Complex Numbers ISC Class 11: Notes, Proofs and Solved Questions

What Complex Numbers is about

Weightage: Complex Numbers belongs to the unit Algebra, which CISCE weights at 26 of the 80 marks in the ISC Class XI Mathematics Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Mathematics syllabus, examination year 2028, Class XI unit weightage

Key concepts

The number i and the form a + ib

i is defined by i² = −1, so √(−k) = i√k for k > 0. 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. A complex number is z = a + ib with a, b real; a is the real part and b the imaginary part (b itself is real, not ib). z is purely real if b = 0 and purely imaginary if a = 0, b ≠ 0. Two complex numbers are equal only when their real parts are equal and their imaginary parts are equal; this turns one complex equation into two real equations. Complex numbers cannot be ordered: statements such as 2 + i > 1 + 3i have no meaning.

Worked example

Simplify i⁹⁹ + i¹⁰⁰ + i¹⁰¹ + i¹⁰², and find real x, y if (x + iy)(2 − 3i) = 4 + i.

Try it first, then show the solutionSolution

1. 99 = 4 × 24 + 3, so i⁹⁹ = i³ = −i; similarly i¹⁰⁰ = 1, i¹⁰¹ = i, i¹⁰² = −1. Sum = −i + 1 + i − 1 = 0
2. x + iy = (4 + i)/(2 − 3i) = (4 + i)(2 + 3i)/(4 + 9) = (8 + 12i + 2i − 3)/13 = (5 + 14i)/13
3. Equating real and imaginary parts: x = 5/13, y = 14/13

Answer: 0; x = 5/13, y = 14/13

Sum, difference, product, quotient and inverses

For z₁ = a + ib and z₂ = c + id: z₁ ± z₂ = (a ± c) + i(b ± d); z₁z₂ = (ac − bd) + i(ad + bc). Addition and multiplication are commutative and associative, multiplication distributes over addition, 0 is the additive identity and 1 the multiplicative identity. The additive inverse of z is −z = −a − ib. The multiplicative inverse of z ≠ 0 is 1/z = z̄/|z|² = (a − ib)/(a² + b²). To divide, rationalise: z₁/z₂ = z₁z̄₂/|z₂|², i.e. multiply numerator and denominator by the conjugate of the denominator.

Worked example

Write (3 + 2i)/(1 − i) in the form a + ib, and find the multiplicative inverse of 3 − 4i.

Try it first, then show the solutionSolution

1. (3 + 2i)(1 + i)/((1 − i)(1 + i)) = (3 + 3i + 2i + 2i²)/2 = (1 + 5i)/2
2. Inverse of 3 − 4i = (3 + 4i)/(3² + 4²) = (3 + 4i)/25
3. Check: (3 − 4i)(3 + 4i)/25 = 25/25 = 1

Answer: 1/2 + (5/2)i; 3/25 + (4/25)i

Conjugate and its properties

The conjugate of z = a + ib is z̄ = a − ib: the mirror image of z in the real axis. Properties: (z̄)‾ = z; z + z̄ = 2 Re z; z − z̄ = 2i Im z; z z̄ = a² + b² = |z|²; conj(z₁ ± z₂) = z̄₁ ± z̄₂; conj(z₁z₂) = z̄₁ z̄₂; conj(z₁/z₂) = z̄₁/z̄₂. Two useful tests follow: z is real ⇔ z = z̄, and z is purely imaginary (or zero) ⇔ z + z̄ = 0. These tests often prove results without splitting into a and b.

Worked example

Find the conjugate of (2 + i)/(1 − 2i).

