Hard Further Maths Complex numbers: de Moivre Questions

Challenging, exam-style Further Maths Complex numbers: de Moivre questions with worked solutions. Stretch yourself on the hardest de-moivre, powers, cart, modarg problems.

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Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
De Moivre's theorem is being extended to negative integers. For a positive integer nn, which of the following is the simplified value of (cos⁡θ+isin⁡θ)−n\left(\cos\theta+i\sin\theta\right)^{-n}?
Show worked solution

Worked solution

  1. Write the negative power as a reciprocal

    (cos⁡θ+isin⁡θ)−n=1(cos⁡θ+isin⁡θ)n\left(\cos\theta+i\sin\theta\right)^{-n}=\frac{1}{\left(\cos\theta+i\sin\theta\right)^{n}}

    A negative index means the reciprocal of the positive power.

  2. Use de Moivre's theorem on the positive power

    =1cos⁡nθ+isin⁡nθ=\frac{1}{\cos n\theta+i\sin n\theta}

    The theorem has already been proved for positive integers.

  3. Multiply the numerator and the denominator by the conjugate

    =cos⁡nθ−isin⁡nθ(cos⁡nθ+isin⁡nθ)(cos⁡nθ−isin⁡nθ)=\frac{\cos n\theta-i\sin n\theta}{\left(\cos n\theta+i\sin n\theta\right)\left(\cos n\theta-i\sin n\theta\right)}

    This is the standard way of dividing by a complex number.

  4. Simplify the denominator

    cos⁡2nθ+sin⁡2nθ=1\cos^{2}n\theta+\sin^{2}n\theta=1

    The denominator is the squared modulus, which equals 11.

  5. Note the modulus of the base

    ∣cos⁡θ+isin⁡θ∣=1\left|\cos\theta+i\sin\theta\right|=1

    A unit modulus is what makes the denominator collapse to 11.

  6. Recall the conjugate of a unit complex number

    cos⁡α+isin⁡α‾=cos⁡α−isin⁡α\overline{\cos\alpha+i\sin\alpha}=\cos\alpha-i\sin\alpha

    The conjugate reflects the number in the real axis.

  7. Rewrite using even and odd symmetry

    cos⁡(−nθ)=cos⁡nθ,sin⁡(−nθ)=−sin⁡nθ\cos\left(-n\theta\right)=\cos n\theta,\quad\sin\left(-n\theta\right)=-\sin n\theta

    This shows the answer is exactly the de Moivre form with −n-n in place of nn.

  8. Conclude that the theorem holds for negative integers

    (cos⁡θ+isin⁡θ)m=cos⁡mθ+isin⁡mθ  ∀m∈Z\left(\cos\theta+i\sin\theta\right)^{m}=\cos m\theta+i\sin m\theta\ \ \forall m\in\mathbb{Z}

    The result therefore holds for every integer index.

  9. Quote de Moivre's theorem

    (cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ\left(\cos\theta+i\sin\theta\right)^{n}=\cos n\theta+i\sin n\theta

    This is the result quoted in the formula book; it holds for every integer nn.

  10. Recall modulus-argument form

    z=r(cos⁡θ+isin⁡θ)z=r\left(\cos\theta+i\sin\theta\right)

    Here r=∣z∣r=\left|z\right| and θ=arg⁡z\theta=\arg z.

  11. Recall the power rule in modulus-argument form

    [r(cos⁡θ+isin⁡θ)]n=rn(cos⁡nθ+isin⁡nθ)\left[r\left(\cos\theta+i\sin\theta\right)\right]^{n}=r^{n}\left(\cos n\theta+i\sin n\theta\right)

    The modulus is raised to the power and the argument is multiplied by it.

  12. Recall Euler's relation

    eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta

    Exponential form is the compact way of writing modulus-argument form.

  13. Recall the power rule in exponential form

    (reiθ)n=rneinθ\left(re^{i\theta}\right)^{n}=r^{n}e^{in\theta}

    A power of an exponential simply multiplies the exponent.

  14. Recall the effect of a power on the modulus

    ∣zn∣=∣z∣n\left|z^{n}\right|=\left|z\right|^{n}

    Moduli multiply, so a power of zz raises the modulus to that power.

