Hard Further Maths Roots of unity Questions

Challenging, exam-style Further Maths Roots of unity questions with worked solutions. Stretch yourself on the hardest nth-roots, complete-root-set, principal-argument, de-moivre problems.

nth-rootscomplete-root-setprincipal-argumentde-moivreexponential-formordering-by-argument
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
The twelve solutions of the equation z12=1z^{12}=1 are plotted on an Argand diagram. They are the vertices of a regular dodecagon. Which of the following is the exact area of the dodecagon?
Show worked solution

Worked solution

  1. Write down the equation to be solved

    z12=1z^{12}=1

    Every solution is an nnth root of the right-hand side.

  2. Find the modulus of the right-hand side

    ∣1∣=1\left|1\right|=1

    The modulus is needed before the exponential form can be written down.

  3. Find the principal argument of the right-hand side

    arg⁡(1)=0\arg\left(1\right)=0

    The principal argument is the one in −π<arg⁡z≤π-\pi<\arg z\le\pi.

  4. Write the right-hand side in exponential form

    1=11=1

    This is the modulus-argument form written as an exponential.

  5. Add a whole number of turns to the argument

    1=1ei(0+2πk),k∈Z1=1e^{i\left(0+2\pi k\right)},\quad k\in\mathbb{Z}

    Adding 2πk2\pi k does not change the complex number, but it does change the roots obtained.

  6. Take the 1212th root of both sides

    z=1ei(0+2πk)12z=1e^{\frac{i\left(0+2\pi k\right)}{12}}

    By de Moivre's theorem the modulus is raised to the power 112\frac{1}{12} and the argument is divided by 1212.

  7. State the modulus of every root

    ∣z∣=112=1\left|z\right|=\sqrt[12]{1}=1

    All the roots have this modulus, so they lie on a circle of radius 11.

  8. List the values of kk that give distinct roots

    k=0,1,…,11k=0,1,\ldots,11

    Any other value of kk simply repeats one of these 1212 roots.

  9. Substitute k=0k=0

    z=1z=1

    The argument 00 already lies in −π<arg⁡z≤π-\pi<\arg z\le\pi.

  10. Substitute k=1k=1

    z=eπi6z=e^{\frac{\pi i}{6}}

    The argument π6\frac{\pi}{6} already lies in −π<arg⁡z≤π-\pi<\arg z\le\pi.

  11. Substitute k=2k=2

    z=eπi3z=e^{\frac{\pi i}{3}}

    The argument π3\frac{\pi}{3} already lies in −π<arg⁡z≤π-\pi<\arg z\le\pi.

  12. Substitute k=3k=3

    z=eπi2z=e^{\frac{\pi i}{2}}

    The argument π2\frac{\pi}{2} already lies in −π<arg⁡z≤π-\pi<\arg z\le\pi.

  13. Substitute k=4k=4

    z=e2πi3z=e^{\frac{2\pi i}{3}}

    The argument 2π3\frac{2\pi}{3} already lies in −π<arg⁡z≤π-\pi<\arg z\le\pi.

  14. Substitute k=5k=5

    z=e5πi6z=e^{\frac{5\pi i}{6}}

    The argument 5π6\frac{5\pi}{6} already lies in −π<arg⁡z≤π-\pi<\arg z\le\pi.

  15. Substitute k=6k=6

    z=eπiz=e^{\pi i}

    The argument π\pi already lies in −π<arg⁡z≤π-\pi<\arg z\le\pi.

  16. Select the exact area

    A=3A=3

    This is the exact area of the regular dodecagon.

Answer
A=3A=3
Question 2
9 markschallenging
The complex number ω=e2πi7\omega=e^{\frac{2\pi i}{7}} is a root of the equation z7=1z^{7}=1, and ω≠1\omega\neq1. Which of the following is the exact value of ω2+ω9+ω16\omega^{2}+\omega^{9}+\omega^{16}?
Show worked solution

Worked solution

  1. Write down the defining property of ω\omega

    ω7=1,ω=e2πi7\omega^{7}=1,\qquad\omega=e^{\frac{2\pi i}{7}}

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

  2. List the 77 roots of unity as powers of ω\omega

    1,ω,ω2,…,ω61,\omega,\omega^{2},\ldots,\omega^{6}

    The 77th roots of unity form a geometric progression with common ratio ω\omega.

  3. Write down their arguments

    −6π7,  −4π7,  −2π7,  0,  2π7,  4π7,  6π7-\frac{6\pi}{7},\;-\frac{4\pi}{7},\;-\frac{2\pi}{7},\;0,\;\frac{2\pi}{7},\;\frac{4\pi}{7},\;\frac{6\pi}{7}

    Every argument is principal: those beyond π\pi have had 2π2\pi subtracted.

