De Moivre's theorem: 7 exam questions solved step by step

← Previous9 of 12Next →

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

7 questions, one theorem

  • 1. Polar form first: sketch it.
  • 2. (r cis θ)ⁿ = rⁿ cis(nθ).
  • 3. Roots: nth root of r, (θ + 2πk)/n.
  • 4. Bring every argument into range.
Your turn

Question 1 · Achieved

  • Write (3 cis(π/12))⁴
  • in the form a + bi, exactly.
Worked example

Q1: (3 cis(π/12))⁴

  • Raise r, multiply θ.
  • Then exact values.
  1. (3 cis(π/12))⁴
  2. = 3⁴ cis(4 × π/12)de Moivre
  3. = 81 cis(π/3)
  4. cos(π/3) = 1/2, sin(π/3) = √3/2
  5. 81(1/2) + 81(√3/2)i
  6. = 81/2 + (81√3/2)i
Your turn

Question 2 · Achieved

  • Use de Moivre to write (√3 − i)⁴
  • in the form a + bi, exactly.
Worked example

Q2: (√3 − i)⁴

  • Below the axis: θ < 0.
  • Exact cos and sin.
  1. √3 − i: x > 0, y < 0fourth quadrant
  2. r = √(3 + 1) = 2modulus
  3. θ = −tan⁻¹(1/√3) = −π/6below the axis
  4. (2 cis(−π/6))⁴ = 2⁴ cis(−4π/6)
  5. = 16 cis(−2π/3)
  6. cos(−2π/3) = −1/2
  7. sin(−2π/3) = −√3/2
  8. 16(−1/2) + 16(−√3/2)i
  9. = −8 − 8√3i
Your turn

Question 3 · Merit

  • Use de Moivre to write (−2 + 2i)⁵
  • in the form a + bi.
Worked example

Q3: (−2 + 2i)⁵

  • Second quadrant.
  • Subtract 2π twice.
  1. −2 + 2i: x < 0, y > 0second quadrant
  2. r = √(4 + 4) = √8 = 2√2modulus
  3. θ = π − π/4 = 3π/4reference π/4
  4. (2√2)⁵ = 32 × 4√2 = 128√2
  5. arg: 5 × 3π/4 = 15π/4de Moivre
  6. 15π/4 − 4π = −π/4subtract 2π twice
  7. = 128√2 cis(−π/4)
  8. 128√2(√2/2) − 128√2(√2/2)i
  9. = 128 − 128i
Your turn

Question 4 · Merit

  • Solve z⁴ = −16.
  • Give the roots in the form a + bi,
  • plot them, and find the side of
  • the shape they make.
Worked example

Q4: the four arguments

  • Let z = s cis α.
  • Four values of k.
  1. −16 = 16 cis πnegative real axis
  2. s⁴ = 16, so s = 2modulus
  3. 4α = π + 2πk
  4. α = π/4 + πk/2divide by 4
  5. k = 0: π/4, k = 1: 3π/4
  6. k = 2: 5π/4 − 2π = −3π/4into range
  7. k = 3: 7π/4 − 2π = −π/4into range
  8. z = 2 cis α, all four α
Worked example

Q4: check on the Argand diagram

  • All on the circle |z| = 2.
  • A quarter turn apart.
Q4: check on the Argand diagram
Worked example

Q4: a + bi, and the square

  • Exact cos and sin.
  • Answer with units.
  1. 2 cis(π/4) = √2 + √2i
  2. 2 cis(3π/4) = −√2 + √2i
  3. 2 cis(−3π/4) = −√2 − √2i
  4. 2 cis(−π/4) = √2 − √2i
  5. (√2 + √2i)² = 2 + 4i − 2 = 4i
  6. (4i)² = 16i² = −16
  7. side: √2 − (−√2) = 2√2top two roots
  8. a square of side 2√2 units
Your turn

Question 5 · Merit

  • Solve z³ = −4√3 + 4i.
  • Give every root in polar form.
  • Careful: sketch before tan⁻¹.
Worked example

Q5: the starting argument

  • Second quadrant.
  • Not the calculator angle.
  1. −4√3 + 4i: x < 0, y > 0second quadrant
  2. r = √(48 + 16) = √64 = 8modulus
  3. tan⁻¹(4/(−4√3)) = −π/6fourth quadrant
  4. reference: tan⁻¹(1/√3) = π/6
  5. θ = π − π/6 = 5π/6second quadrant
  6. z³ = 8 cis(5π/6)
Worked example

