Math · Precalculus · Worked example

Find powers and roots with De Moivre’s theorem

Use De Moivre’s theorem to find (1 + i)⁸ and the three cube roots of 8.

Write 1 + i in polar form

Its modulus is √2 and its argument π/4.

1 + i = \sqrt{2} (cos \frac{π}{4} + i sin \frac{π}{4})

\begin{array}{c} 1 + i \\ = \sqrt{2} (cos \frac{π}{4} + i sin \frac{π}{4}) \end{array}

Raise it to the 8th power

Raise the modulus to the 8th power and multiply the argument by 8. An angle of 2π is a full turn, so the result is real.

\begin{array}{l} {(\sqrt{2})}^{8} = 16 \\ 8 \cdot \frac{π}{4} = 2 π \end{array}

Find the cube roots of 8

8 has modulus 8 and argument 0. Each cube root has modulus ∛8 = 2 and argument (0 + 2πk)/3: 0, 2π/3 or 4π/3.

Convert to a + bi

The root at 0 is 2. At 2π/3 the coordinates are 2 cos(2π/3) = −1 and 2 sin(2π/3) = √3, giving −1 + i√3; at 4π/3 the sine changes sign, giving −1 − i√3.

\begin{array}{l} 2 cos \frac{2 π}{3} = - 1 \\ 2 sin \frac{2 π}{3} = \sqrt{3} \end{array}

Result

(1 + i)⁸ = 16. The cube roots of 8 are 2, −1 + i√3 and −1 − i√3.

Your turn

Find the two square roots of −9 using polar form.

Show the answer and explanation

3i and −3i.

−9 = 9(cos π + i sin π). The square roots have modulus 3 and arguments π/2 and (π + 2π)/2 = 3π/2, which give 3i and −3i.

\begin{array}{l} \sqrt{9} = 3 \\ \frac{π + 2 π}{2} = \frac{3 π}{2} \end{array}

Keep exploring

In Complex plane & roots, the three roots sit 120° apart on a circle of radius 2. Change the root degree to 6 and six roots appear, 60° apart.

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