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De Moivre's theorem

Numbers · level 8, lesson 2 of 6

A question to start

Work out \((1 + i)^8\).

The honest way is to multiply eight brackets together, collecting terms and replacing \(i^2\) each time. With the binomial theorem it is 9 terms with coefficients 1, 8, 28, 56, 70, 56, 28, 8, 1 and powers of \(i\) to sort out. It can be done, but it is slow, and \((1 + i)^{100}\) would be hopeless that way.

Now think in polar form. \(1 + i\) has length \(\sqrt2\) and angle 45°. Multiplying multiplies lengths and adds angles, so eight copies give length \((\sqrt2)^8 = 16\) and angle \(8 \times 45° = 360°\), a full turn. The answer is simply 16. This lesson turns that thought into a theorem, proves it, and uses it to produce trigonometry formulas almost for free.

The idea

The theorem

De Moivre's theorem. For any real \(\theta\) and any integer \(n\),

\[(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta\]

and so, for \(z = r\operatorname{cis}\theta\),

\[z^n = r^n\operatorname{cis}(n\theta).\]

In words: to raise a complex number to the \(n\)th power, raise its length to the \(n\)th power and multiply its angle by \(n\). It is named after Abraham de Moivre, a French-born mathematician who worked in London in the early 1700s; the formula in the form written here was set down by Euler.

Proof for positive powers: induction

This is a natural place for proof by induction, which proves a statement for every whole number \(n \ge 1\) in two steps, like knocking over a line of dominoes:

  1. Base case: show it is true for \(n = 1\) (the first domino falls).
  2. Inductive step: show that if it is true for some \(n = k\), then it is true for \(n = k + 1\) (each domino knocks over the next).

Base case. For \(n = 1\) the statement reads \(\operatorname{cis}\theta = \operatorname{cis}\theta\). True.

Inductive step. Suppose \((\operatorname{cis}\theta)^k = \operatorname{cis}(k\theta)\) for some \(k \ge 1\). Then

\[(\operatorname{cis}\theta)^{k+1} = (\operatorname{cis}\theta)^k \cdot \operatorname{cis}\theta = \operatorname{cis}(k\theta)\cdot\operatorname{cis}\theta = \operatorname{cis}(k\theta + \theta) = \operatorname{cis}\big((k + 1)\theta\big).\]

The third equals sign is the multiplication rule from lesson 1: angles add. So the statement for \(k\) forces the statement for \(k + 1\). By induction it holds for every \(n \ge 1\). ∎

Zero and negative powers

For \(n = 0\): \((\operatorname{cis}\theta)^0 = 1\) and \(\operatorname{cis}0 = \cos0 + i\sin0 = 1\). True.

For a negative power \(n = -m\) (with \(m \ge 1\)), use the reciprocal rule from lesson 1, \(\dfrac{1}{\operatorname{cis}\phi} = \operatorname{cis}(-\phi)\):

\[(\operatorname{cis}\theta)^{-m} = \frac{1}{(\operatorname{cis}\theta)^m} = \frac{1}{\operatorname{cis}(m\theta)} = \operatorname{cis}(-m\theta).\]

So the theorem holds for all integers \(n\). (For fractional powers, such as square roots, there is a catch, since a number has several roots; that is lesson 3.)

A picture: powers spiral

Since each multiplication by \(z\) turns by \(\theta\) and stretches by \(r\), the powers \(1, z, z^2, z^3, \ldots\) step round 0 like a spiral staircase seen from above. If \(r > 1\) the spiral winds outwards; if \(r < 1\) it winds in towards 0; if \(r = 1\) the powers stay on the unit circle forever.

1 1 + i 2i −2 + 2i −4 −4 − 4i −8i

The dots are \(1, z, z^2, \ldots, z^6\) for \(z = 1 + i\). Two more steps reach \(z^7 = 8 - 8i\) and \(z^8 = 16\), on the positive real axis again, as the opening question found.

Trigonometry for free

Here is the most surprising use. Expand the left side of De Moivre's theorem with ordinary algebra, then compare real and imaginary parts with the right side. Write \(c = \cos\theta\) and \(s = \sin\theta\) for short.