Try it first, then show the solutionSolution

1. (2 + i)(1 + 2i)/(1 + 4) = (2 + 4i + i + 2i²)/5 = (0 + 5i)/5 = i
2. Conjugate of i is −i

Answer: −i

Modulus and its properties

The modulus |z| = √(a² + b²) is the distance of the point (a, b) from the origin. Properties: |z| ≥ 0, with |z| = 0 only for z = 0; |z̄| = |−z| = |z|; |z₁z₂| = |z₁||z₂|; |z₁/z₂| = |z₁|/|z₂|; |zⁿ| = |z|ⁿ. The triangle inequality |z₁ + z₂| ≤ |z₁| + |z₂| holds, with equality only when z₁ and z₂ point the same way. Also |z₁ − z₂| is the distance between the points z₁ and z₂. Because modulus respects products and quotients, you can find the modulus of a messy fraction without simplifying it.

Worked example

Find |(3 + 4i)(1 − i)/(2 + i)| without expanding.

Try it first, then show the solutionSolution

1. |3 + 4i| = 5, |1 − i| = √2, |2 + i| = √5
2. Modulus = 5√2/√5 = √10

Answer: √10 ≈ 3.16

Argand plane, argument and polar form

In the Argand plane the horizontal axis is the real axis and the vertical axis the imaginary axis; z = a + ib is the point (a, b). If r = |z| and θ is the angle from the positive real axis to the line joining O to z, then a = r cos θ, b = r sin θ, so z = r(cos θ + i sin θ): the polar form. θ is an argument of z; the principal argument is the one with −π < θ ≤ π. To find it, compute the acute angle α = tan⁻¹|b/a| and then place it by quadrant: Q I θ = α; Q II θ = π − α; Q III θ = −(π − α); Q IV θ = −α. Useful properties: arg(z₁z₂) = arg z₁ + arg z₂ and arg(z₁/z₂) = arg z₁ − arg z₂ (adjusting by 2π to stay in the principal range); arg z̄ = −arg z.

Worked example

Write −1 + i√3 and 1 − i in polar form.

Try it first, then show the solutionSolution

1. −1 + i√3: r = √(1 + 3) = 2; α = tan⁻¹√3 = π/3; Q II, so θ = π − π/3 = 2π/3
2. −1 + i√3 = 2(cos 2π/3 + i sin 2π/3)
3. 1 − i: r = √2; α = π/4; Q IV, so θ = −π/4; 1 − i = √2(cos(−π/4) + i sin(−π/4))

Answer: 2(cos 2π/3 + i sin 2π/3); √2(cos(−π/4) + i sin(−π/4))

Square root of a complex number

To find √(a + ib), let it be x + iy with x, y real. Squaring: x² − y² + 2ixy = a + ib, so x² − y² = a and 2xy = b. Also |x + iy|² = |a + ib|, so x² + y² = √(a² + b²). Adding and subtracting: x² = (|z| + a)/2 and y² = (|z| − a)/2. Finally, choose signs using 2xy = b: if b > 0, x and y have the same sign; if b < 0, opposite signs. There are always exactly two square roots, negatives of each other. Always check by squaring your answer.

Worked example

Find the square roots of −5 + 12i.

Try it first, then show the solutionSolution

1. |z| = √(25 + 144) = 13
2. x² = (13 − 5)/2 = 4, y² = (13 + 5)/2 = 9
3. b = 12 > 0, so x, y have the same sign: x + iy = ±(2 + 3i)
4. Check: (2 + 3i)² = 4 + 12i − 9 = −5 + 12i ✓

Answer: ±(2 + 3i)

Cube roots of unity

Solving z³ = 1: z³ − 1 = (z − 1)(z² + z + 1) = 0, so z = 1 or z = (−1 ± i√3)/2. Writing ω = (−1 + i√3)/2, the other non-real root is ω² = (−1 − i√3)/2 = ω̄. Properties: ω³ = 1; 1 + ω + ω² = 0; ω² = 1/ω; |ω| = 1; the three roots form an equilateral triangle on the unit circle with arguments 0, 2π/3 and −2π/3. To simplify ωⁿ, divide n by 3 and keep the remainder. In expressions like a + bω + cω², replace a pair such as 1 + ω² by −ω.

