Roots of unity Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Roots of unity questions. See exactly how to solve problems on nth-roots, modulus, circle-of-roots, roots-of-unity.

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.

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 z3=1z^{3}=1 are z1,z2,…,z3z_{1},z_{2},\ldots,z_{3}. Find the exact value of ∑k=13zk\sum_{k=1}^{3}z_{k}.

Worked solution

  1. Write down the equation

    z3=1z^{3}=1

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

  2. Write the equation as a polynomial equation

    z3−1=0z^{3}-1=0

    The 33 roots are the roots of this polynomial.

  3. Compare with the general polynomial of degree 33

    z3+a2z2+⋯+a0=0z^{3}+a_{2}z^{2}+\cdots+a_{0}=0

    The sum of the roots is −a2-a_{2}.

  4. State the value of the sum

    ∑k=13zk=0\sum_{k=1}^{3}z_{k}=0

    This is the required exact value.

Answer
∑k=13zk=0\sum_{k=1}^{3}z_{k}=0
Question 3
2 markseasy
The equation z3=1z^{3}=1 has exactly three solutions. Which of the following is the complete set of solutions, written in the form a+bia+bi?

Worked solution

  1. Write down the equation to be solved

    z3=1z^{3}=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. Select the option giving the complete set of 3 solutions

    {−12−3i2,  1,  −12+3i2}\left\{-\frac{1}{2}-\frac{\sqrt{3}i}{2},\;1,\;-\frac{1}{2}+\frac{\sqrt{3}i}{2}\right\}

    There are exactly 33 distinct roots and every argument is the principal one.

Answer
{−12−3i2,  1,  −12+3i2}\left\{-\frac{1}{2}-\frac{\sqrt{3}i}{2},\;1,\;-\frac{1}{2}+\frac{\sqrt{3}i}{2}\right\}
Question 4
2 markseasy
The solutions of the equation z4=1z^{4}=1 are z1,z2,…,z4z_{1},z_{2},\ldots,z_{4}. Find the exact value of ∏k=14zk\prod_{k=1}^{4}z_{k}.

Worked solution

  1. Write down the equation

    z4=1z^{4}=1

    The product of the roots follows from the constant term.

  2. Write the equation as a polynomial equation

    z4−1=0z^{4}-1=0

    The 44 roots are the roots of this polynomial.

  3. State the value of the product

    ∏k=14zk=−1\prod_{k=1}^{4}z_{k}=-1

    This is the required exact value.

Answer
∏k=14zk=−1\prod_{k=1}^{4}z_{k}=-1
Question 5
2 markseasy
The solutions of the equation z3=1z^{3}=1 are labelled z1,z2,…,z3z_{1},z_{2},\ldots,z_{3} so that arg⁡z1<arg⁡z2<⋯<arg⁡z3\arg z_{1}<\arg z_{2}<\cdots<\arg z_{3}, where every argument is the principal argument, −π<arg⁡z≤π-\pi<\arg z\le\pi. Find arg⁡z3\arg z_{3}, giving your answer as an exact multiple of π\pi.

Worked solution

  1. Write down the equation to be solved

    z3=1z^{3}=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. Take the 33th root of both sides

    z=1ei(0+2πk)3,k=0,1,…,2z=1e^{\frac{i\left(0+2\pi k\right)}{3}},\quad k=0,1,\ldots,2

    The 33 values k=0,1,…,2k=0,1,\ldots,2 give the 33 distinct roots.

  4. State the argument

    arg⁡z3=2π3\arg z_{3}=\frac{2\pi}{3}

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

Answer
arg⁡z3=2π3\arg z_{3}=\frac{2\pi}{3}

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