Complex Numbers: PECTAA Class 11 Maths, Punjab Board

Solve x² + 1 = 0. No real number works, because a real number squared is never negative. So mathematicians made a new number, i, with i² = −1. Learn five ideas in this unit: the form a + ib, the conjugate, the modulus, division and equality. The rest follows.

Start with the problem. Every real number you try gives x² + 1 an answer of 1 or more, yet the equation says x² = −1. There is a way out: allow a new kind of number. This is how Unit 1 of the new PECTAA Mathematics 11 book begins.

If you study FSc Part 1 (1st year, Class 11) in Lahore, Faisalabad, Sargodha or Gujrat, this is your first unit. PECTAA writes the Punjab books. Your own board sets your paper. Gujrat district comes under the Gujranwala board.

What to know

The number i. From x² + 1 = 0 we get x² = −1. The book calls √−1 an imaginary number and writes it as i. So i² = −1.

Complex number. A number of the form z = a + ib, where a and b are real numbers. Book examples: 3 + 4i, 2 − (5/7)i and −7 − 2i. The set of all complex numbers is written ℂ.

• The real part is a. It is written Re z.
• The imaginary part is b, a real number without the i. It is written Im z.
• For z = 3 + 4i: Re z = 3 and Im z = 4.
• Every real number is a complex number. Its imaginary part is 0.

Conjugate. Change the sign of the imaginary part. The conjugate of a + ib is a − ib. The book gives 5 − 4i as the conjugate of 5 + 4i. A real number is its own conjugate.

Argand diagram and modulus. Every complex number x + iy is one point (x, y) on the complex plane. The x-axis is the real axis. The y-axis is the imaginary axis. The modulus is the distance from the origin:

• |x + iy| = √(x² + y²)
• Example: |3 + 4i| = √(9 + 16) = 5.

Key points

Operations.

• Add: (a + ib) + (c + id) = (a + c) + i(b + d).
• Multiply every term by every term, then use i² = −1. The book gives (a + ib)(c + id) = (ac − bd) + i(ad + bc).
• Divide: multiply top and bottom by the conjugate of the bottom.

Ordered pairs. The book also writes a + ib as the pair (a, b). Then (a, b) + (c, d) = (a + c, b + d) and (a, b)(c, d) = (ac − bd, ad + bc). The multiplicative identity is (1, 0). The inverse of (a, b) is (a/(a² + b²), −b/(a² + b²)). Complex numbers cannot be put in order.

Equality. a + ib = c + id exactly when a = c and b = d. One complex equation gives two real equations.

Roots and factors. A polynomial of degree n ≥ 1 has exactly n roots in ℂ. So z² + 4 = (z + 2i)(z − 2i).

Roots of unity. From x³ = 1 we get (x − 1)(x² + x + 1) = 0. The roots are 1 and (−1 ± √3 i)/2. Call the two imaginary roots ω and ω². Then 1 + ω + ω² = 0 and ω³ = 1. The fourth roots of 1 are 1, −1, i and −i. Their sum is 0 and their product is −1.

Worked examples

Example 1 (book Example 2). If z₁ = (4, 2) and z₂ = (3, −1), find z₁ ÷ z₂.

1. Write z₁ = 4 + 2i and z₂ = 3 − i.
2. Multiply top and bottom by 3 + i.
3. Top: (4 + 2i)(3 + i) = 12 + 4i + 6i + 2i² = 10 + 10i.
4. Bottom: (3 − i)(3 + i) = 9 − i² = 10.
5. Answer: (10 + 10i) ÷ 10 = 1 + i.

Example 2 (book Example 4). If (3 + 2i)(x + iy) = 5 + 12i, find x and y.

1. Left side: (3x − 2y) + (2x + 3y)i.
2. Real parts: 3x − 2y = 5. Imaginary parts: 2x + 3y = 12.
3. Multiply the first by 3 and the second by 2, then add: 13x = 39, so x = 3.
4. Then 9 − 2y = 5, so y = 2.

Example 3 (book Example 5). Find the square roots of 5 + 12i.

1. |z| = √(25 + 144) = 13.
2. p = √((13 + 5)/2) = 3 and q = √((13 − 5)/2) = 2.
3. The roots are 3 + 2i and −3 − 2i.

Example 4 (book Example 3). Write z = (1 + 2i)² ÷ (2 − i) as a + ib.

1. (1 + 2i)² = 1 + 4i + 4i² = −3 + 4i.
2. Multiply by (2 + i) over (2 + i). Top: −10 + 5i. Bottom: 5.
3. z = −2 + i. Its modulus is √5.

Common mistakes in the exam

1. Treating i² as +1. It is −1. So 4i² = −4.
2. Squaring a sum part by part. (1 + 2i)² = 1 + 4i + 4i², not 1 + 4i².
3. Writing the imaginary part as 4i. For 3 + 4i, Im z = 4.
4. Changing both signs for the conjugate. The conjugate of −2 + 3i is −2 − 3i.
5. Mixing real and imaginary parts in equality. Write two separate equations.
6. Giving one square root. A non-zero complex number has two, one the negative of the other.
7. Saying the product of the four fourth roots of unity is 1. It is −1.

Exam technique

• Write the formula, then the numbers, then the answer.
• For division, say "multiply by the conjugate" in your first line. It shows your method.
• The PECTAA Smart Syllabus for Mathematics 11, notified for the Annual 2026 exam, gave Unit 1 two MCQs, four of the twelve short questions in the question 2 section, and one five-mark part of a long question. It left out polar form (section 1.4.1, Examples 12 to 16) and section 1.5. We found no notice for the 2027 exam, so read your board's notice before you skip anything.

Practice questions

1. Write Re z and Im z for z = 3 + 4i. Answer: Re z = 3 and Im z = 4.
2. Find the conjugate of −2 + 3i, and find |3 + 4i|. Answer: −2 − 3i, and 5.
3. Find the sum, difference and product of (8, 9) and (5, −6). Answer: Sum (13, 3). Difference (3, 15). Product (94, −3).
4. Write (1 + 3i) ÷ (1 − i) as a + ib. Answer: Multiply by (1 + i) over (1 + i). Top: 1 + i + 3i + 3i² = −2 + 4i. Bottom: 2. The answer is −1 + 2i.
5. Find the multiplicative inverse of (3, 4). Answer: (3/25, −4/25). Check: (3 + 4i)(3 − 4i) = 25.

Quick revision

• i² = −1. z = a + ib. Conjugate: flip the sign of the imaginary part.
• |x + iy| = √(x² + y²).
• Divide with the conjugate of the bottom.
• Equal numbers: equal real and imaginary parts.
• Cube roots of unity: 1, ω, ω². Sum 0. Product 1.

Learn it step by step in the lessons Complex Numbers Basics (/learn/fsc-1/maths/complex-numbers-basics) and Operations on Complex Numbers (/learn/fsc-1/maths/operations-on-complex-numbers).

Ready to practise this? Continue in 1st Year Mathematics.