  15. Select the correct simplified form

    (cos⁡θ+isin⁡θ)−n=cos⁡(nθ)−isin⁡(nθ)\left(\cos\theta+i\sin\theta\right)^{-n}=\cos\left(n\theta\right)-i\sin\left(n\theta\right)

    This is the reciprocal written in de Moivre form.

Answer
cos⁡(nθ)−isin⁡(nθ)\cos\left(n\theta\right)-i\sin\left(n\theta\right)
Question 2
9 markschallenging
The equation z5=32z^{5}=32 has 5 roots. Which of the following best describes the points representing these roots on an Argand diagram?
Show worked solution

Worked solution

  1. Find the modulus of every root

    ∣z∣=3215=2\left|z\right|=32^{\frac{1}{5}}=2

    All 55 roots share the same modulus, so they lie on a circle of radius 22.

  2. Find the spacing of the arguments

    2π5\frac{2\pi}{5}

    Consecutive roots differ in argument by exactly this angle.

  3. Recall the general root formula

    z=2(cos⁡(0+2kπ5)+isin⁡(0+2kπ5))z=2\left(\cos\left(\frac{0+2k\pi}{5}\right)+i\sin\left(\frac{0+2k\pi}{5}\right)\right)

    The 2kπ2k\pi is what spreads the roots evenly around the circle.

  4. Deduce the shape

    5 equally spaced points on a circle\text{5 equally spaced points on a circle}

    Equal radii and equal angular spacing give a regular polygon.

  5. Reject the option with radius 3232

    ∣z∣5=32 ⇒ ∣z∣=2\left|z\right|^{5}=32\ \Rightarrow\ \left|z\right|=2

    The radius is the 55th root of the modulus, not the modulus itself.

  6. Quote de Moivre's theorem

    (cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ\left(\cos\theta+i\sin\theta\right)^{n}=\cos n\theta+i\sin n\theta

    This is the result quoted in the formula book; it holds for every integer nn.

  7. Recall modulus-argument form

    z=r(cos⁡θ+isin⁡θ)z=r\left(\cos\theta+i\sin\theta\right)

    Here r=∣z∣r=\left|z\right| and θ=arg⁡z\theta=\arg z.

  8. Recall the power rule in modulus-argument form

    [r(cos⁡θ+isin⁡θ)]n=rn(cos⁡nθ+isin⁡nθ)\left[r\left(\cos\theta+i\sin\theta\right)\right]^{n}=r^{n}\left(\cos n\theta+i\sin n\theta\right)

    The modulus is raised to the power and the argument is multiplied by it.

  9. Recall Euler's relation

    eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta

    Exponential form is the compact way of writing modulus-argument form.

  10. Recall the power rule in exponential form

    (reiθ)n=rneinθ\left(re^{i\theta}\right)^{n}=r^{n}e^{in\theta}

    A power of an exponential simply multiplies the exponent.

  11. Recall the effect of a power on the modulus

    ∣zn∣=∣z∣n\left|z^{n}\right|=\left|z\right|^{n}

    Moduli multiply, so a power of zz raises the modulus to that power.

  12. Recall the effect of a power on the argument

    arg⁡(zn)=narg⁡z (mod 2π)\arg\left(z^{n}\right)=n\arg z\ \left(\text{mod }2\pi\right)

    Arguments add, so a power of zz multiplies the argument by nn.

  13. Check the principal argument range

    −π<arg⁡z≤π-\pi<\arg z\le\pi

    Add or subtract multiples of 2π2\pi until the angle lies in this interval.

  14. Recall the formula for the nnth roots

    z1n=r1n[cos⁡(θ+2kπn)+isin⁡(θ+2kπn)]z^{\frac{1}{n}}=r^{\frac{1}{n}}\left[\cos\left(\frac{\theta+2k\pi}{n}\right)+i\sin\left(\frac{\theta+2k\pi}{n}\right)\right]

    Taking k=0,1,…,n−1k=0,1,\dots,n-1 produces all nn distinct roots.

  15. Recall how the roots are arranged

    the n roots are spaced 2πn apart\text{the }n\text{ roots are spaced }\frac{2\pi}{n}\text{ apart}

    They are the vertices of a regular nn-gon centred at the origin.

  16. Select the correct description

    regular pentagon, centre 0, radius 2\text{regular pentagon, centre }0\text{, radius }2

    The roots are the vertices of a regular pentagon centred at the origin.

