Complex numbers as points: modulus, amplitude and trigonometric form
الأسبوع 4 · الأعداد المركبة على مستوى أرجاند والمرافق ومقياس العدد المركب وسعته والسعة الأساسية والصورة المثلثية · 3 دروس
2
The modulus and the amplitude of a complex number, and the principal amplitude by quadrant
Pure Mathematics Sec 3 Term 1, Unit 2, Lesson 1
GoalFind the modulus and the principal amplitude of a complex number, using the quadrant.
New words
modulus |z| · r = √(x² + y²), the distance from the originamplitude · an angle θ with x = r cos θ and y = r sin θ; adding 2π gives anotherprincipal amplitude · the amplitude in ]−π, π]
Explanation
The modulus |z| = r = √(x² + y²) is the distance of the point from the origin, so it is never negative. An amplitude of z is any angle θ with x = r cos θ and y = r sin θ. Adding a full turn 2π gives another amplitude of the same number, so to have a single answer we use the principal amplitude, the one in ]−π, π] (in degrees ]−180°, 180°]). To find it, first find the acute reference angle α = tan⁻¹(|y|/|x|) from the lengths alone, then look at the quadrant of the point: quadrant I gives θ = α, II gives π − α, III gives α − π and IV gives −α (see the table). The book writes tan⁻¹(y/x) with the signs inside and adds or subtracts π; both ways give the same result. On the axes there is no angle to find: a positive real number has amplitude 0, a negative real number π, a positive imaginary number π/2 and a negative imaginary number −π/2. Three facts will be used all unit: |z| = |z̄| = |−z|, z · z̄ = |z|² (because (x + yi)(x − yi) = x² + y²), and if z has principal amplitude θ then z̄ has −θ (except that a negative real number keeps π), while −z has θ − π when θ > 0 and θ + π when θ ≤ 0.
α = tan⁻¹(|y| ÷ |x|), the acute reference angle
Quadrant
signs of x and y
principal amplitude θ
I
x > 0, y > 0
α
II
x < 0, y > 0
π − α
III
x < 0, y < 0
α − π
IV
x > 0, y < 0
−α
Worked examples
Find the modulus and principal amplitude of z = −3 − 3√3 i. |z| = √(9 + 27) = 6. The reference angle is α = tan⁻¹(3√3/3) = tan⁻¹(√3) = π/3. The point is in quadrant III, so θ = α − π = −2π/3 (that is −120°). Check: 6 cos(−120°) = −3 and 6 sin(−120°) = −3√3.
Find the modulus and principal amplitude of 5√3 − 5i and of −4 + 4i. First: |z| = √(75 + 25) = 10, α = tan⁻¹(5/(5√3)) = π/6, quadrant IV, so θ = −π/6. Second: |z| = 4√2, α = tan⁻¹(1) = π/4, quadrant II, so θ = π − π/4 = 3π/4.
Numbers on the axes: −7 has modulus 7 and amplitude π; 6i has modulus 6 and amplitude π/2; −2i has modulus 2 and amplitude −π/2. And if z has principal amplitude 2π/3 (more than 0), then z̄ has −2π/3 and −z has 2π/3 − π = −π/3.
Class activity
Quadrant check. Choose a point in each of the four quadrants, for example (2, 1), (−2, 1), (−2, −1), (2, −1), and find α for each. Then use the table to write down all four principal amplitudes in terms of α.
Practice
1
Find |3 − 4i|, |−5 + 12i| and |2√2 − 2√2 i|.
√(9 + 16) = 5; √(25 + 144) = 13; √(8 + 8) = 4.
2
Find the principal amplitudes of 3 + 3i and of −2 + 2√3 i.