Complex numbers: de Moivre Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Complex numbers: de Moivre questions. See exactly how to solve problems on de-moivre, powers, modarg, cart.

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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.

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
Use de Moivre's theorem to evaluate (cos⁡(π12)+isin⁡(π12))4\left(\cos\left(\frac{\pi}{12}\right)+i\sin\left(\frac{\pi}{12}\right)\right)^{4}, giving your answer in the form a+bia+bi.

Worked solution

  1. Read off the modulus and the argument

    r=1,θ=π12r=1,\qquad\theta=\frac{\pi}{12}

    The base is already written in modulus-argument form.

  2. Apply de Moivre's theorem

    [1(cos⁡(π12)+isin⁡(π12))]4=14(cos⁡(4×π12)+isin⁡(4×π12))\left[1\left(\cos\left(\frac{\pi}{12}\right)+i\sin\left(\frac{\pi}{12}\right)\right)\right]^{4}=1^{4}\left(\cos\left(4\times \frac{\pi}{12}\right)+i\sin\left(4\times \frac{\pi}{12}\right)\right)

    Raise the modulus to the power 44 and multiply the argument by 44.

  3. Multiply out

    1(12+32i)=12+3i21\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)=\frac{1}{2} + \frac{\sqrt{3} i}{2}

    Distribute the modulus across the real and imaginary parts.

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

    (cos⁡(π12)+isin⁡(π12))4=12+3i2\left(\cos\left(\frac{\pi}{12}\right)+i\sin\left(\frac{\pi}{12}\right)\right)^{4}=\frac{1}{2} + \frac{\sqrt{3} i}{2}

    This is the exact value of the power.

Answer
12+3i2\frac{1}{2} + \frac{\sqrt{3} i}{2}
Question 3
2 markseasy
Use de Moivre's theorem to evaluate [2(cos⁡(π3)+isin⁡(π3))]2\left[2\left(\cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{3}\right)\right)\right]^{2}, giving your answer in the form a+bia+bi.

Worked solution

  1. Read off the modulus and the argument

    r=2,θ=π3r=2,\qquad\theta=\frac{\pi}{3}

    The base is already written in modulus-argument form.

  2. Apply de Moivre's theorem

    [2(cos⁡(π3)+isin⁡(π3))]2=22(cos⁡(2×π3)+isin⁡(2×π3))\left[2\left(\cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{3}\right)\right)\right]^{2}=2^{2}\left(\cos\left(2\times \frac{\pi}{3}\right)+i\sin\left(2\times \frac{\pi}{3}\right)\right)

    Raise the modulus to the power 22 and multiply the argument by 22.

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

    [2(cos⁡(π3)+isin⁡(π3))]2=−2+23i\left[2\left(\cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{3}\right)\right)\right]^{2}=-2 + 2 \sqrt{3} i

    This is the exact value of the power.

Answer
−2+23i-2 + 2 \sqrt{3} i
Question 4
2 markseasy
Use de Moivre's theorem to evaluate [3(cos⁡(π4)+isin⁡(π4))]2\left[3\left(\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right)\right]^{2}, giving your answer in the form a+bia+bi.

Worked solution

  1. Read off the modulus and the argument

    r=3,θ=π4r=3,\qquad\theta=\frac{\pi}{4}

    The base is already written in modulus-argument form.

  2. Apply de Moivre's theorem

    [3(cos⁡(π4)+isin⁡(π4))]2=32(cos⁡(2×π4)+isin⁡(2×π4))\left[3\left(\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right)\right]^{2}=3^{2}\left(\cos\left(2\times \frac{\pi}{4}\right)+i\sin\left(2\times \frac{\pi}{4}\right)\right)

    Raise the modulus to the power 22 and multiply the argument by 22.

  3. Multiply out

    9(0+1i)=9i9\left(0+1 i\right)=9 i

    Distribute the modulus across the real and imaginary parts.

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

    [3(cos⁡(π4)+isin⁡(π4))]2=9i\left[3\left(\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right)\right]^{2}=9 i

    This is the exact value of the power.

Answer
9i9 i
Question 5
2 markseasy
Use de Moivre's theorem to evaluate (1+i)2\left(1 + i\right)^{2}, giving your answer in the form a+bia+bi.

Worked solution

  1. Write the base in modulus-argument form

    1+i=2(cos⁡(π4)+isin⁡(π4))1 + i=\sqrt{2}\left(\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right)

    Its modulus is 2\sqrt{2} and its principal argument is π4\frac{\pi}{4}.

  2. Apply de Moivre's theorem

    [2(cos⁡(π4)+isin⁡(π4))]2=22(cos⁡(2×π4)+isin⁡(2×π4))\left[\sqrt{2}\left(\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right)\right]^{2}=\sqrt{2}^{2}\left(\cos\left(2\times \frac{\pi}{4}\right)+i\sin\left(2\times \frac{\pi}{4}\right)\right)

    Raise the modulus to the power 22 and multiply the argument by 22.

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

    (1+i)2=2i\left(1 + i\right)^{2}=2 i

    This is the exact value of the power.

Answer
2i2 i

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