\(n = 2\). \((c + is)^2 = c^2 + 2ics + i^2 s^2 = (c^2 - s^2) + i(2cs)\). De Moivre says this equals \(\cos2\theta + i\sin2\theta\). Comparing parts:

\[\cos2\theta = \cos^2\theta - \sin^2\theta, \qquad \sin2\theta = 2\sin\theta\cos\theta.\]

These are the double-angle formulas, obtained without a single diagram.

\(n = 3\). Using \((x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3\) with \(x = c\), \(y = is\), and \(i^2 = -1\), \(i^3 = -i\):

\[(c + is)^3 = c^3 + 3ic^2s - 3cs^2 - is^3 = (c^3 - 3cs^2) + i(3c^2s - s^3).\]

Comparing with \(\cos3\theta + i\sin3\theta\), and then using \(s^2 = 1 - c^2\) and \(c^2 = 1 - s^2\):

\[\cos3\theta = c^3 - 3c(1 - c^2) = 4\cos^3\theta - 3\cos\theta,\]
\[\sin3\theta = 3(1 - s^2)s - s^3 = 3\sin\theta - 4\sin^3\theta.\]

These triple-angle formulas are tiresome to get from angle-sum formulas and take three lines here. The same method gives \(\cos n\theta\) and \(\sin n\theta\) for any \(n\).

Worked examples

1. The opening question, carefully. Find \((1 + i)^8\).

  1. \(1 + i = \sqrt2\operatorname{cis}\tfrac\pi4\).
  2. \((1 + i)^8 = (\sqrt2)^8\operatorname{cis}\left(8 \times \tfrac\pi4\right) = 16\operatorname{cis}2\pi = 16\).
  3. Check by repeated squaring: \((1 + i)^2 = 2i\), \((2i)^2 = -4\), \((-4)^2 = 16\). ✓

2. A sixth power. Find \((\sqrt3 - i)^6\).

  1. \(\sqrt3 - i\): \(r = \sqrt{3 + 1} = 2\), fourth quadrant, reference angle \(\tfrac\pi6\), so \(\theta = -\tfrac\pi6\).
  2. \((\sqrt3 - i)^6 = 2^6\operatorname{cis}\left(-\pi\right) = 64(\cos(-\pi) + i\sin(-\pi)) = -64\).

3. A tenth power, exactly. Find \((1 + i\sqrt3)^{10}\) as \(a + bi\).

  1. \(1 + i\sqrt3 = 2\operatorname{cis}\tfrac\pi3\).
  2. \((1 + i\sqrt3)^{10} = 1024\operatorname{cis}\tfrac{10\pi}{3}\).
  3. Remove full turns: \(\tfrac{10\pi}{3} - 2\pi = \tfrac{4\pi}{3}\) (that is 240°).
  4. \(\cos\tfrac{4\pi}3 = -\tfrac12\) and \(\sin\tfrac{4\pi}{3} = -\tfrac{\sqrt3}{2}\), so the answer is \(1024\left(-\tfrac12 - \tfrac{\sqrt3}{2}i\right) = -512 - 512\sqrt3\,i\).

4. A negative power. Find \((1 + i)^{-4}\).

\((1 + i)^{-4} = (\sqrt2)^{-4}\operatorname{cis}\left(-4 \times \tfrac\pi4\right) = \tfrac14\operatorname{cis}(-\pi) = -\tfrac14\). Check: \((1 + i)^4 = -4\) (worked example 1, step 3), and \(\dfrac{1}{-4} = -\tfrac14\). ✓

5. Powers of cosine. Write \(\cos^3\theta\) in terms of \(\cos\theta\) and \(\cos3\theta\).

From the triple-angle formula, \(\cos3\theta = 4\cos^3\theta - 3\cos\theta\). Solve for \(\cos^3\theta\):

\[\cos^3\theta = \frac{3\cos\theta + \cos3\theta}{4}.\]

Formulas like this turn powers of cosine into single cosines, which is how \(\displaystyle\int\cos^3\theta\,d\theta\) is done in calculus.

