What is the polar form of a complex number in the PECTAA Class 11 Maths book?

The book writes z = x + iy as r cos θ + i r sin θ, where r = |z| = √(x² + y²) is the modulus and θ is an argument of z.

What is the principal argument?

The book defines it as the angle between the positive real axis and the line joining the point (a, b) to the origin. It lies in the range (−π, π].

How do I multiply two complex numbers in polar form?

Multiply the moduli and add the arguments. The book proves arg(z₁z₂) = arg(z₁) + arg(z₂).

How do I divide two complex numbers in polar form?

Divide the moduli and subtract the arguments. The book proves arg(z₁ ÷ z₂) = arg(z₁) − arg(z₂).

Which board sets my paper if I study in Gujrat?

Gujrat district is listed under the Board of Intermediate and Secondary Education, Gujranwala. PECTAA writes the books. Your board sets the paper and the dates, so follow its notices.

Polar Form of Complex Numbers: PECTAA Class 11 Maths, Punjab Board

A point on a map can be given by two distances: how far east and how far north. It can also be given by one distance and one direction. Polar form does this for a complex number. Learn it in four steps: r, θ, the polar form, and the two rules for multiplying and dividing.

In Unit 1 of the PECTAA Mathematics 11 book, a complex number z = x + iy is a point (x, y) on the Argand plane. Section 1.4 shows a second way to name that point. This note covers that section for FSc Part 1 (1st year, Class 11) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

Polar coordinates. The book fixes a point O called the pole and a line from it called the polar axis, usually the positive x-axis. A point P is written (r, θ). Here r is the directed distance from the pole, and θ is the angle between the polar axis and the line OP.

Usually r ≥ 0.
The book also allows r < 0. Then the point lies |r| units from the pole in the opposite direction, that is, at angle θ + π.
Book note: (−5, π/4) and (5, 5π/4) are the same point.

Polar form (section 1.4.1). Take z = x + iy. From the diagram, x = r cos θ and y = r sin θ. So

x + iy = r cos θ + i r sin θ, which the book calls the polar form of z
r = |z| = √(x² + y²) is the modulus
θ = tan⁻¹(y/x) is an argument of z

Special cases from the book.

x = 0 and y > 0: θ = 90°
x = 0 and y < 0: θ = −90°
x = 0 and y = 0: θ is not defined
y = 0 and x > 0: θ = 0°
y = 0 and x < 0: θ = 180°

Principal argument. The principal argument Arg z is the angle between the positive real axis and the line joining (a, b) to the origin in the Argand plane. The book writes Arg z ∈ (−π, π]. It is one chosen value of the argument.

Key points

Multiplication. For z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂), the book multiplies out and uses the sine and cosine sum identities. The result is

z₁z₂ = r₁r₂[cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]

In words: multiply the moduli and add the arguments.

Division. The book multiplies top and bottom by the conjugate of cos θ₂ + i sin θ₂ and gets

z₁ ÷ z₂ = (r₁ ÷ r₂)[cos(θ₁ − θ₂) + i sin(θ₁ − θ₂)]

In words: divide the moduli and subtract the arguments.

Addition and subtraction. The book writes both in polar form but gives no shortcut. For these two, go back to x + iy form.

Worked examples

Example 1 (book Example 12). Express 1 + i√3 in polar form.

Put r cos θ = 1 and r sin θ = √3.
r² = 1² + (√3)² = 1 + 3 = 4, so r = 2.
θ = tan⁻¹(√3 ÷ 1) = tan⁻¹ √3 = 60°.
So 1 + i√3 = 2 cos 60° + i 2 sin 60°.

Example 2 (book Example 13). Find the product of 5(cos π/6 + i sin π/6) and 4(cos 3π/2 + i sin 3π/2).

r₁ = 5, θ₁ = π/6, r₂ = 4, θ₂ = 3π/2.
Moduli: 5 × 4 = 20.
Arguments: π/6 + 3π/2 = π/6 + 9π/6 = 10π/6 = 5π/3.
Product: 20(cos 5π/3 + i sin 5π/3).

Example 3 (book Example 14). Divide (2/7)(cos 7π/6 + i sin 7π/6) by (3/5)(cos(−π/2) + i sin(−π/2)).

r₁ = 2/7, θ₁ = 7π/6, r₂ = 3/5, θ₂ = −π/2.
Moduli: (2/7) ÷ (3/5) = (2/7)(5/3) = 10/21.
Arguments: 7π/6 − (−π/2) = 7π/6 + 3π/6 = 10π/6 = 5π/3.
Quotient: (10/21)(cos 5π/3 + i sin 5π/3).

Example 4 (check yourself). Write −1 + i in polar form.

r = √(1 + 1) = √2.
The point (−1, 1) is in the second quadrant, so the angle is 135°, not tan⁻¹(−1) = −45°.
−1 + i = √2(cos 135° + i sin 135°).

Common mistakes in the exam

Using tan⁻¹(y/x) without looking at the quadrant. Draw the point first. Example 4 shows why.
Forgetting that r is a modulus. In the standard form r ≥ 0.
Adding the moduli when multiplying. Moduli are multiplied. Arguments are added.
Dividing and adding the arguments. In division the arguments are subtracted.
Mixing degrees and radians. The book uses 60° in Example 12 and π-fractions in Examples 13 and 14. Stay in one unit inside a question.
Giving θ = 0 for y = 0 and x < 0. The book gives 180° for that case.

Exam technique

Write the formula first, then the numbers, then the answer.
Show r and θ as separate steps, as the book does in Steps I to III.
Check the sum or difference of arguments on a separate line. In Examples 13 and 14 both come to 5π/3.
The PECTAA syllabus scheme for Mathematics 11, notified for the Annual 2026 exam, left out polar form (section 1.4.1, Examples 12 to 16). We found no notice for the 2027 exam, so read your board's notice before you skip it.

Practice questions

Find the modulus of 3 + 4i. Answer: √(9 + 16) = 5.
Express 1 + i in polar form. Answer: r = √2, θ = tan⁻¹ 1 = 45°. So 1 + i = √2(cos 45° + i sin 45°).
Find the product of 2(cos 30° + i sin 30°) and 3(cos 60° + i sin 60°). Answer: 6(cos 90° + i sin 90°) = 6i.
Divide 8(cos 150° + i sin 150°) by 2(cos 30° + i sin 30°). Answer: 4(cos 120° + i sin 120°).
For z = −3 (a point on the negative real axis), what is θ? Answer: y = 0 and x < 0, so θ = 180°.

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