Further Maths Roots of unity Practice Questions

Free Further Maths Roots of unity practice questions with full step-by-step worked solutions. Covers nth-roots, modulus, circle-of-roots, roots-of-unity. Practise exam-style problems and check your method.

nth-rootsmoduluscircle-of-rootsroots-of-unitysum-of-rootsgeometric-series
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Every solution of the equation z4=16z^{4}=16 has the same modulus ∣z∣\left|z\right|. Find its exact value.
Show worked solution

Worked solution

  1. Write down the equation

    z4=16z^{4}=16

    The modulus of each root can be found without finding the roots themselves.

  2. Take the modulus of both sides

    ∣z4∣=∣16∣=16\left|z^{4}\right|=\left|16\right|=16

    The modulus of a product is the product of the moduli.

  3. State the modulus

    ∣z∣=2\left|z\right|=2

    This is the common modulus of all 44 roots.

Answer
∣z∣=2\left|z\right|=2
Question 2
2 markseasy
The solutions of the equation z6=1z^{6}=1 are labelled z1,z2,…,z6z_{1},z_{2},\ldots,z_{6} so that arg⁡z1<arg⁡z2<⋯<arg⁡z6\arg z_{1}<\arg z_{2}<\cdots<\arg z_{6}, where every argument is the principal argument, −π<arg⁡z≤π-\pi<\arg z\le\pi. Find arg⁡z5\arg z_{5}, giving your answer as an exact multiple of π\pi.
Show worked solution

Worked solution

  1. Write down the equation to be solved

    z6=1z^{6}=1

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

  2. Write the right-hand side in exponential form, allowing a whole number of extra turns

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

    Adding 2πk2\pi k leaves the number unchanged but produces the other roots.

  3. State the argument

    arg⁡z5=2π3\arg z_{5}=\frac{2\pi}{3}

    The value lies in −π<arg⁡z≤π-\pi<\arg z\le\pi, as required.

Answer
arg⁡z5=2π3\arg z_{5}=\frac{2\pi}{3}
Question 3
4 marksintermediate
The solutions of the equation z4=16z^{4}=16 are labelled z1,z2,…,z4z_{1},z_{2},\ldots,z_{4} so that arg⁡z1<arg⁡z2<⋯<arg⁡z4\arg z_{1}<\arg z_{2}<\cdots<\arg z_{4}, where every argument is the principal argument, −π<arg⁡z≤π-\pi<\arg z\le\pi. Find z3z_{3}, giving your answer in exponential form reiθre^{i\theta} with −π<θ≤π-\pi<\theta\le\pi.
Show worked solution

Worked solution

  1. Write down the equation to be solved

    z4=16z^{4}=16

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

  2. 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.

  3. 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.

  4. Write the right-hand side in exponential form

    16=1616=16

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

  5. 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.

  6. State z3z_{3}

    z3=2eπi2z_{3}=2e^{\frac{\pi i}{2}}

    This is the root whose argument is the 33th smallest.

Answer
z3=2eπi2z_{3}=2e^{\frac{\pi i}{2}}
Question 4
6 markshard
The solutions of the equation z5=−32z^{5}=-32 are labelled z1,z2,…,z5z_{1},z_{2},\ldots,z_{5} so that arg⁡z1<arg⁡z2<⋯<arg⁡z5\arg z_{1}<\arg z_{2}<\cdots<\arg z_{5}, where every argument is the principal argument, −π<arg⁡z≤π-\pi<\arg z\le\pi. Find z4z_{4}, giving your answer in exponential form reiθre^{i\theta} with −π<θ≤π-\pi<\theta\le\pi.
Show worked solution

Worked solution

  1. Write down the equation to be solved

    z5=−32z^{5}=-32

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

  2. Find the modulus of the right-hand side

    ∣−32∣=32\left|-32\right|=32

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

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

    arg⁡(−32)=π\arg\left(-32\right)=\pi

    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

    −32=32eπi-32=32e^{\pi i}

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

  5. Add a whole number of turns to the argument

    −32=32ei(π+2πk),k∈Z-32=32e^{i\left(\pi+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 55th root of both sides

    z=2ei(π+2πk)5z=2e^{\frac{i\left(\pi+2\pi k\right)}{5}}

    By de Moivre's theorem the modulus is raised to the power 15\frac{1}{5} and the argument is divided by 55.

  7. State the modulus of every root

    ∣z∣=325=2\left|z\right|=\sqrt[5]{32}=2

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

  8. List the values of kk that give distinct roots

    k=0,1,…,4k=0,1,\ldots,4

    Any other value of kk simply repeats one of these 55 roots.

  9. Substitute k=0k=0

    z=2eπi5z=2e^{\frac{\pi i}{5}}

    The argument π5\frac{\pi}{5} already lies in −π<arg⁡z≤π-\pi<\arg z\le\pi.

  10. Substitute k=1k=1

    z=2e3πi5z=2e^{\frac{3\pi i}{5}}

    The argument 3π5\frac{3\pi}{5} already lies in −π<arg⁡z≤π-\pi<\arg z\le\pi.

  11. State z4z_{4}

    z4=2e3πi5z_{4}=2e^{\frac{3\pi i}{5}}

    This is the root whose argument is the 44th smallest.

Answer
z4=2e3πi5z_{4}=2e^{\frac{3\pi i}{5}}
Question 5
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

Unlock 65 more Roots of unity questions

Create a free account to work through every Further Maths Roots of unity question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Roots of unity practice

Hard QuestionsWorked Solutions

Related Pure Maths topics

Argand diagramsComplex numbers: arithmeticModelling with differential equationsComplex numbers: de MoivreFirst-order differential equationsFurther calculus