Worked example

Evaluate (1 + ω − ω²)(1 − ω + ω²).

Try it first, then show the solutionSolution

1. 1 + ω = −ω², so 1 + ω − ω² = −2ω²
2. 1 + ω² = −ω, so 1 − ω + ω² = −2ω
3. Product = (−2ω²)(−2ω) = 4ω³ = 4

Answer: 4

Maths in real life: Complex Numbers

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Algebra, Conjugate and Modulus of Complex Numbers
• Polar Form, Square Roots and Cube Roots of Unity

Complex Numbers formula sheet

Powers of i

$i^2=-1,i^3=-i,i^4=1,i^{4k+r}=i^r$

Divide the power by 4 and keep the remainder.

Product

$(a+ib)(c+id)=(ac-bd)+i(ad+bc)$

Expand, then replace i² by −1.

Quotient and inverse

$\frac{z_1}{z_2}=\frac{z_1{\bar{z}}_2}{|z_2{|}^2};z^{-1}=\frac{a-ib}{a^2+b^2}$

Multiply top and bottom by the conjugate of the denominator.

Conjugate

$z\bar{z}=|z{|}^2;z+\bar{z}=2\mathrm{Re}z;\overline{z_1{z}_2}={\bar{z}}_1{\bar{z}}_2$

z real ⇔ z = z̄.

Modulus

$|z|=\sqrt{a^2+b^2};|z_1{z}_2|=|z_1||z_2|;|z_1+z_2|\le |z_1|+|z_2|$

Modulus of a quotient is the quotient of the moduli.

Polar form

$z=r(\cos \theta +i\sin \theta ),r=|z|,-\pi <\theta \le \pi$

Place the acute angle by quadrant.

Square root

$x^2=\frac{|z|+a}{2},y^2=\frac{|z|-a}{2},2xy=b$

Same signs if b > 0, opposite signs if b < 0.

Cube roots of unity

$\omega =\frac{-1+i\sqrt{3}}{2};{\omega }^3=1;1+\omega +{\omega }^2=0$

ω² = ω̄ = 1/ω.

Complex Numbers: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Complex Numbers

The question patterns to expect, and the method that works for each.

Tricks to remember Complex Numbers

Short memory aids. Say each one aloud once, then use it on a question the same day.

• i goes round in fours

  i, −1, −i, 1, then repeat: only the remainder on division by 4 matters.

  Use it when: Powers of i.

• Conjugate kills i

  (a + ib)(a − ib) = a² + b²: multiplying by the conjugate makes the denominator real.

  Use it when: Division and inverses.

• All Students Take Calculus for arguments

  Q I: α; Q II: π − α; Q III: −(π − α); Q IV: −α. Never trust a calculator's tan⁻¹ alone.

  Use it when: Principal argument.

• Sign of b sets the signs

  b positive: same signs (x, y) both + or both −; b negative: opposite signs.

  Use it when: Square roots.

• One plus omega plus omega squared is nothing

  1 + ω + ω² = 0, so any two of them add to minus the third.

  Use it when: Cube roots of unity simplification.

Complex Numbers: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Complex Numbers

The slips examiners see most often, and what to write instead.

• ✗Mistake: Taking the argument straight from a calculator

  tan⁻¹(b/a) only returns angles between −π/2 and π/2, so Q II and Q III points get the wrong angle.

  Wrong working

  arg(−1 + i) = tan⁻¹(−1) = −π/4.

  Right working

  −1 + i is in Q II: arg = π − π/4 = 3π/4.

  ✓Do this: Sketch the point first, then place the acute angle.

  Usually costs: 1-2 marks.

• ✗Mistake: Multiplying square roots of negatives

  √a √b = √(ab) is only valid when a, b are not both negative.

  Wrong working

  √(−4) × √(−9) = √36 = 6.

  Right working

  2i × 3i = 6i² = −6.

  ✓Do this: Convert to i form first: √(−k) = i√k.