Answer
The vertices of a regular pentagon centred at the origin, inscribed in a circle of radius 22
Question 3
9 markschallenging
The complex number ω=e2iπ7\omega=e^{\frac{2 i \pi}{7}}. Find the exact value of ∑r=16ωr\sum_{r=1}^{6}\omega^{r}, giving your answer in exact form.
Show worked solution

Worked solution

  1. Identify ω\omega as a root of unity

    ω=e2iπ7,ω7=1\omega=e^{\frac{2 i \pi}{7}},\qquad \omega^{7}=1

    ω\omega is a primitive 77th root of unity.

  2. Recognise a geometric series with a complex common ratio

    ∑r=16wrwithw complex\sum_{r=1}^{6}w^{r}\quad\text{with}\quad w\ \text{complex}

    De Moivre's theorem turns each term into a power of a single complex number.

  3. Write out the terms of the series

    e2iπ7,e4iπ7,e6iπ7, …, e−2iπ7e^{\frac{2 i \pi}{7}},\quad e^{\frac{4 i \pi}{7}},\quad e^{\frac{6 i \pi}{7}},\ \dots,\ e^{- \frac{2 i \pi}{7}}

    There are 66 terms in the sum.

  4. Add the terms

    ∑r=16ωr=−1\sum_{r=1}^{6}\omega^{r}=-1

    The imaginary parts cancel in pairs where the arguments are symmetric.

  5. Recall that the roots of unity sum to zero

    1+ω+ω2+⋯+ω6=01+\omega+\omega^{2}+\cdots+\omega^{6}=0

    They are the vertices of a regular 77-gon centred at the origin.

  6. Recall the geometric sum formula

    ∑r=0mwr=wm+1−1w−1\sum_{r=0}^{m}w^{r}=\frac{w^{m+1}-1}{w-1}

    This is valid for a complex ratio ww provided w≠1w\neq1.

  7. Use de Moivre's theorem on the common ratio

    wr=∣w∣r(cos⁡rθ+isin⁡rθ)w^{r}=\left|w\right|^{r}\left(\cos r\theta+i\sin r\theta\right)

    Each power is found by raising the modulus and multiplying the argument.

  8. Note the symmetry of the arguments

    the arguments are equally spaced around the origin\text{the arguments are equally spaced around the origin}

    Equally spaced unit vectors sum to zero, which is why so much cancels.

  9. Check the result numerically

    ∑r=16ωr≈−1.0\sum_{r=1}^{6}\omega^{r}\approx -1.0

    The decimal value agrees with the exact answer.

  10. Take the real part if only a cosine sum is required

    ∑cos⁡rθ=Re⁡∑wr\sum\cos r\theta=\operatorname{Re}\sum w^{r}

    The cosine series is the real part of the complex geometric series.

  11. Quote de Moivre's theorem

    (cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ\left(\cos\theta+i\sin\theta\right)^{n}=\cos n\theta+i\sin n\theta

    This is the result quoted in the formula book; it holds for every integer nn.

  12. Recall modulus-argument form

    z=r(cos⁡θ+isin⁡θ)z=r\left(\cos\theta+i\sin\theta\right)

    Here r=∣z∣r=\left|z\right| and θ=arg⁡z\theta=\arg z.

  13. Recall the power rule in modulus-argument form

    [r(cos⁡θ+isin⁡θ)]n=rn(cos⁡nθ+isin⁡nθ)\left[r\left(\cos\theta+i\sin\theta\right)\right]^{n}=r^{n}\left(\cos n\theta+i\sin n\theta\right)

    The modulus is raised to the power and the argument is multiplied by it.

  14. Recall Euler's relation

    eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta

    Exponential form is the compact way of writing modulus-argument form.

  15. State the exact value

    ∑r=16ωr=−1\sum_{r=1}^{6}\omega^{r}=-1

    This is the exact sum of the series.

Answer
−1-1
Question 4
9 markschallenging
Find the exact value of ∑r=05cos⁡(rπ3)2r\sum_{r=0}^{5}\frac{\cos\left(\frac{r\pi}{3}\right)}{2^{r}}, giving your answer in exact form.
Show worked solution

Worked solution

  1. Recognise a geometric series with a complex common ratio

    ∑r=05wrwithw complex\sum_{r=0}^{5}w^{r}\quad\text{with}\quad w\ \text{complex}

    De Moivre's theorem turns each term into a power of a single complex number.