Q5: the three roots

  • Cube root of 8.
  • 2π/3 = 12π/18 apart.
  1. z³ = 8 cis(5π/6)
  2. s³ = 8, so s = 2modulus
  3. 3α = 5π/6 + 2πk
  4. α = 5π/18 + 2πk/3divide by 3
  5. k = 0: α = 5π/18
  6. k = 1: 5π/18 + 12π/18 = 17π/18
  7. k = −1: 5π/18 − 12π/18 = −7π/18
  8. z = 2 cis(5π/18), 2 cis(17π/18)
  9. or z = 2 cis(−7π/18)
Your turn

Question 6 · Excellence

  • Find the smallest positive integer n
  • for which (−√3 + i)ⁿ is real.
  • Give the value of that power.
Worked example

Q6: when is it real?

  • Real: imaginary part 0.
  • Reason, do not guess.
  1. −√3 + i: x < 0, y > 0second quadrant
  2. r = √(3 + 1) = 2modulus
  3. θ = π − π/6 = 5π/6reference π/6
  4. (2 cis(5π/6))ⁿ = 2ⁿ cis(5nπ/6)
  5. sin(5nπ/6) = 0imaginary part 0
  6. 5nπ/6 = kπ, so 5n = 6k
  7. 6 divides 5n, so 6 divides n5 and 6: no common factor
  8. smallest n = 6
Worked example

Q6: the value

  • n = 6 in de Moivre.
  • Smaller n: not real.
  1. 2⁶ cis(6 × 5π/6) = 64 cis(5π)
  2. 5π − 4π = πsubtract 2π twice
  3. 64 cis π = 64(−1 + 0i) = −64
  4. n = 3: 8 cis(5π/2) = 8iimaginary, not real
Your turn

Question 7 · Excellence

  • Solve (z + 2i)³ = −27i.
  • Give every solution in the form a + bi.
Worked example

Q7: substitute, then roots

  • Let w = z + 2i.
  • Cube roots of −27i.
  1. w = z + 2i: w³ = −27i
  2. −27i = 27 cis(−π/2)negative Im axis
  3. s³ = 27, so s = 3modulus
  4. 3α = −π/2 + 2πk
  5. α = −π/6 + 2πk/3divide by 3
  6. k = 0: α = −π/6
  7. k = 1: −π/6 + 4π/6 = π/2
  8. k = −1: −π/6 − 4π/6 = −5π/6
Worked example

Q7: back to z

  • w first, in a + bi.
  • Then z = w − 2i.
  1. w = 3 cis(−π/6) = 3√3/2 − (3/2)i
  2. w = 3 cis(π/2) = 3i
  3. w = 3 cis(−5π/6) = −3√3/2 − (3/2)i
  4. z = w − 2iundo the substitution
  5. z = 3√3/2 − (7/2)i
  6. z = i
  7. z = −3√3/2 − (7/2)i
  8. z = i: (3i)³ = 27i³ = −27i
Summary

The method that solved all seven

  • Polar form from a sketch, de Moivre, every k, into range.
StepHow
Polar formsketch, then r and θ
Powerrⁿ cis(nθ)
Into range± 2π until −π < θ ≤ π
Rootsnth root of r
Arguments(θ + 2πk)/n, all n
Checkcircle, equal spacing

More in AS 91577

Complex arithmetic, conjugates and equating parts7 minPolar form by hand: modulus, argument and quadrants7 minDe Moivre's theorem: powers and roots of complex numbers6 minComplex roots of polynomials: factor theorem and conjugates6 minAlgebra with k: quadratics, discriminants and surd equations6 minComplex loci, purely real conditions and proofs7 minComplex arithmetic: 7 exam questions solved step by step11 minPolar form: 7 exam questions solved step by step11 minDe Moivre's theorem: 7 exam questions solved step by step11 minComplex polynomials: 7 exam questions solved step by step12 minAlgebra with constants: 7 exam questions solved step by step12 minComplex loci and proof: 7 exam questions solved step by step12 min

Is your child on track?

Take the free placement test