  Usually costs: Full marks for the part.

• ✗Mistake: Calling ib the imaginary part

  Im(a + ib) is the real number b.

  Wrong working

  Im(4 − 7i) = −7i.

  Right working

  Im(4 − 7i) = −7.

  ✓Do this: Both Re z and Im z are real numbers.

  Usually costs: 1 mark.

• ✗Mistake: Wrong pairing of signs in a square root

  Students write ±4 ± 3i (four answers) or ignore the sign of b.

  Wrong working

  √(7 − 24i) = ±(4 + 3i).

  Right working

  b < 0, so opposite signs: ±(4 − 3i).

  ✓Do this: Check by squaring before you write the final answer.

  Usually costs: 1-2 marks.

• ✗Mistake: Giving an argument outside (−π, π]

  Students give 5π/3 or 7π/4 for Q IV points.

  Wrong working

  arg(1 − i) = 7π/4.

  Right working

  Principal argument = −π/4.

  ✓Do this: If the question says principal argument, subtract 2π from any angle above π.

  Usually costs: 1 mark.

Practice questions on Complex Numbers

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The conjugate of (1 + i)² is:
   (a) 2i
   (b) −2i
   (c) 2
   (d) 0

   Hint

   Square first.

   Answer

   (b): (1 + i)² = 2i, whose conjugate is −2i

2. Q2 · 1 mark · MCQExam level

   The modulus of 5 − 12i is:
   (a) 7
   (b) 13
   (c) 17
   (d) √119

   Hint

   √(a² + b²).

   Answer

   (b)

3. Q3 · 1 mark · Very short answerExam level

   Find the value of i⁻³⁵.

   Hint

   1/i³.

   Answer

   i⁻³⁵ = 1/i³⁵ = 1/i³ = 1/(−i) = i

4. Q4 · 1 mark · Very short answerExam level

   Write the additive inverse of −3 + 7i.

   Hint

   Change both signs.

   Answer

   3 − 7i

5. Q5 · 2 marks · Short answerExam level

   Express (2 − 3i)(−1 + 4i) in the form a + ib.

   Hint

   Expand; i² = −1.

   Answer

   −2 + 8i + 3i + 12 = 10 + 11i

6. Q6 · 2 marks · Short answerHarder exam level

   Find real x and y if (x + iy)/(2 + i) = 1 − i.

   Hint

   Multiply across.

   Answer

   x + iy = (1 − i)(2 + i) = 2 + i − 2i + 1 = 3 − i, so x = 3, y = −1

7. Q7 · 2 marks · Short answerHarder exam level

   Find the modulus and principal argument of −2 − 2i.

   Hint

   Q III.

   Answer

   |z| = 2√2; arg = −(π − π/4) = −3π/4

8. Q8 · 3 marks · Short answerHarder exam level

   Find the square roots of −15 − 8i.

   Hint

   |z| = 17.

   Answer

   x² = (17 − 15)/2 = 1, y² = (17 + 15)/2 = 16, opposite signs: ±(1 − 4i). Check: (1 − 4i)² = 1 − 8i − 16 = −15 − 8i.

9. Q9 · 3 marks · Short answerHarder exam level

   If (a + ib)/(c + id) = x + iy, prove that x² + y² = (a² + b²)/(c² + d²).

   Hint

   Take modulus of both sides.

   Answer

   |x + iy| = |a + ib|/|c + id|, so √(x² + y²) = √(a² + b²)/√(c² + d²); square both sides.

10. Q10 · 2 marks · Short answerHarder exam level

    Show that ω¹⁰⁰ + ω²⁰⁰ + 1 = 0, where ω is a non-real cube root of unity.

    Hint

    Reduce the powers mod 3.

    Answer

    100 = 3 × 33 + 1 and 200 = 3 × 66 + 2, so the sum is ω + ω² + 1 = 0.

11. Q11 · 3 marks · Short answerHarder exam level

    Find the real number k for which (k + 2i)/(1 − i) is purely imaginary, and find the value of the expression.

