Why this is a geometry chapter
Consider the condition |z − 1| = |z + 1|. Algebraically you would square both sides, expand, and simplify. Geometrically it says the point z is equidistant from 1 and −1, so it lies on the perpendicular bisector of the segment joining them - the imaginary axis. One line, no algebra.
The same shift applies throughout. |z| = 2 is a circle, arg(z) = π/4 is a ray, |z − z₁| + |z − z₂| = k is an ellipse. JEE constructs questions knowing that students who manipulate will spend three minutes where students who visualise spend twenty seconds.
How JEE actually asks Complex Numbers
NTA does not publish chapter-wise weightage, so any figure online is a coaching estimate from past papers; confirm your session's syllabus and pattern in the official NTA information bulletin.
Three families recur. Algebraic manipulation - powers of i, conjugates, division - supplies the easiest marks. De Moivre's theorem and the roots of unity supply short, highly patterned questions. Argand-plane loci and rotation supply the harder ones, and those are where the geometric reading pays off most.
Key concepts, compressed
- ✓A complex number z = x + iy corresponds to the point (x, y), so the whole plane becomes available as a tool.
- ✓The modulus |z| is the distance from the origin; the argument is the angle from the positive real axis.
- ✓The conjugate reflects a point in the real axis, and z times its conjugate gives |z|², which is always real.
- ✓Polar form z = r(cos θ + i sin θ) = re^(iθ) turns multiplication into multiplying moduli and adding arguments.
- ✓De Moivre's theorem makes powers easy: raise the modulus to the power and multiply the argument by it.
- ✓The n-th roots of unity are equally spaced on the unit circle and always sum to zero.
Results you need before attempting the questions
| Quantity | Result | Note |
|---|---|---|
| Powers of i | i² = −1, i³ = −i, i⁴ = 1 | cycle of 4 |
| Modulus | |x + iy| = √(x² + y²) | |
| Conjugate product | z · z̄ = |z|² | always real |
| Division | 1/z = z̄/|z|² | rationalise by the conjugate |
| Polar form | z = r(cos θ + i sin θ) = re^(iθ) | |
| Product of moduli | |z₁z₂| = |z₁||z₂| | |
| Sum of arguments | arg(z₁z₂) = arg z₁ + arg z₂ | |
| De Moivre | (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ | |
| Cube roots of unity | 1 + ω + ω² = 0, ω³ = 1 | |
| Triangle inequality | |z₁ + z₂| ≤ |z₁| + |z₂| | equality when args match |
| Rotation | multiply by e^(iθ) | by i means 90° |
| Circle | |z − a| = r | centre a, radius r |
The five mistakes that cost the most marks
- ✓Taking the argument as arctan(y/x) without checking the quadrant. The signs of x and y decide which of two candidate angles is correct.
- ✓Assuming |z₁ + z₂| = |z₁| + |z₂|. Equality holds only when the two have the same argument; otherwise it is a strict inequality.
- ✓Applying De Moivre's theorem to a fractional power and keeping only one value. Fractional powers have multiple roots.
- ✓Forgetting 1 + ω + ω² = 0. Nearly every roots-of-unity question is a direct application of it together with ω³ = 1.
- ✓Writing inequalities between complex numbers. There is no ordering on the complex field, so only real quantities such as moduli can be compared.
Practice set 1: algebra of complex numbers
1. Evaluate i², i³ and i⁴.
i² = −1, i³ = −i, i⁴ = 1. The powers cycle with period four, which is the whole content of the next question. Note that i³ = i²·i = −i, and i⁴ = (i²)² = 1 - deriving them rather than memorising them means you cannot get the signs wrong.
2. Evaluate i⁵⁰.
−1. Divide the exponent by 4 and keep the remainder: 50 = 4(12) + 2, so i⁵⁰ = i² = −1. Any power of i reduces this way, which is why the only thing to remember is the four-step cycle. For a negative exponent, add multiples of 4 until the exponent is non-negative.
3. Compute z₁ + z₂ where z₁ = 2 + 3i and z₂ = 1 − i.
3 + 2i. Add real parts and imaginary parts separately: (2 + 1) + (3 − 1)i = 3 + 2i. Geometrically this is vector addition in the Argand plane, which is why the triangle inequality for complex numbers is the same statement as the triangle inequality for vectors.
4. Compute z₁z₂ for the same numbers.
5 + i. Expand: (2 + 3i)(1 − i) = 2 − 2i + 3i − 3i². Since i² = −1, the last term is +3, giving 2 + 3 + i = 5 + i. The most common error is treating −3i² as −3 - the two minus signs make it positive.