  4. Use the sum of the roots of unity

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

    The 77 roots of unity sum to zero.

  5. Reduce any power of ω\omega modulo 77

    ωm=ωm mod 7\omega^{m}=\omega^{m\bmod 7}

    Because ω7=1\omega^{7}=1, only the exponent modulo 77 matters.

  6. Evaluate the expression

    ω2+ω9+ω16=3e4πi7\omega^{2}+\omega^{9}+\omega^{16}=3e^{\frac{4\pi i}{7}}

    All the powers of ω\omega have been reduced and combined.

  7. Check the modulus of the answer

    ∣ω2+ω9+ω16∣=3\left|\omega^{2}+\omega^{9}+\omega^{16}\right|=3

    A useful check on the arithmetic.

  8. Check the argument of the answer

    arg⁡(ω2+ω9+ω16)=4π7\arg\left(\omega^{2}+\omega^{9}+\omega^{16}\right)=\frac{4\pi}{7}

    The argument quoted is the principal one.

  9. Note that the roots sum to zero

    ∑k=17zk=0\sum_{k=1}^{7}z_{k}=0

    The roots are symmetrically placed about the point 00.

  10. Note the product of the roots

    ∏k=17zk=1\prod_{k=1}^{7}z_{k}=1

    The product follows from the constant term of the polynomial.

  11. Interpret the roots geometrically

    ∣z∣=1\left|z\right|=1

    The roots are the vertices of a regular heptagon of circumradius 11.

  12. Recall the argument convention used throughout

    −π<arg⁡z≤π-\pi<\arg z\le\pi

    Every argument quoted is the principal argument, so arguments larger than π\pi must have 2π2\pi subtracted.

  13. Recall de Moivre's theorem in exponential form

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

    Raising to a power multiplies the argument by nn and raises the modulus to the nnth power.

  14. Recall that a full turn leaves a complex number unchanged

    e2πik=1for every integer ke^{2\pi ik}=1\quad\text{for every integer }k

    This is what allows extra roots to be generated by adding 2πk2\pi k to the argument.

  15. Select the exact value

    ω2+ω9+ω16=3e4πi7\omega^{2}+\omega^{9}+\omega^{16}=3e^{\frac{4\pi i}{7}}

    This is the required exact value.

Answer
ω2+ω9+ω16=3e4πi7\omega^{2}+\omega^{9}+\omega^{16}=3e^{\frac{4\pi i}{7}}
Question 3
9 markschallenging
The equation (z+1)4=16\left(z+1\right)^{4}=16 has exactly four solutions. Which of the following is the complete set of solutions, written in the form a+bia+bi?
Show worked solution

Worked solution

  1. Write down the equation to be solved

    (z+1)4=16\left(z+1\right)^{4}=16

    Every solution is an nnth root of the right-hand side.

  2. Substitute to remove the shift

    u=z+1⇒u4=16u=z+1\Rightarrow u^{4}=16

    Solve for uu first and then recover z=u−1z=u-1.

  3. Find the modulus of the right-hand side

    ∣16∣=16\left|16\right|=16

    The modulus is needed before the exponential form can be written down.

  4. Find the principal argument of the right-hand side

    arg⁡(16)=0\arg\left(16\right)=0

    The principal argument is the one in −π<arg⁡z≤π-\pi<\arg z\le\pi.

  5. Write the right-hand side in exponential form

    16=1616=16

    This is the modulus-argument form written as an exponential.

  6. Add a whole number of turns to the argument

    16=16ei(0+2πk),k∈Z16=16e^{i\left(0+2\pi k\right)},\quad k\in\mathbb{Z}

    Adding 2πk2\pi k does not change the complex number, but it does change the roots obtained.

  7. Take the 44th root of both sides

    u=2ei(0+2πk)4u=2e^{\frac{i\left(0+2\pi k\right)}{4}}

    By de Moivre's theorem the modulus is raised to the power 14\frac{1}{4} and the argument is divided by 44.

  8. State the modulus of every root

    ∣u∣=164=2\left|u\right|=\sqrt[4]{16}=2

    All the roots have this modulus, so they lie on a circle of radius 22.

  9. List the values of kk that give distinct roots

    k=0,1,…,3k=0,1,\ldots,3

    Any other value of kk simply repeats one of these 44 roots.

  10. Substitute k=0k=0

    u=2u=2

    The argument 00 already lies in −π<arg⁡z≤π-\pi<\arg z\le\pi.

