Complex numbers as points: modulus, amplitude and trigonometric form
Week 4 · Complex numbers on the Argand plane, conjugates, modulus and amplitude, the principal amplitude and the trigonometric form · 3 lessons
3
Writing a complex number in trigonometric form
Pure Mathematics Sec 3 Term 1, Unit 2, Lesson 1
GoalWrite a complex number in trigonometric form r(cos θ + i sin θ).
New words
trigonometric form · z = r(cos θ + i sin θ), with r = |z| and θ the principal amplitudestandard form · r > 0, the same angle in cos and sin, and a plus sign between them
Explanation
Since x = r cos θ and y = r sin θ, the number z = x + yi equals r cos θ + i r sin θ = r(cos θ + i sin θ). This is the trigonometric (polar) form (in Arabic الصورة المثلثية). To write a number in this form: (1) find r = |z|; (2) find the reference angle and the quadrant; (3) write θ, the principal amplitude; (4) put r, cos θ and sin θ together, and check one part by multiplying back. For the standard form we need r positive, the same angle in cos and sin, and a plus sign between them. Expressions that only look like it must be rewritten: r(cos θ − i sin θ) = r(cos(−θ) + i sin(−θ)); a minus in front, −r(cos θ + i sin θ), adds or subtracts π to the angle; and sin θ + i cos θ swaps sine and cosine, which means turning to the angle 90° − θ (because sin θ = cos(90° − θ) and cos θ = sin(90° − θ)). The table repeats the exact values you need. Numbers on the axes are easy: 6 = 6(cos 0 + i sin 0), −6 = 6(cos π + i sin π), 4i = 4(cos π/2 + i sin π/2) and −2i = 2(cos(−π/2) + i sin(−π/2)).
θ (radians)
θ (degrees)
cos θ
sin θ
0
0°
1
0
π/6
30°
√3/2
1/2
π/4
45°
√2/2
√2/2
π/3
60°
1/2
√3/2
π/2
90°
0
1
π
180°
−1
0
Worked examples
Write 3 − 3i in trigonometric form. r = √(9 + 9) = 3√2. The reference angle is π/4 and the point is in quadrant IV, so θ = −π/4. So 3 − 3i = 3√2 (cos(−π/4) + i sin(−π/4)). Check: 3√2 cos(−π/4) = 3√2 × √2/2 = 3, and the sine part gives −3.
Write −5√3 + 5i in trigonometric form. r = √(75 + 25) = 10. The reference angle is tan⁻¹(5/(5√3)) = π/6; quadrant II gives θ = π − π/6 = 5π/6. So the form is 10(cos 5π/6 + i sin 5π/6). Check: 10 × (−√3/2) = −5√3 and 10 × 1/2 = 5.
Bring these to standard form. (a) 2(sin 20° + i cos 20°) = 2(cos 70° + i sin 70°). (b) −3(cos 40° + i sin 40°): the minus subtracts 180° from the angle, giving 3(cos(−140°) + i sin(−140°)). (c) 5(cos 30° − i sin 30°) = 5(cos(−30°) + i sin(−30°)).
Go back to x + yi: 6(cos 2π/3 + i sin 2π/3) = 6(−1/2 + (√3/2) i) = −3 + 3√3 i.
Class activity
Spot the faults. For each of these, say what stops it being in standard form: 4(cos 20° − i sin 20°), −2(cos 50° + i sin 50°), 3(cos 10° + i sin 70°), 7(sin 40° + i cos 40°). Then repair each one.
Practice
1
Write 2 + 2√3 i in trigonometric form.
r = √(4 + 12) = 4, α = tan⁻¹(√3) = π/3, quadrant I: 4(cos π/3 + i sin π/3).
2
Write −√3 − i in trigonometric form.
r = √(3 + 1) = 2, α = tan⁻¹(1/√3) = π/6, quadrant III: θ = π/6 − π = −5π/6. So 2(cos(−5π/6) + i sin(−5π/6)).
3
Write −9 and 7i in trigonometric form.
−9 = 9(cos π + i sin π). 7i = 7(cos π/2 + i sin π/2).
4
Put 4(cos 60° − i sin 60°) in standard form and then in the form x + yi.
4(cos(−60°) + i sin(−60°)) = 4(1/2 − (√3/2) i) = 2 − 2√3 i.
5
Put −2(cos π/6 + i sin π/6) in standard form and then in the form x + yi.
The minus subtracts π: 2(cos(−5π/6) + i sin(−5π/6)) = 2(−√3/2 − i/2) = −√3 − i.