5. What is the conjugate of 3 + 4i, and what does conjugation do geometrically?
3 − 4i, and geometrically it reflects the point in the real axis. Conjugation negates the imaginary part, leaves the modulus unchanged and negates the argument. It is the operation that makes division possible, as question 7 shows.
6. Compute (3 + 4i)(3 − 4i).
25. Expanding gives 9 − 12i + 12i − 16i² = 9 + 16 = 25. This is the general result z·z̄ = |z|², and it is always a non-negative real number. It is the reason multiplying by the conjugate clears i from a denominator.
7. Express 1/(3 + 4i) in the form a + ib.
(3 − 4i)/25, that is 0.12 − 0.16i. Multiply numerator and denominator by the conjugate 3 − 4i: the denominator becomes |3 + 4i|² = 25 by question 6, and the numerator becomes 3 − 4i. Dividing complex numbers is always this single move - rationalise by the conjugate of the denominator.
Practice this now
Practice set 2: modulus and argument
8. Find |3 + 4i|.
5. |x + iy| = √(x² + y²) = √(9 + 16) = √25 = 5. This is the distance from the origin to the point (3, 4), and the 3-4-5 triangle is why the number comes out whole. The modulus is always a non-negative real number, whatever the complex number.
9. Find |1 + i|.
√2, approximately 1.414. Here x = y = 1, so the modulus is √(1 + 1) = √2. This particular value appears constantly because 1 + i sits at 45° on the unit diagonal, making it the standard example for polar form and De Moivre.
10. Find the argument of 1 + i.
π/4, or 45°. Both the real and imaginary parts are positive, so the point lies in the first quadrant, and arctan(1/1) = π/4 is therefore correct without adjustment. This is the case where naive arctan happens to work - the next question is the case where it does not.
11. Find the argument of −1 + i.
3π/4, not −π/4. Computing arctan(1/(−1)) = arctan(−1) gives −π/4 on a calculator, but the point (−1, 1) lies in the second quadrant where the argument must be between π/2 and π. Adding π gives 3π/4. Always locate the quadrant from the signs before trusting an arctangent.
12. Write 1 + i in polar form.
√2(cos π/4 + i sin π/4), equivalently √2 e^(iπ/4). Combine the modulus from question 9 with the argument from question 10. Polar form is what makes powers and roots tractable, so converting early is usually the right move whenever an exponent appears.
13. If |z₁| = 3 and |z₂| = 4, what is |z₁z₂|?
12. Moduli multiply: |z₁z₂| = |z₁||z₂| = 3 × 4 = 12. Arguments, by contrast, add. This pairing - multiply the moduli, add the arguments - is the entire reason polar form simplifies multiplication, and it generalises directly into De Moivre's theorem.
14. When does |z₁ + z₂| equal |z₁| + |z₂|?
Only when z₁ and z₂ have the same argument, that is when they point in the same direction from the origin. Otherwise the triangle inequality is strict. Geometrically, the sum of two vectors has length equal to the sum of their lengths only when they are collinear and codirectional - and questions test exactly this condition.
Practice set 3: De Moivre and roots of unity
15. State De Moivre's theorem.
(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. In polar form this says that raising a complex number to a power raises the modulus to that power and multiplies the argument by it. It converts an n-fold binomial expansion into a single multiplication, which is why it is the standard tool for high powers.
16. Evaluate (1 + i)⁸.
16. In polar form 1 + i has modulus √2 and argument π/4, so the eighth power has modulus (√2)⁸ = 16 and argument 8 × π/4 = 2π. Since cos 2π = 1 and sin 2π = 0, the result is 16 - a real number. A fast check: (1 + i)² = 2i, so (1 + i)⁸ = (2i)⁴ = 16 i⁴ = 16, which agrees.
17. Why must you be careful applying De Moivre's theorem to fractional powers?
Because a fractional power has several values, not one. The n-th roots of a complex number are n distinct points, obtained by adding 2πk to the argument before dividing by n, for k = 0 to n − 1. Taking only the principal value silently discards the others, and questions asking for 'all values' are testing precisely this.
18. What are the cube roots of unity and what identity do they satisfy?
1, ω and ω², where ω = cos(2π/3) + i sin(2π/3). They satisfy ω³ = 1 and 1 + ω + ω² = 0. Geometrically they are three equally spaced points on the unit circle, and the sum being zero is why their centroid is the origin. Almost every question in this area reduces to those two facts.
19. Simplify ω⁴.
ω. Since ω³ = 1, write ω⁴ = ω³ · ω = 1 · ω = ω. Any power of ω reduces by taking the exponent modulo 3, exactly as powers of i reduce modulo 4. So ω¹⁰⁰ = ω^(99+1) = ω, since 99 is divisible by 3.
