CP 2.8 - De Moivre's theorem

CP 2.8 - De Moivre's theorem

Find integer powers of a complex number by raising its modulus and multiplying its angle. You will see why de Moivre's theorem works, use it for large positive powers and extend it carefully to zero and negative powers.

A power repeats the same rotation

Can you find (1+i)20(1 + \mathrm{i})^{20} exactly without expanding twenty brackets? Its answer is a small-looking real number, even though most of the intermediate powers have imaginary parts. The useful picture is repeated rotation.

You already know

Multiplying complex numbers multiplies their moduli and adds their arguments. The number r(cos⁡θ+isin⁡θ)r(\cos\theta + \mathrm{i}\sin\theta) has modulus rr and an argument θ\theta; angles differing by 2π2\pi represent the same direction.

Start with w=cos⁡π3+isin⁡π3w = \cos\frac{\pi}{3} + \mathrm{i}\sin\frac{\pi}{3}. Multiplying by ww rotates a point anticlockwise through π3\frac{\pi}{3} without changing its distance from the origin. Thus

w2=cos⁡2π3+isin⁡2π3add two anglesw3=cos⁡π+isin⁡π=−1add a third anglew6=cos⁡2π+isin⁡2π=1one complete turn\begin{aligned} w^2 &= \cos\frac{2\pi}{3} + \mathrm{i}\sin\frac{2\pi}{3} && \text{add two angles} \\ w^3 &= \cos\pi + \mathrm{i}\sin\pi = -1 && \text{add a third angle} \\ w^6 &= \cos 2\pi + \mathrm{i}\sin 2\pi = 1 && \text{one complete turn} \end{aligned}

Here is a pleasing consequence: ww is not real, but its cube is real and its sixth power is 11. Powers need not become more complicated as they grow.

[DIAGRAM: asset_name: Successive powers as equal rotations; asset_slug: 9fm0-cp08-equal-rotations; caption: Multiplying by w adds the same angle each time; description: A unit circle on an Argand diagram has the six points 1, w, w squared, w cubed equals minus 1, w to the fourth and w to the fifth spaced anticlockwise by pi over 3. Blue radial lines and sage curved arrows show the repeated rotation, with w to the sixth returning to 1. A side panel states modulus remains 1 and each multiplication adds pi over 3.]
Multiplying by w adds the same angle each time

For a general number z=r(cos⁡θ+isin⁡θ)z = r(\cos\theta + \mathrm{i}\sin\theta), a product of nn copies of zz has modulus rr multiplied by itself nn times, and angle θ\theta added nn times. This proves the positive-integer case directly from the product rule.

De Moivre's theorem

For real θ\theta, r>0r > 0 and any integer nn,

{r(cos⁡θ+isin⁡θ)}n=rn(cos⁡nθ+isin⁡nθ)\bigl\{r(\cos\theta + \mathrm{i}\sin\theta)\bigr\}^{n} = r^n\bigl(\cos n\theta + \mathrm{i}\sin n\theta\bigr)

For positive nn, the formula also works when r=0r = 0. Zero and negative powers require z≠0z \ne 0.

In the formula booklet.

We have explained positive powers; the last part will justify the other integer powers. For now, notice what is being multiplied: nθn\theta is the angle, while rnr^n is the modulus. The theorem says nothing about replacing these by θn\theta^n or nrnr.

Key idea

A complex power has two separate jobs: raise the modulus to the power and multiply the angle by the power.

The following choices separate those two jobs.

Finding positive integer powers

Write the number in modulus-argument form before applying the power; then simplify the trigonometric values and return to Cartesian form if requested.

Take z=3+iz = \sqrt{3} + \mathrm{i}. Its modulus is 3+1=2\sqrt{3 + 1} = 2, and its argument is π6\frac{\pi}{6}, so

z=2(cos⁡π6+isin⁡π6)z = 2\left(\cos\frac{\pi}{6} + \mathrm{i}\sin\frac{\pi}{6}\right)

An angle such as 7π6\frac{7\pi}{6} is perfectly usable inside sine and cosine. Only if you are asked for the principal argument must you reduce it to −π<arg⁡z≤π-\pi < \arg z \le \pi. This distinction lets you keep the power calculation tidy.

Worked example 1: a power on an axis

Question. Use de Moivre's theorem to find (3+i)3(\sqrt{3} + \mathrm{i})^3.

Plan. Use modulus 22 and angle π6\frac{\pi}{6}, raise the modulus and multiply the angle.