Try it

  1. Find \((1 - i)^6\) and \((-1 + i)^5\) in the form \(a + bi\).
  2. Find \((\sqrt3 + i)^9\).
  3. Find the smallest positive whole number \(n\) for which \((1 + i\sqrt3)^n\) is a real number, and find that real number.
  4. Find \(\left(\tfrac{\sqrt3}{2} + \tfrac12 i\right)^{-3}\).
  5. Find \((1 + i)^{2026}\) in the form \(2^k \times (\text{something simple})\).
  6. Use De Moivre with \(n = 4\) to show that \(\cos4\theta = 8\cos^4\theta - 8\cos^2\theta + 1\).
  7. Show that \((\cos\theta - i\sin\theta)^n = \cos n\theta - i\sin n\theta\) for every whole number \(n\).
  8. Show that if \(|z| = 1\) then \(z^n + \bar z^{\,n} = 2\cos n\theta\), where \(\theta = \arg z\).
Answers
  1. \(1 - i = \sqrt2\operatorname{cis}\left(-\tfrac\pi4\right)\), so \((1 - i)^6 = 8\operatorname{cis}\left(-\tfrac{3\pi}{2}\right) = 8\operatorname{cis}\tfrac\pi2 = 8i\). And \(-1 + i = \sqrt2\operatorname{cis}\tfrac{3\pi}{4}\), so \((-1 + i)^5 = 4\sqrt2\operatorname{cis}\tfrac{15\pi}{4} = 4\sqrt2\operatorname{cis}\left(-\tfrac\pi4\right) = 4\sqrt2\left(\tfrac{\sqrt2}2 - \tfrac{\sqrt2}{2}i\right) = 4 - 4i\).
  2. \(\sqrt3 + i = 2\operatorname{cis}\tfrac\pi6\), so the ninth power is \(512\operatorname{cis}\tfrac{3\pi}{2} = -512i\).
  3. The angle is \(n \times 60°\), which must be a multiple of 180°: smallest \(n = 3\). Then \((1 + i\sqrt3)^3 = 8\operatorname{cis}\pi = -8\).
  4. The number is \(\operatorname{cis}\tfrac\pi6\), so its \(-3\)rd power is \(\operatorname{cis}\left(-\tfrac\pi2\right) = -i\).
  5. \((\sqrt2)^{2026} = 2^{1013}\), and the angle is \(2026 \times 45° = 91\,170°\). Dividing by 360 leaves a remainder of \(91\,170 - 253 \times 360 = 90°\). So \((1 + i)^{2026} = 2^{1013}\,i\).
  6. \((c + is)^4 = c^4 + 4c^3(is) + 6c^2(is)^2 + 4c(is)^3 + (is)^4 = (c^4 - 6c^2s^2 + s^4) + i(4c^3s - 4cs^3)\). So \(\cos4\theta = c^4 - 6c^2s^2 + s^4\). Put \(s^2 = 1 - c^2\): \(c^4 - 6c^2 + 6c^4 + 1 - 2c^2 + c^4 = 8c^4 - 8c^2 + 1\).
  7. \(\cos\theta - i\sin\theta = \cos(-\theta) + i\sin(-\theta)\), because cosine is even and sine is odd. By De Moivre its \(n\)th power is \(\cos(-n\theta) + i\sin(-n\theta) = \cos n\theta - i\sin n\theta\).
  8. \(z = \operatorname{cis}\theta\), \(\bar z = \operatorname{cis}(-\theta)\). So \(z^n + \bar z^{\,n} = \operatorname{cis}n\theta + \operatorname{cis}(-n\theta) = (\cos n\theta + i\sin n\theta) + (\cos n\theta - i\sin n\theta) = 2\cos n\theta\).

Common mistake

Raising the angle to the power instead of multiplying it. \((2\operatorname{cis}30°)^3\) is \(8\operatorname{cis}90°\), not \(8\operatorname{cis}27\,000°\). The length is powered; the angle is multiplied by \(n\).

Forgetting to bring the angle back into range. \(\operatorname{cis}\tfrac{10\pi}{3}\) is correct but unhelpful; take off whole turns (\(2\pi\) each) to reach \(\tfrac{4\pi}{3}\) or the principal value \(-\tfrac{2\pi}{3}\) before reading off cosine and sine.

Where it leads