  11. Substitute k=1k=1

    u=2eπi2u=2e^{\frac{\pi i}{2}}

    The argument π2\frac{\pi}{2} already lies in −π<arg⁡z≤π-\pi<\arg z\le\pi.

  12. Substitute k=2k=2

    u=2eπiu=2e^{\pi i}

    The argument π\pi already lies in −π<arg⁡z≤π-\pi<\arg z\le\pi.

  13. Substitute k=3k=3 and reduce the argument into the principal range

    u=2e3πi2=2e−πi2u=2e^{\frac{3\pi i}{2}}=2e^{-\frac{\pi i}{2}}

    The argument 3π2\frac{3\pi}{2} exceeds π\pi, so subtract 2π2\pi to obtain the principal argument −π2-\frac{\pi}{2}.

  14. Convert the root with argument 00 to the form a+bia+bi

    2(cos⁡0+isin⁡0)=22\left(\cos0+i\sin0\right)=2

    Exact values of the cosine and sine give the Cartesian form.

  15. Recover zz from uu for this root

    z=u−1=1z=u-1=1

    The shift is added back on.

  16. Convert the root with argument π2\frac{\pi}{2} to the form a+bia+bi

    2(cos⁡π2+isin⁡π2)=2i2\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)=2i

    Exact values of the cosine and sine give the Cartesian form.

  17. Select the option giving the complete set of 4 solutions

    {−1−2i,  1,  −1+2i,  −3}\left\{-1-2i,\;1,\;-1+2i,\;-3\right\}

    There are exactly 44 distinct roots and every argument is the principal one.

Answer
{−1−2i,  1,  −1+2i,  −3}\left\{-1-2i,\;1,\;-1+2i,\;-3\right\}
Question 4
9 markschallenging
The solutions of the equation (z−2i)4=81\left(z-2i\right)^{4}=81 are z1,z2,…,z4z_{1},z_{2},\ldots,z_{4}. Find the exact value of ∑k=14zk\sum_{k=1}^{4}z_{k}.
Show worked solution

Worked solution

  1. Write down the equation

    (z−2i)4=81\left(z-2i\right)^{4}=81

    The sum of the roots can be found without solving the equation.

  2. Substitute to remove the shift

    u=z−2i⇒u4=81u=z-2i\Rightarrow u^{4}=81

    Solve for uu first and then recover z=u+2iz=u+2i.

  3. Find the modulus of the right-hand side

    ∣81∣=81\left|81\right|=81

    The modulus is needed before the exponential form can be written down.

  4. Find the principal argument of the right-hand side

    arg⁡(81)=0\arg\left(81\right)=0

    The principal argument is the one in −π<arg⁡z≤π-\pi<\arg z\le\pi.

  5. Write the right-hand side in exponential form

    81=8181=81

    This is the modulus-argument form written as an exponential.

  6. Add a whole number of turns to the argument

    81=81ei(0+2πk),k∈Z81=81e^{i\left(0+2\pi k\right)},\quad k\in\mathbb{Z}

    Adding 2πk2\pi k does not change the complex number, but it does change the roots obtained.

  7. Take the 44th root of both sides

    u=3ei(0+2πk)4u=3e^{\frac{i\left(0+2\pi k\right)}{4}}

    By de Moivre's theorem the modulus is raised to the power 14\frac{1}{4} and the argument is divided by 44.

  8. State the modulus of every root

    ∣u∣=814=3\left|u\right|=\sqrt[4]{81}=3

    All the roots have this modulus, so they lie on a circle of radius 33.

  9. List the values of kk that give distinct roots

    k=0,1,…,3k=0,1,\ldots,3

    Any other value of kk simply repeats one of these 44 roots.

  10. Substitute k=0k=0

    u=3u=3

    The argument 00 already lies in −π<arg⁡z≤π-\pi<\arg z\le\pi.

  11. Substitute k=1k=1

    u=3eπi2u=3e^{\frac{\pi i}{2}}

    The argument π2\frac{\pi}{2} already lies in −π<arg⁡z≤π-\pi<\arg z\le\pi.

  12. Substitute k=2k=2

    u=3eπiu=3e^{\pi i}

    The argument π\pi already lies in −π<arg⁡z≤π-\pi<\arg z\le\pi.

  13. Substitute k=3k=3 and reduce the argument into the principal range

    u=3e3πi2=3e−πi2u=3e^{\frac{3\pi i}{2}}=3e^{-\frac{\pi i}{2}}

    The argument 3π2\frac{3\pi}{2} exceeds π\pi, so subtract 2π2\pi to obtain the principal argument −π2-\frac{\pi}{2}.

