NCEA Level 3De Moivre's theoremAS 91577

De Moivre's theorem: powers and roots of complex numbers

What this lesson covers

  • Raising r cis θ to the power n raises the modulus to n and multiplies the argument by n.
  • The n roots of a complex number are equally spaced around a circle, so one wrong starting argument makes every root wrong.
Key idea

De Moivre's theorem

  • (r cis θ)ⁿ = rⁿ cis(nθ).
  • Raise the modulus, multiply the argument.
  • It is in the formula booklet.
Worked example

(1 + √3i)⁵

  • Polar form first.
  • Raise r, multiply θ.
  • Bring θ into range.
  1. 1 + √3i: x > 0, y > 0first quadrant
  2. r = √(1 + 3) = 2modulus
  3. θ = tan⁻¹(√3/1) = π/3argument
  4. (2 cis(π/3))⁵ = 2⁵ cis(5π/3)
  5. = 32 cis(5π/3)
  6. 5π/3 − 2π = −π/3more than π
  7. = 32 cis(−π/3)
Worked example

Back to a + bi

  • Exact cos and sin.
  • Check the modulus.
  1. (1 + √3i)⁵ = 32 cis(−π/3)
  2. cos(−π/3) = 1/2
  3. sin(−π/3) = −√3/2below the axis
  4. 32(1/2) + 32(−√3/2)i
  5. = 16 − 16√3i
  6. 16² + (16√3)² = 256 + 768check
  7. √1024 = 32
Key idea

Solving zⁿ = r cis θ

  • Let z = s cis α.
  • s is the positive nth root of r.
  • nα = θ + 2πk: n roots, 2π/n apart.
Worked example

z³ = 8i

  • Polar form of 8i.
  • Modulus: the cube root.
  • Arguments 2π/3 apart.
  1. 8i = 8 cis(π/2)positive Im axis
  2. s³ = 8, so s = ∛8 = 2modulus
  3. 3α = π/2 + 2πk
  4. α = π/6 + 2πk/3divide by 3
  5. k = 0: α = π/6
  6. k = 1: π/6 + 2π/3 = 5π/6
  7. k = 2: π/6 + 4π/3 = 3π/2
  8. 3π/2 − 2π = −π/2more than π
  9. 2cis(π/6), 2cis(5π/6), 2cis(−π/2)
Worked example

The cube roots of 8i

  • All on the circle of radius 2.
  • A third of a turn apart.
  • Equally spaced: a triangle.
The cube roots of 8i
Worked example

The 2025 error: the starting argument

  • Sketch: second quadrant.
  • Calculator: −π/3, wrong.
  • π − π/3 = 2π/3.
  1. −8 + 8√3i: x < 0, y > 0second quadrant
  2. r = √((−8)² + (8√3)²)modulus
  3. = √(64 + 192) = √256 = 16
  4. tan⁻¹(−√3) = −π/3fourth quadrant
  5. reference: tan⁻¹(√3) = π/3
  6. θ = π − π/3 = 2π/3second quadrant
  7. z⁴ = 16 cis(2π/3)
Worked example

The four roots

  • Modulus: ⁴√16 = 2.
  • A quarter turn apart.
  • Keep each one in range.
  1. z⁴ = 16 cis(2π/3)
  2. s⁴ = 16, so s = ⁴√16 = 2modulus
  3. 4α = 2π/3 + 2πk, α = π/6 + πk/2
  4. k = 0: α = π/6
  5. k = 1: π/6 + π/2 = 2π/3
  6. k = 2: π/6 + π = 7π/6
  7. 7π/6 − 2π = −5π/6more than π
  8. k = 3: π/6 + 3π/2 = 5π/3
  9. 5π/3 − 2π = −π/3more than π
Worked example

Check on the Argand diagram

  • Four roots on the circle of radius 2.
  • A quarter turn apart: a square.
  • A wrong start turns the whole square.
Check on the Argand diagram
Worked example

The roots in rectangular form

  • Exact cos and sin.
  • Answer in the form asked.
  1. 2 cis(π/6) = 2(√3/2 + i/2)
  2. = √3 + i
  3. 2 cis(2π/3) = 2(−1/2 + (√3/2)i)
  4. = −1 + √3i
  5. 2 cis(−5π/6) = 2(−√3/2 − i/2)
  6. = −√3 − i
  7. 2 cis(−π/3) = 2(1/2 − (√3/2)i)
  8. = 1 − √3i
Worked example

Check: a root to the fourth

  • Square, then square again.
  • It gives back the number.
  1. z⁴ = −8 + 8√3i
  2. (√3 + i)² = 3 + 2√3i + i²expand
  3. = 2 + 2√3ii² = −1
  4. (√3 + i)⁴ = (2 + 2√3i)²square again
  5. = 4 + 8√3i + 12i²expand
  6. = 4 + 8√3i − 12i² = −1
  7. = −8 + 8√3i
Key idea

Setting out a roots question

  • Sketch, then write it in polar form.
  • nth root of the modulus.
  • Argument with k; keep each in range.
Key idea

Where the marks are

  • Achieved: de Moivre for a power.
  • Merit: all the roots of zⁿ = a + bi.
  • Excellence: roots inside longer problems.
Summary

What to take away

  • Sketch first, get the starting argument right, then add 2π/n.
Powerrⁿ cis(nθ)
Sketch firstthe right argument
Root moduluspositive nth root
Arguments(θ + 2πk)/n
n rootsspaced 2π/n
Answerin the form asked
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • Sketch the number in 3 and 4 first.
  • Give every root.
GivenFind
1(−1 + i)⁶a + bi
2(2 cis(π/8))⁴a + bi
3z³ = −27all roots, polar
4z² = −2 − 2√3iboth roots, a + bi
Show answer

1. −1 + i = √2 cis(3π/4), so the power is 8 cis(9π/2) = 8 cis(π/2) = 8i. 2. 16 cis(π/2) = 16i. 3. −27 = 27 cis π: 3 cis(π/3), 3 cis π and 3 cis(−π/3). 4. −2 − 2√3i = 4 cis(−2π/3): 1 − √3i and −1 + √3i.

Worked example

The answers: questions 1 and 2

  1. 1. −1 + i: second quadrant
  2. r = √2, θ = π − π/4 = 3π/4
  3. (√2)⁶ cis(18π/4) = 8 cis(9π/2)de Moivre
  4. 9π/2 − 4π = π/2into range
  5. 8 cis(π/2) = 8(0 + i) = 8i
  6. 2. 2⁴ cis(4π/8) = 16 cis(π/2)de Moivre
  7. = 16(0 + i) = 16i
Worked example

The answers: question 3

  1. 3. −27 = 27 cis πnegative real axis
  2. s³ = 27, so s = 3modulus
  3. 3α = π + 2πk, α = π/3 + 2πk/3
  4. k = 0, 1, −1: π/3, π, −π/3all in range
  5. 3 cis(π/3), 3 cis π, 3 cis(−π/3)
  6. 3 cis π = −3
Worked example

The answers: question 4

  1. 4. −2 − 2√3i: third quadrant
  2. r = √(4 + 12) = 4modulus
  3. reference: tan⁻¹(2√3/2) = π/3
  4. θ = −(π − π/3) = −2π/3below the axis
  5. s² = 4, s = 2; 2α = −2π/3 + 2πk
  6. α = −π/3 + πk: −π/3, 2π/3k = 0, 1
  7. 2 cis(−π/3) = 1 − √3i
  8. 2 cis(2π/3) = −1 + √3i

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