(3+i)3=23(cos⁡3π6+isin⁡3π6)de Moivre=8(cos⁡π2+isin⁡π2)simplify the angle=8(0+i)exact values\begin{aligned} (\sqrt{3} + \mathrm{i})^3 &= 2^3\left(\cos\frac{3\pi}{6} + \mathrm{i}\sin\frac{3\pi}{6}\right) && \text{de Moivre} \\ &= 8\left(\cos\frac{\pi}{2} + \mathrm{i}\sin\frac{\pi}{2}\right) && \text{simplify the angle} \\ &= 8(0 + \mathrm{i}) && \text{exact values} \end{aligned}

Why this step? Multiplying the angle first reveals that the answer lies on the positive imaginary axis, so its real part must be zero.

Answer. (3+i)3=8i(\sqrt{3} + \mathrm{i})^3 = 8\mathrm{i}.

Check. (3+i)2=2+2i3(\sqrt{3} + \mathrm{i})^2 = 2 + 2\mathrm{i}\sqrt{3}, and multiplying once more gives 23+2i+6i−23=8i2\sqrt{3} + 2\mathrm{i} + 6\mathrm{i} - 2\sqrt{3} = 8\mathrm{i}.

The next example changes only the exponent, from 33 to 55. The answer now lies in the second quadrant, so neither component vanishes.

Worked example 2: an exact Cartesian answer

Question. Use de Moivre's theorem to find (3+i)5(\sqrt{3} + \mathrm{i})^5, giving your answer in the form a+iba + \mathrm{i}b, where aa and bb are exact real numbers.

Plan. Keep the same modulus-argument form, apply the fifth power and evaluate the angle 5π6\frac{5\pi}{6}.

(3+i)5=25(cos⁡5π6+isin⁡5π6)de Moivre=32(−32+i2)second quadrant=−163+16imultiply both terms\begin{aligned} (\sqrt{3} + \mathrm{i})^5 &= 2^5\left(\cos\frac{5\pi}{6} + \mathrm{i}\sin\frac{5\pi}{6}\right) && \text{de Moivre} \\ &= 32\left(-\frac{\sqrt{3}}{2} + \frac{\mathrm{i}}{2}\right) && \text{second quadrant} \\ &= -16\sqrt{3} + 16\mathrm{i} && \text{multiply both terms} \end{aligned}

Why this step? The reference angle is π6\frac{\pi}{6}, but the cosine is negative in the second quadrant. Using the first-quadrant sign would change the answer.

Answer. (3+i)5=−163+16i(\sqrt{3} + \mathrm{i})^5 = -16\sqrt{3} + 16\mathrm{i}.

Check. The modulus is (163)2+162=1024=32=25\sqrt{(16\sqrt{3})^2 + 16^2} = \sqrt{1024} = 32 = 2^5, and the signs place the answer in the second quadrant.

Exact trigonometric values turn the polar calculation into the prescribed final form. Decimal approximations would throw away information the question asks you to retain.

Watch out

Wrong: (3+i)5=25(cos⁡5π6+isin⁡5π6)(\sqrt{3} + \mathrm{i})^5 = 2^5(\cos^5\frac{\pi}{6} + \mathrm{i}\sin^5\frac{\pi}{6}). The power applies to the entire complex number. Right: 25(cos⁡5π6+isin⁡5π6)2^5(\cos\frac{5\pi}{6} + \mathrm{i}\sin\frac{5\pi}{6}).

Show the modulus and multiplied angle in your working, especially when the question requires all stages. They make the method visible even if an exact-value slip follows.

Zero and negative integer powers

A negative power means a reciprocal. In modulus-argument form, taking a reciprocal replaces rr by 1r\frac{1}{r} and reverses the angle.

To see why, multiply r(cos⁡θ+isin⁡θ)r(\cos\theta + \mathrm{i}\sin\theta) by 1r(cos⁡(−θ)+isin⁡(−θ))\frac{1}{r}(\cos(-\theta) + \mathrm{i}\sin(-\theta)). The moduli multiply to 11, and the angles add to 00, so the product is 11.

Reciprocals and negative powers

If z=r(cos⁡θ+isin⁡θ)≠0z = r(\cos\theta + \mathrm{i}\sin\theta) \ne 0, then

z−m=r−m(cos⁡(−mθ)+isin⁡(−mθ))z^{-m} = r^{-m}\bigl(\cos(-m\theta) + \mathrm{i}\sin(-m\theta)\bigr)

for a positive integer mm. Also, z0=1=cos⁡0+isin⁡0z^0 = 1 = \cos 0 + \mathrm{i}\sin 0.

These are the negative- and zero-integer cases of de Moivre's theorem. In the formula booklet.

This completes the justification for every integer exponent. It does not make fractional powers single-valued: for example, both i\mathrm{i} and −i-\mathrm{i} square to −1-1. Finding all roots is a later lesson.

Worked example 3: a negative power

Question. Find (3+i)−2(\sqrt{3} + \mathrm{i})^{-2} in exact Cartesian form.