  2. Write out the terms of the series

    1,14,−18, …, 1641,\quad \frac{1}{4},\quad - \frac{1}{8},\ \dots,\ \frac{1}{64}

    There are 66 terms in the sum.

  3. Add the terms

    ∑r=05cos⁡(rπ3)2r=6364\sum_{r=0}^{5}\frac{\cos\left(\frac{r\pi}{3}\right)}{2^{r}}=\frac{63}{64}

    The imaginary parts cancel in pairs where the arguments are symmetric.

  4. Recall the geometric sum formula

    ∑r=0mwr=wm+1−1w−1\sum_{r=0}^{m}w^{r}=\frac{w^{m+1}-1}{w-1}

    This is valid for a complex ratio ww provided w≠1w\neq1.

  5. Use de Moivre's theorem on the common ratio

    wr=∣w∣r(cos⁡rθ+isin⁡rθ)w^{r}=\left|w\right|^{r}\left(\cos r\theta+i\sin r\theta\right)

    Each power is found by raising the modulus and multiplying the argument.

  6. Note the symmetry of the arguments

    the arguments are equally spaced around the origin\text{the arguments are equally spaced around the origin}

    Equally spaced unit vectors sum to zero, which is why so much cancels.

  7. Check the result numerically

    ∑r=05cos⁡(rπ3)2r≈0.984375\sum_{r=0}^{5}\frac{\cos\left(\frac{r\pi}{3}\right)}{2^{r}}\approx 0.984375

    The decimal value agrees with the exact answer.

  8. Take the real part if only a cosine sum is required

    ∑cos⁡rθ=Re⁡∑wr\sum\cos r\theta=\operatorname{Re}\sum w^{r}

    The cosine series is the real part of the complex geometric series.

  9. Quote de Moivre's theorem

    (cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ\left(\cos\theta+i\sin\theta\right)^{n}=\cos n\theta+i\sin n\theta

    This is the result quoted in the formula book; it holds for every integer nn.

  10. Recall modulus-argument form

    z=r(cos⁡θ+isin⁡θ)z=r\left(\cos\theta+i\sin\theta\right)

    Here r=∣z∣r=\left|z\right| and θ=arg⁡z\theta=\arg z.

  11. Recall the power rule in modulus-argument form

    [r(cos⁡θ+isin⁡θ)]n=rn(cos⁡nθ+isin⁡nθ)\left[r\left(\cos\theta+i\sin\theta\right)\right]^{n}=r^{n}\left(\cos n\theta+i\sin n\theta\right)

    The modulus is raised to the power and the argument is multiplied by it.

  12. Recall Euler's relation

    eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta

    Exponential form is the compact way of writing modulus-argument form.

  13. Recall the power rule in exponential form

    (reiθ)n=rneinθ\left(re^{i\theta}\right)^{n}=r^{n}e^{in\theta}

    A power of an exponential simply multiplies the exponent.

  14. Recall the effect of a power on the modulus

    ∣zn∣=∣z∣n\left|z^{n}\right|=\left|z\right|^{n}

    Moduli multiply, so a power of zz raises the modulus to that power.

  15. Recall the effect of a power on the argument

    arg⁡(zn)=narg⁡z (mod 2π)\arg\left(z^{n}\right)=n\arg z\ \left(\text{mod }2\pi\right)

    Arguments add, so a power of zz multiplies the argument by nn.

  16. State the exact value

    ∑r=05cos⁡(rπ3)2r=6364\sum_{r=0}^{5}\frac{\cos\left(\frac{r\pi}{3}\right)}{2^{r}}=\frac{63}{64}

    This is the exact sum of the series.

