Multiplying, powers and De Moivre's theorem

When you reach for it

In the form a + bi, multiplying complex numbers means distributing and replacing i^2 by -1. That is fine once. It is slow for a product of several factors and hopeless for (1 + i)^{12}, and it hides what multiplication does to a point in the plane.

Polar form fixes all of this. Once each factor is written as r\operatorname{cis}\theta = r(\cos\theta + i\sin\theta), a product, a quotient or a power comes from multiplying lengths and adding angles. Reach for it when the numbers are already in polar form, when a power is high or negative, when a point is to be turned about 0, and, turned around, when you want a formula for \cos 3\theta or \sin 4\theta.

The lesson runs over two days. Day 1 is multiplying and dividing in polar form, and the picture of a product as a stretch and a turn. Day 2 is De Moivre’s theorem for powers, positive and negative, and its use for multiple-angle formulas.

The procedure

Throughout, \operatorname{Arg} z is the principal argument in (-\pi, \pi], as in Modulus, argument and polar form, and every modulus is positive.

Day 1: multiplying in polar form

Let z_1 = r_1(\cos\theta_1 + i\sin\theta_1) and z_2 = r_2(\cos\theta_2 + i\sin\theta_2). The moduli are real numbers, so they come out to the front (real numbers commute with everything), and the two brackets are multiplied by distributing:

With i^2 = -1, the real part is \cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2, which is \cos(\theta_1 + \theta_2) by the cosine sum formula. The imaginary part is \sin\theta_1\cos\theta_2 + \cos\theta_1\sin\theta_2, which is \sin(\theta_1 + \theta_2) by the sine sum formula. So

The moduli multiply and the arguments add. The sum formulas, which looked like a fact about triangles, are exactly the rule for multiplying directions.

Dividing. If z_2 \ne 0, then

To see why, multiply the right side by z_2 using the product rule: the moduli give \displaystyle \frac{r_1}{r_2}\cdot r_2 = r_1 and the arguments give (\theta_1 - \theta_2) + \theta_2 = \theta_1, so the product is z_1. That is what it means to be z_1 \div z_2. In particular, taking z_1 = 1 = 1\operatorname{cis}0, the reciprocal of r\operatorname{cis}\theta is \displaystyle \frac{1}{r}\operatorname{cis}(-\theta).

The picture. Multiplying z by w = s\operatorname{cis}\varphi multiplies the distance from 0 by s and adds \varphi to the direction: a stretch by s and a turn by \varphi about 0 (Figure 2, with Example 3). Multiplying by \displaystyle i = 1\operatorname{cis}\frac{\pi}{2} is a quarter turn counterclockwise with no stretch; multiplying by -1 = 1\operatorname{cis}\pi is a half turn. Dividing by w undoes it: shrink by s and turn back by \varphi.

Arguments that leave the range. If \theta_1 and \theta_2 are principal arguments, their sum is an argument of z_1z_2 but may fall outside (-\pi, \pi]. Add or subtract 2\pi to bring it back; the number does not change, only the name of its angle.

Day 2: De Moivre’s theorem

A power is a product of equal factors, so the product rule applies again and again. Each factor of z = r\operatorname{cis}\theta multiplies the modulus by r and adds \theta to the argument:

De Moivre’s theorem. For every whole number n (that is, n = 0, 1, 2, \ldots),

The pattern above is made airtight by induction. The case n = 0 holds, since both sides are 1. For the inductive step, suppose z^k = r^k\operatorname{cis}k\theta for some whole number k. Then, by the product rule,

so the statement for k forces the statement for k + 1, and it holds for every whole number.

Negative powers. For z \ne 0 and n a positive whole number, z^{-n} means \displaystyle \frac{1}{z^n}. The reciprocal rule turns r^n\operatorname{cis}n\theta into r^{-n}\operatorname{cis}(-n\theta), which is the same formula with -n in place of n. So for z \ne 0 the theorem holds for every integer n.

Using it. Write the base in polar form, raise the modulus to the power, multiply the argument by the power, reduce the angle by whole turns (2\pi k, k an integer) to a familiar one, and convert back to a + bi if asked. Figure 1 shows what happens: the powers of one number march around 0 by equal turns, moving outward if r > 1.

Run backwards, it makes identities. With r = 1 the theorem says (\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta. Expanding the left side by plain algebra and matching real and imaginary parts writes \cos n\theta and \sin n\theta in terms of \cos\theta and \sin\theta (Example 7).

Worked examples

Examples 1 to 3 go with Day 1, Examples 4 to 7 with Day 2.

Boundaries

Division and negative powers need a nonzero number. 0 has no argument and no reciprocal, so \displaystyle \frac{z_1}{z_2} needs z_2 \ne 0 and z^{-n} needs z \ne 0. For n = 0, z^0 = 1 is taken for z \ne 0 only.

The added argument need not be the principal one. \operatorname{Arg}(z_1z_2) equals \operatorname{Arg}z_1 + \operatorname{Arg}z_2 only when that sum lies in (-\pi, \pi]; otherwise it differs by 2\pi. The rule “arguments add” is a statement about some argument, not about the principal one.