Plan. Use the same modulus 22 and angle π6\frac{\pi}{6}, with exponent −2-2.

(3+i)−2=2−2(cos⁡(−π3)+isin⁡(−π3))multiply by −2=14(12−i32)clockwise angle=18−i38simplify\begin{aligned} (\sqrt{3} + \mathrm{i})^{-2} &= 2^{-2}\left(\cos\left(-\frac{\pi}{3}\right) + \mathrm{i}\sin\left(-\frac{\pi}{3}\right)\right) && \text{multiply by } -2 \\ &= \frac14\left(\frac12 - \frac{\mathrm{i}\sqrt{3}}2\right) && \text{clockwise angle} \\ &= \frac18 - \frac{\mathrm{i}\sqrt{3}}8 && \text{simplify} \end{aligned}

Why this step? The exponent changes both the size and the direction. Keeping the angle positive would give the conjugate of the required answer.

Answer. (3+i)−2=18−i38(\sqrt{3} + \mathrm{i})^{-2} = \frac18 - \frac{\mathrm{i}\sqrt{3}}8.

Check. Multiply by the square: (2+2i3)(18−i38)=14+34=1(2 + 2\mathrm{i}\sqrt{3})(\frac18 - \frac{\mathrm{i}\sqrt{3}}8) = \frac14 + \frac34 = 1.

Change only the exponent from −2-2 to −3-3. This puts the reciprocal power on an axis and provides a quick sign check.

Worked example 4: a reciprocal on an axis

Question. Find (3+i)−3(\sqrt{3} + \mathrm{i})^{-3}.

Plan. Apply de Moivre's theorem with n=−3n = -3, then evaluate the angle −π2-\frac{\pi}{2}.

(3+i)−3=2−3(cos⁡(−π2)+isin⁡(−π2))de Moivre=18(0−i)negative imaginary axis\begin{aligned} (\sqrt{3} + \mathrm{i})^{-3} &= 2^{-3}\left(\cos\left(-\frac{\pi}{2}\right) + \mathrm{i}\sin\left(-\frac{\pi}{2}\right)\right) && \text{de Moivre} \\ &= \frac18(0 - \mathrm{i}) && \text{negative imaginary axis} \end{aligned}

Why this step? A reciprocal reverses the direction of the positive-power angle. Since the cube is 8i8\mathrm{i}, the reciprocal must point downwards.

Answer. (3+i)−3=−i8(\sqrt{3} + \mathrm{i})^{-3} = -\frac{\mathrm{i}}8.

Check. (8i)(−i8)=−i2=1(8\mathrm{i})(-\frac{\mathrm{i}}8) = -\mathrm{i}^2 = 1.

Now return to the opening question. Since 1+i=2(cos⁡π4+isin⁡π4)1 + \mathrm{i} = \sqrt{2}(\cos\frac{\pi}{4} + \mathrm{i}\sin\frac{\pi}{4}),

(1+i)20=(2)20(cos⁡5π+isin⁡5π)raise size, multiply angle=210(−1+0i)an odd half-turn=−1024\begin{aligned} (1 + \mathrm{i})^{20} &= (\sqrt{2})^{20}(\cos 5\pi + \mathrm{i}\sin 5\pi) && \text{raise size, multiply angle} \\ &= 2^{10}(-1 + 0\mathrm{i}) && \text{an odd half-turn} \\ &= -1024 \end{aligned}

The rotation picture explains the real answer: twenty turns of half a right angle make five half-turns. As an algebraic check, (1+i)2=2i(1 + \mathrm{i})^2 = 2\mathrm{i}, so (1+i)4=−4(1 + \mathrm{i})^4 = -4 and (1+i)20=(−4)5=−1024(1 + \mathrm{i})^{20} = (-4)^5 = -1024.

Watch out

Wrong: 2−3=−82^{-3} = -8. A negative exponent means reciprocal, so 2−3=182^{-3} = \frac18. The negative sign in nθn\theta controls the angle; it does not make the modulus negative.

A useful final check is that a negative power of a number with modulus greater than 11 must have modulus less than 11.

Quick Recap

  • De Moivre's theorem raises the modulus to rnr^n and multiplies the angle to nθn\theta.
  • Use exact sine and cosine values to give a Cartesian answer.
  • Angles may make several turns; reduce them only as needed, and use −π<arg⁡z≤π-\pi < \arg z \le \pi when a principal argument is requested.
  • Zero and negative integer powers require z≠0z \ne 0; a negative exponent means reciprocal modulus and reversed angle.

The opening power is (1+i)20=−1024(1 + \mathrm{i})^{20} = -1024: the multiplied angle lands on the negative real axis.