Answer
6364\frac{63}{64}
Question 5
9 markschallenging
Use the substitution z=cos⁡θ+isin⁡θz=\cos\theta+i\sin\theta to express sin⁡6θ\sin^{6}\theta in terms of cosines of multiples of θ\theta.
Show worked solution

Worked solution

  1. Set z=cos⁡θ+isin⁡θz=\cos\theta+i\sin\theta and use de Moivre's theorem

    zm+1zm=2cos⁡mθ,zm−1zm=2isin⁡mθz^{m}+\frac{1}{z^{m}}=2\cos m\theta,\qquad z^{m}-\frac{1}{z^{m}}=2i\sin m\theta

    De Moivre's theorem gives zm=cos⁡mθ+isin⁡mθz^{m}=\cos m\theta+i\sin m\theta and z−m=cos⁡mθ−isin⁡mθz^{-m}=\cos m\theta-i\sin m\theta.

  2. Take the case m=1m=1

    z−1z=2isin⁡θz-\frac{1}{z}=2i\sin\theta

    This is the expression that will be raised to the power 66.

  3. Raise both sides to the power 66

    (z−1z)6=(2isin⁡θ)6\left(z-\frac{1}{z}\right)^{6}=\left(2i\sin\theta\right)^{6}

    The left-hand side is expanded with the binomial theorem.

  4. Expand the left-hand side with the binomial theorem

    (z−1z)6=z6−6z4+15z2−20+151z2−61z4+1z6\left(z-\frac{1}{z}\right)^{6}=z^{6}-6z^{4}+15z^{2}-20+15\frac{1}{z^{2}}-6\frac{1}{z^{4}}+\frac{1}{z^{6}}

    Row 66 of Pascal's triangle supplies the coefficients.

  5. Pair the terms zmz^{m} and z−mz^{-m}

    z6+1z6=2cos⁡6θ,z4+1z4=2cos⁡4θz^{6}+\frac{1}{z^{6}}=2\cos6\theta,\qquad z^{4}+\frac{1}{z^{4}}=2\cos4\theta

    Each pair collapses to a multiple angle by the result quoted above.

  6. Divide by the constant on the right-hand side

    sin⁡6θ=132(10−15cos⁡2θ+6cos⁡4θ−cos⁡6θ)\sin^{6}\theta=\frac{1}{32}\left(10-15\cos2\theta+6\cos4\theta-\cos6\theta\right)

    Dividing by (2i)6\left(2i\right)^{6} isolates the required power.

  7. Note the number of terms in the expansion

    7 terms7\ \text{terms}

    A 66th power gives 77 terms before any pairing.

  8. Note that only cosines survive

    the answer contains only cosines of multiples of θ\text{the answer contains only cosines of multiples of }\theta

    The symmetry of the pairing removes the other trigonometric function.

  9. Check the identity at θ=0\theta=0

    sin⁡60=0\sin^{6}0=0

    Both sides agree at θ=0\theta=0.

  10. Check the identity at θ=π2\theta=\frac{\pi}{2}

    sin⁡6π2=1\sin^{6}\frac{\pi}{2}=1

    The multiple-angle form reproduces the same value.

  11. Note the unpaired middle term

    (63)=20\binom{6}{3}=20

    An even power leaves an unpaired middle term, which becomes the constant.

  12. Quote de Moivre's theorem

    (cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ\left(\cos\theta+i\sin\theta\right)^{n}=\cos n\theta+i\sin n\theta

    This is the result quoted in the formula book; it holds for every integer nn.

  13. Recall modulus-argument form

    z=r(cos⁡θ+isin⁡θ)z=r\left(\cos\theta+i\sin\theta\right)

    Here r=∣z∣r=\left|z\right| and θ=arg⁡z\theta=\arg z.

  14. Recall the power rule in modulus-argument form

    [r(cos⁡θ+isin⁡θ)]n=rn(cos⁡nθ+isin⁡nθ)\left[r\left(\cos\theta+i\sin\theta\right)\right]^{n}=r^{n}\left(\cos n\theta+i\sin n\theta\right)

    The modulus is raised to the power and the argument is multiplied by it.

  15. State the identity

    sin⁡6θ=132(10−15cos⁡2θ+6cos⁡4θ−cos⁡6θ)\sin^{6}\theta=\frac{1}{32}\left(10-15\cos2\theta+6\cos4\theta-\cos6\theta\right)

    This is sin⁡6θ\sin^{6}\theta written using cosines of multiples of θ\theta.

Answer
132(10−15cos⁡2θ+6cos⁡4θ−cos⁡6θ)\frac{1}{32}\left(10-15\cos2\theta+6\cos4\theta-\cos6\theta\right)

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