    Hint

    Rationalise and set the real part to zero.

    Answer

    (k + 2i)(1 + i)/2 = [(k − 2) + (k + 2)i]/2. Real part zero gives k = 2, and the value is 2i.

12. Q12 · 5 marks · Long answerHarder exam level

    (a) Express z = (1 + 2i)/(1 − 3i) in the form a + ib. [2]
    (b) Find |z| and the principal argument of z. [2]
    (c) Write z in polar form. [1]

    Hint

    Multiply by 1 + 3i.

    Answer

    (a) (1 + 2i)(1 + 3i)/10 = (−5 + 5i)/10 = −1/2 + (1/2)i. (b) |z| = 1/√2; Q II, so arg = 3π/4. (c) z = (1/√2)(cos 3π/4 + i sin 3π/4).

13. Q13 · 3 marks · Short answerHOTS / competency

    If |z₁| = |z₂| = 1 and z₁z₂ ≠ −1, prove that (z₁ + z₂)/(1 + z₁z₂) is real.

    Hint

    Use z̄ = 1/z for |z| = 1.

    Answer

    Let w be the expression. w̄ = (1/z₁ + 1/z₂)/(1 + 1/(z₁z₂)); multiply top and bottom by z₁z₂ to get (z₂ + z₁)/(z₁z₂ + 1) = w. Since w̄ = w, w is real.

14. Q14 · 5 marks · Long answerHOTS / competency

    (a) Solve z³ = 1 and show that the non-real roots are (−1 ± i√3)/2. [2]
    (b) Prove that 1 + ω + ω² = 0. [1]
    (c) Evaluate (1 − ω)(1 − ω²)(1 − ω⁴)(1 − ω⁵). [2]

    Hint

    ω⁴ = ω and ω⁵ = ω².

    Answer

    (a) (z − 1)(z² + z + 1) = 0, and z² + z + 1 = 0 gives z = (−1 ± √(−3))/2 = (−1 ± i√3)/2. (b) ω is a root of z² + z + 1 = 0. (c) The product is [(1 − ω)(1 − ω²)]² = (1 − ω − ω² + ω³)² = (1 + 1 + 1)² = 9.

15. Q15 · 3 marks · Short answerHarder exam level

    Find all complex numbers z for which z̄ = z².

    Hint

    Put z = x + iy and equate parts.

    Answer

    x − iy = x² − y² + 2ixy gives x = x² − y² and −y = 2xy. If y = 0: x = 0 or 1. If y ≠ 0: x = −1/2, then y² = 3/4. So z = 0, 1, −1/2 ± (√3/2)i (zero and the three cube roots of unity).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.i⁴ⁿ⁺³?Show answer

   −i.

2. 2.z z̄ for z = a + ib?Show answer

   a² + b² = |z|².

3. 3.Multiplicative inverse of a + ib?Show answer

   (a − ib)/(a² + b²).

4. 4.Range of the principal argument?Show answer

   −π < θ ≤ π.

5. 5.Polar form?Show answer

   z = r(cos θ + i sin θ), r = |z|, θ = arg z.

6. 6.|z₁z₂|?Show answer

   |z₁||z₂|.

7. 7.Square root signs when b < 0?Show answer

   x and y have opposite signs.

8. 8.1 + ω + ω²? ω³?Show answer

   0; 1.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you find iⁿ for any integer n, including negative n?
• ☐Can you divide two complex numbers and give a + ib?
• ☐Can you state five properties of the conjugate?
• ☐Can you find the modulus of a quotient without simplifying it?
• ☐Can you find the principal argument of a point in Q II or Q III?
• ☐Can you write any complex number in polar form?
• ☐Can you find both square roots of a + ib with correct signs?
• ☐Can you simplify (a + bω + cω²)ⁿ using 1 + ω + ω² = 0?

Complex Numbers: frequently asked questions