
Master these six lines and the entire complex-numbers section of your NDA paper becomes routine scoring. At The Cavalier, our defence aspirants treat this chapter as a guaranteed-marks zone, because the formulas are few, the patterns repeat year after year, and a little daily practice locks them into memory. Spend ten focused minutes a day for a week and you will rarely lose a single mark here again.
Frequently asked questions
What exactly is iota (i) in complex numbers?
Iota is the imaginary unit defined as the square root of −1, so i² = −1. It lets us write square roots of negative numbers and forms the imaginary part of any complex number.
How do I find the modulus of a complex number?
For z = a + ib, the modulus is |z| = √(a² + b²), which is its distance from the origin on the Argand plane. The modulus is always zero or positive, never negative.
What is the difference between argument and modulus?
The modulus is the length (distance from the origin), while the argument is the angle θ the number makes with the positive real axis, given by tanθ = b/a. Together they give the polar form of z.
Why are cube roots of unity important for NDA?
Questions on ω appear almost every year. The two facts 1 + ω + ω² = 0 and ω³ = 1 let you simplify long expressions in seconds, making them quick guaranteed marks.
How many marks does Complex Numbers carry in NDA Maths?
Typically 2 to 3 questions appear from this chapter in each NDA Maths paper. Because the topic is short and formula-driven, it offers an excellent marks-per-effort ratio for serious aspirants.
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