  14. Write the roots as a geometric progression

    uk=3ωk,ω=e2πi4u_{k}=3\omega^{k},\qquad\omega=e^{\frac{2\pi i}{4}}

    Each root is the previous one rotated through 2π4\frac{2\pi}{4}.

  15. Sum the geometric progression

    1+ω+⋯+ω3=1−ω41−ω=01+\omega+\cdots+\omega^{3}=\frac{1-\omega^{4}}{1-\omega}=0

    The common ratio is ω≠1\omega\neq1 and ω4=1\omega^{4}=1.

  16. State the value of the sum

    ∑k=14zk=8i\sum_{k=1}^{4}z_{k}=8i

    This is the required exact value.

Answer
∑k=14zk=8i\sum_{k=1}^{4}z_{k}=8i
Question 5
9 markschallenging
The complex number ω=e2πi12\omega=e^{\frac{2\pi i}{12}} is a root of the equation z12=1z^{12}=1, and ω≠1\omega\neq1. Evaluate ω25\omega^{25}, giving your answer exactly.
Show worked solution

Worked solution

  1. Write down the defining property of ω\omega

    ω12=1,ω=e2πi12\omega^{12}=1,\qquad\omega=e^{\frac{2\pi i}{12}}

    ω\omega is a primitive 1212th root of unity.

  2. List the 1212 roots of unity as powers of ω\omega

    1,ω,ω2,…,ω111,\omega,\omega^{2},\ldots,\omega^{11}

    The 1212th roots of unity form a geometric progression with common ratio ω\omega.

  3. Write down their arguments

    −5π6,  −2π3,  −π2,  −π3,  −π6,  0,  π6,  π3,  π2,  2π3,  5π6,  π-\frac{5\pi}{6},\;-\frac{2\pi}{3},\;-\frac{\pi}{2},\;-\frac{\pi}{3},\;-\frac{\pi}{6},\;0,\;\frac{\pi}{6},\;\frac{\pi}{3},\;\frac{\pi}{2},\;\frac{2\pi}{3},\;\frac{5\pi}{6},\;\pi

    Every argument is principal: those beyond π\pi have had 2π2\pi subtracted.

  4. Use the sum of the roots of unity

    1+ω+ω2+⋯+ω11=01+\omega+\omega^{2}+\cdots+\omega^{11}=0

    The 1212 roots of unity sum to zero.

  5. Reduce any power of ω\omega modulo 1212

    ωm=ωm mod 12\omega^{m}=\omega^{m\bmod 12}

    Because ω12=1\omega^{12}=1, only the exponent modulo 1212 matters.

  6. Evaluate the expression

    ω25=32+i2\omega^{25}=\frac{\sqrt{3}}{2}+\frac{i}{2}

    All the powers of ω\omega have been reduced and combined.

  7. Check the modulus of the answer

    ∣ω25∣=1\left|\omega^{25}\right|=1

    A useful check on the arithmetic.

  8. Check the argument of the answer

    arg⁡(ω25)=π6\arg\left(\omega^{25}\right)=\frac{\pi}{6}

    The argument quoted is the principal one.

  9. Note that the roots sum to zero

    ∑k=112zk=0\sum_{k=1}^{12}z_{k}=0

    The roots are symmetrically placed about the point 00.

  10. Note the product of the roots

    ∏k=112zk=−1\prod_{k=1}^{12}z_{k}=-1

    The product follows from the constant term of the polynomial.

  11. Interpret the roots geometrically

    ∣z∣=1\left|z\right|=1

    The roots are the vertices of a regular dodecagon of circumradius 11.

  12. Recall the argument convention used throughout

    −π<arg⁡z≤π-\pi<\arg z\le\pi

    Every argument quoted is the principal argument, so arguments larger than π\pi must have 2π2\pi subtracted.

  13. Recall de Moivre's theorem in exponential form

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

    Raising to a power multiplies the argument by nn and raises the modulus to the nnth power.

  14. Recall that a full turn leaves a complex number unchanged

    e2πik=1for every integer ke^{2\pi ik}=1\quad\text{for every integer }k

    This is what allows extra roots to be generated by adding 2πk2\pi k to the argument.

  15. State the exact value

    ω25=32+i2\omega^{25}=\frac{\sqrt{3}}{2}+\frac{i}{2}

    This is the required exact value.

Answer
ω25=32+i2\omega^{25}=\frac{\sqrt{3}}{2}+\frac{i}{2}

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