The theorem here is for integer powers. For a fraction such as \displaystyle \frac{1}{2} the formula \displaystyle r^{1/2}\operatorname{cis}\frac{\theta}{2} gives one square root, but adding 2\pi to \theta first gives a different one: a fractional power has several values, and the formula alone does not choose. That case belongs to roots.

The form must really be polar. The rules apply to r(\cos\theta + i\sin\theta) with r > 0 and the same angle in both places. A number such as \cos\theta - i\sin\theta must first be rewritten as \operatorname{cis}(-\theta).

Matching parts needs a real angle. Example 7 splits the expansion into real and imaginary parts because \cos\theta and \sin\theta are real when \theta is real.

Where this goes next

Running De Moivre’s theorem backwards answers the question “which numbers have z^n = w?” Each nth root must have modulus \displaystyle \sqrt[n]{|w|} and an argument that becomes an argument of w when multiplied by n, and the extra whole turns that were a nuisance here now give n different answers. That is roots of a complex number and roots of unity. The same multiply-and-turn rule is the start of the exponential form re^{i\theta}, in which “arguments add” becomes the familiar law e^{a}e^{b} = e^{a + b}.

Exercises

Exercises 1 to 10 go with Day 1 (Examples 1 to 3), Exercises 11 to 21 with Day 2 (Examples 4 to 7), and 22 to 25 draw on both.

Day 1: products and quotients in polar form

Give each answer in polar form with the principal argument, and as a + bi when the angle is a special one.

1. \displaystyle 5\operatorname{cis}\frac{\pi}{8} \cdot 4\operatorname{cis}\frac{3\pi}{8}
2. \displaystyle 7\operatorname{cis}\frac{5\pi}{9} \cdot 2\operatorname{cis}\frac{7\pi}{9}
3. \displaystyle \dfrac{12\operatorname{cis}\frac{11\pi}{12}}{4\operatorname{cis}\frac{\pi}{4}}
4. \displaystyle \dfrac{9\operatorname{cis}\left(-\frac{3\pi}{4}\right)}{6\operatorname{cis}\frac{\pi}{2}}
5. The reciprocal of \displaystyle 5\operatorname{cis}\frac{2\pi}{7}.

Day 1: converting first, and turning points

6. Write \displaystyle \sqrt{2} + \sqrt{2}\,i and \displaystyle -4\sqrt{3} + 4i in polar form, multiply them in polar form, and check against the product in rectangular form.
7. Find \displaystyle \dfrac{6 + 6\sqrt{3}\,i}{2 - 2i} in polar form, then in rectangular form, and use the two to find \displaystyle \cos\frac{7\pi}{12} exactly.
8. Which complex number w turns every point a quarter turn clockwise about 0 and halves its distance from 0? Where does it send 10 + 4i?
9. Turn 3 + i counterclockwise by \displaystyle \frac{\pi}{4} about 0, keeping its distance from 0.
10. If \displaystyle zw = 15\operatorname{cis}\frac{\pi}{3} and \displaystyle z = 5\operatorname{cis}\frac{4\pi}{5}, find w.

Day 2: powers

Give each answer as a + bi.

11. \displaystyle \left(2\operatorname{cis}\frac{\pi}{10}\right)^5
12. \displaystyle \left(\sqrt{3}\operatorname{cis}\frac{2\pi}{9}\right)^6
13. \displaystyle \left(3 - \sqrt{3}\,i\right)^6
14. (-2 + 2i)^5
15. \displaystyle \left(-\frac{\sqrt{3}}{2} + \frac{1}{2}i\right)^{20}
16. \displaystyle \left(1 + \sqrt{3}\,i\right)^{-4}
17. (2i)^{-5}
18. Find the smallest positive whole number n for which \displaystyle \left(\sqrt{2} - \sqrt{6}\,i\right)^n is a positive real number, and find that number.

Day 2: identities and repeated turns

19. Use De Moivre’s theorem with n = 4 to write \cos 4\theta and \sin 4\theta in terms of \cos\theta and \sin\theta.
20. Given \displaystyle \cos\theta = \frac{2}{5}, find \cos 3\theta exactly.
21. Yamuna starts at the number 1 and multiplies by \displaystyle w = 0.9\operatorname{cis}\frac{\pi}{6} again and again. Where is she after 6 multiplications, and after 12? Give each to the nearest hundredth.

Reasoning

22. Show that |zw| = |z||w| for nonzero z and w. Then give an example where \operatorname{Arg}(zw) \ne \operatorname{Arg} z + \operatorname{Arg} w.
23. Ranjani says the powers z, z^2, z^3, \ldots stay on one circle when |z| = 1, spiral outward when |z| > 1, and spiral in toward 0 when 0 < |z| < 1. Explain why she is right.
24. Show that if z^n = 1 for some positive whole number n, then |z| = 1.
25. Use De Moivre’s theorem with n = 2 to prove the double-angle formulas \cos 2\theta = \cos^2\theta - \sin^2\theta and \sin 2\theta = 2\sin\theta\cos\theta.