20. Evaluate (1 + ω)³.
−1. From 1 + ω + ω² = 0 we get 1 + ω = −ω². Cubing gives (−ω²)³ = −ω⁶ = −(ω³)² = −(1)² = −1. Substituting the identity first, rather than expanding the binomial, is what makes this a one-line problem - and that substitution is the intended method for every question of this shape.
21. What is the sum of all n-th roots of unity?
Zero, for every n greater than 1. The roots are equally spaced around the unit circle, so they cancel by symmetry - their vector sum returns to the origin. The cube-roots identity 1 + ω + ω² = 0 is the case n = 3, and the same reasoning gives the general result.
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Practice set 4: Argand-plane geometry
22. What locus does |z − 1| = 2 describe?
A circle of radius 2 centred at the point 1, that is at (1, 0). The expression |z − a| is the distance from z to a, so setting it equal to a constant traces all points at that fixed distance - the definition of a circle. Recognising this immediately is worth more than any amount of algebraic expansion.
23. What locus does |z − 1| = |z + 1| describe?
The imaginary axis. The condition says z is equidistant from 1 and −1, so it lies on the perpendicular bisector of the segment joining them - the vertical line through the origin, which is the imaginary axis. Squaring and expanding gives the same answer in four lines instead of one.
24. What is the geometric effect of multiplying a complex number by i?
A rotation of 90° anticlockwise about the origin, with the modulus unchanged. Since i has modulus 1 and argument π/2, multiplying adds π/2 to the argument and leaves the modulus alone. Multiplying by i twice gives a 180° rotation, which is consistent with i² = −1.
25. How do you rotate a point about the origin by an arbitrary angle θ?
Multiply by e^(iθ) = cos θ + i sin θ, which has modulus 1 so only the argument changes. To rotate about a point a rather than the origin, first subtract a, then rotate, then add a back: z′ = a + (z − a)e^(iθ). That three-step pattern answers most rotation questions in the chapter.
26. What does |z₁ − z₂| represent?
The distance between the points z₁ and z₂ in the Argand plane. This single interpretation underlies every locus question: a circle is a fixed distance from one point, a perpendicular bisector is equal distances from two points, and an ellipse is a fixed sum of distances from two points.
27. What locus is described by arg(z) = π/4?
A ray from the origin at 45° to the positive real axis, excluding the origin itself. It is a ray rather than a full line, because the points on the opposite side would have argument −3π/4. The origin is excluded because its argument is undefined - a detail questions sometimes hinge on.
Practice set 5: mixed and conceptual
28. Is the statement z₁ > z₂ meaningful for complex numbers?
No. There is no ordering on the complex numbers consistent with their arithmetic, so inequalities between them are undefined. Only real quantities derived from them - moduli, real parts, imaginary parts, arguments - can be compared. An option offering an inequality between two complex numbers is always wrong.
29. What are the roots of x² + x + 1 = 0?
ω and ω², the non-real cube roots of unity. The discriminant is 1 − 4 = −3, so the roots are complex conjugates, and they satisfy the equation precisely because 1 + ω + ω² = 0. This is the bridge between this chapter and Quadratic Equations, and questions frequently cross it.
30. What does Euler's identity e^(iπ) + 1 = 0 express?
That rotating the point 1 by π radians about the origin lands you at −1. Substituting θ = π into e^(iθ) = cos θ + i sin θ gives cos π + i sin π = −1. Read geometrically it is almost obvious; its fame comes from linking e, i, π, 1 and 0 in a single relation.
How to study this chapter efficiently
- ✓Sketch the Argand plane for any question containing a modulus. Most locus problems are answered by the picture alone.
- ✓Convert to polar form the moment an exponent appears. Binomial expansion is almost never the intended route.
- ✓Learn 1 + ω + ω² = 0 and ω³ = 1 as a pair, and try substituting the first identity before expanding anything.
- ✓Check the quadrant on every argument. Write down the signs of the real and imaginary parts before quoting an angle.
- ✓Remember |z − a| means distance. That one reading turns circles, bisectors and ellipses into recognitions rather than derivations.
- ✓Study this chapter alongside Vector Algebra - addition, modulus and rotation behave identically in both, so the geometric intuition transfers directly.
Turn this into active practice
The gap in this chapter is between knowing that |z − a| = r is a circle and actually seeing it under time pressure on an unfamiliar expression. That recognition only forms by meeting varied loci where nothing announces which shape is coming.
The JEE Complex Numbers quiz on QUFF generates fresh questions across algebra, modulus and argument, De Moivre, roots of unity and Argand geometry, marks them instantly and explains each answer. Do a mixed set and note whether each error came from algebra or from failing to see the geometry - the second is the one worth training.