Further Maths Complex numbers: de Moivre Practice Questions

Free Further Maths Complex numbers: de Moivre practice questions with full step-by-step worked solutions. Covers de-moivre, powers, modarg, cart. Practise exam-style problems and check your method.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Use de Moivre's theorem to evaluate (cos⁡(π6)+isin⁡(π6))3\left(\cos\left(\frac{\pi}{6}\right)+i\sin\left(\frac{\pi}{6}\right)\right)^{3}, giving your answer in the form a+bia+bi.
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Worked solution

  1. Read off the modulus and the argument

    r=1,θ=π6r=1,\qquad\theta=\frac{\pi}{6}

    The base is already written in modulus-argument form.

  2. Apply de Moivre's theorem

    [1(cos⁡(π6)+isin⁡(π6))]3=13(cos⁡(3×π6)+isin⁡(3×π6))\left[1\left(\cos\left(\frac{\pi}{6}\right)+i\sin\left(\frac{\pi}{6}\right)\right)\right]^{3}=1^{3}\left(\cos\left(3\times \frac{\pi}{6}\right)+i\sin\left(3\times \frac{\pi}{6}\right)\right)

    Raise the modulus to the power 33 and multiply the argument by 33.

  3. State the answer in the form a+bia+bi

    (cos⁡(π6)+isin⁡(π6))3=i\left(\cos\left(\frac{\pi}{6}\right)+i\sin\left(\frac{\pi}{6}\right)\right)^{3}=i

    This is the exact value of the power.

Answer
i
Question 2
2 markseasy
De Moivre's theorem is being proved by induction. Assuming that (cos⁡θ+isin⁡θ)k=cos⁡kθ+isin⁡kθ\left(\cos\theta+i\sin\theta\right)^{k}=\cos k\theta+i\sin k\theta, which of the following is the correct simplified value of (cos⁡θ+isin⁡θ)k+1\left(\cos\theta+i\sin\theta\right)^{k+1}?
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Worked solution

  1. State the inductive assumption

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

    Assume the result is true for the positive integer n=kn=k.

  2. Write the (k+1)\left(k+1\right)th power as a product

    (cos⁡θ+isin⁡θ)k+1=(cos⁡θ+isin⁡θ)k(cos⁡θ+isin⁡θ)\left(\cos\theta+i\sin\theta\right)^{k+1}=\left(\cos\theta+i\sin\theta\right)^{k}\left(\cos\theta+i\sin\theta\right)

    Split off one factor so that the assumption can be used.

  3. Substitute the inductive assumption

    =(cos⁡kθ+isin⁡kθ)(cos⁡θ+isin⁡θ)=\left(\cos k\theta+i\sin k\theta\right)\left(\cos\theta+i\sin\theta\right)

    Replace the kkth power by the assumed expression.

  4. Select the correct final line

    (cos⁡θ+isin⁡θ)k+1=cos⁡(k+1)θ+isin⁡(k+1)θ\left(\cos\theta+i\sin\theta\right)^{k+1}=\cos\left(k+1\right)\theta+i\sin\left(k+1\right)\theta

    This is the statement for n=k+1n=k+1, which completes the inductive step.

Answer
cos⁡(k+1)θ+isin⁡(k+1)θ\cos\left(k+1\right)\theta+i\sin\left(k+1\right)\theta
Question 3
4 marksintermediate
Which of the following is a correct expression for cos⁡4θ\cos 4\theta?
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Worked solution

  1. Apply de Moivre's theorem

    (cos⁡θ+isin⁡θ)4=cos⁡4θ+isin⁡4θ\left(\cos\theta+i\sin\theta\right)^{4}=\cos4\theta+i\sin4\theta

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

  2. Take the real part of the binomial expansion

    cos⁡4θ=cos⁡4θ−6cos⁡2θsin⁡2θ+sin⁡4θ\cos4\theta=\cos^{4}\theta-6\cos^{2}\theta\sin^{2}\theta+\sin^{4}\theta

    Equating parts isolates the multiple angle.

  3. Remove the unwanted function with the Pythagorean identity

    sin⁡2θ=1−cos⁡2θ\sin^{2}\theta=1-\cos^{2}\theta

    Only even powers occur, so the substitution is legitimate.

  4. Collect like terms

    cos⁡4θ=8cos⁡4θ−8cos⁡2θ+1\cos4\theta=8\cos^{4}\theta-8\cos^{2}\theta+1

    This is the polynomial form of the multiple angle.

  5. Check at θ=0\theta=0 to eliminate options

    cos⁡0=1\cos 0=1

    Any option that fails this test can be discarded immediately.

  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. Select the matching option

    cos⁡4θ=8cos⁡4θ−8cos⁡2θ+1\cos4\theta=8\cos^{4}\theta-8\cos^{2}\theta+1

    This is the only option that is a correct identity.

Answer
8cos⁡4θ−8cos⁡2θ+18\cos^{4}\theta-8\cos^{2}\theta+1
Question 4
6 markshard
Which of the following is a correct expression for cos⁡5θ\cos 5\theta?
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Worked solution

  1. Apply de Moivre's theorem

    (cos⁡θ+isin⁡θ)5=cos⁡5θ+isin⁡5θ\left(\cos\theta+i\sin\theta\right)^{5}=\cos5\theta+i\sin5\theta

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

  2. Take the real part of the binomial expansion

    cos⁡5θ=cos⁡5θ−10cos⁡3θsin⁡2θ+5cos⁡θsin⁡4θ\cos5\theta=\cos^{5}\theta-10\cos^{3}\theta\sin^{2}\theta+5\cos\theta\sin^{4}\theta

    Equating parts isolates the multiple angle.

  3. Remove the unwanted function with the Pythagorean identity

    sin⁡2θ=1−cos⁡2θ\sin^{2}\theta=1-\cos^{2}\theta

    Only even powers occur, so the substitution is legitimate.

  4. Collect like terms

    cos⁡5θ=16cos⁡5θ−20cos⁡3θ+5cos⁡θ\cos5\theta=16\cos^{5}\theta-20\cos^{3}\theta+5\cos\theta

    This is the polynomial form of the multiple angle.

  5. Check at θ=0\theta=0 to eliminate options

    cos⁡0=1\cos 0=1

    Any option that fails this test can be discarded immediately.

  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. Select the matching option

    cos⁡5θ=16cos⁡5θ−20cos⁡3θ+5cos⁡θ\cos5\theta=16\cos^{5}\theta-20\cos^{3}\theta+5\cos\theta

    This is the only option that is a correct identity.

Answer
16cos⁡5θ−20cos⁡3θ+5cos⁡θ16\cos^{5}\theta-20\cos^{3}\theta+5\cos\theta
Question 5
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}?
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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)

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