10.4 Polar Form for Complex Numbers

Introduction

In Section 10.3 we represented the sum of two complex numbers graphically as a vector addition. Is there a way to visualize the product or quotient of two complex numbers? One way to explore a new idea is to consider a simple case. What happens to the vector representing a complex number when we multiply the number by $i$ ?

Let $z = 1 + i,$ and calculate $i z, i^{2} z,$ and $i^{3} z$ .
Plot $z, i z, i^{2} z,$ and $i^{3} z$ as points on the complex plane.

$i z = i - 1, i^{2} z = - 1 - i, i^{3} z = - i + 1$

The previous example suggests that multiplication by a complex number results in a rotation. Polar coordinates are well suited to processes that involve rotation, because they use angles to specify location. Thus, we will next represent complex numbers in an alternate polar form.

Polar Form

The figure below shows the complex number $z = 3 + 3 i$ , represented as a vector in the complex plane. The distance $r$ from the origin to $z$ is

$r = \sqrt{3^{2} + 3^{2}} = 3 \sqrt{2}$

and the angle from the real axis to the vector is $θ = \frac{π}{4}$ .

Using right triangle trigonometry, we see that

$\begin{aligned} a & = 3 \sqrt{2} \cos (\frac{π}{4}) = 3 \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 3 \\ b & = 3 \sqrt{2} \sin (\frac{π}{4}) = 3 \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 3 \end{aligned}$

In general, we can write the real and imaginary parts of $z$ in terms of $r$ and $θ$ as

$a = r \cos (θ) and b = r \sin (θ)$

Thus, the complex number $z = a + b i$ can also be written as

$z = (r \cos (θ)) + (r \sin (θ)) i or z = r (\cos (θ) + i \sin (θ))$

Example 2

Find the polar form for $z = \sqrt{3} + i$ .

We compute

$r = \sqrt{a^{2} + b^{2}} = \sqrt{3 + 1} = 2 and \tan (θ) = \frac{1}{\sqrt{3}}$

The angle $θ$ is in the first quadrant, so

$θ = \tan^{- 1} (\frac{1}{\sqrt{3}}) = \frac{π}{6}$

Thus, $z = 2 (\cos (\frac{π}{6}) + i \sin (\frac{π}{6}))$

Find the polar form for $w = 1 + i \sqrt{3}$ .

$w = 2 (\cos \frac{π}{3} + i \sin \frac{π}{3})$

Of course, we can always recover the Cartesian form of a complex number from its polar form by evaluating the trigonometric functions. We'll check the result of the previous example:

$\begin{aligned} z & = 2 (\cos (\frac{π}{6}) + i \sin (\frac{π}{6})) \\ = 2 (\frac{\sqrt{3}}{2} + i \cdot \frac{1}{2}) = \sqrt{3} + i \end{aligned}$

Products and Quotients in Polar Form

The polar form is especially convenient for computing the product or quotient of two complex numbers.

This formula, which you will prove in the Homework Problems, says that the product of two complex numbers in polar form is the complex number with modulus $r R$ and argument $α + β$ . Thus, to find the product of two complex numbers, we multiply their lengths and add their arguments.

Example 3

Find the product of $z = 2 (\cos (\frac{π}{6}) + i \sin (\frac{π}{6}))$ and $w = 2 (\cos (\frac{π}{3}) + i \sin (\frac{π}{3}))$ .

We multiply the moduli, $2 \cdot 2 = 4$ , and add the arguments, $\frac{π}{6} + \frac{π}{3} = \frac{π}{2}$ . The polar form of the product is thus

$z w = 4 (\cos (\frac{π}{2}) + i \sin (\frac{π}{2})) = 4 i$

Note that $z$ and $w$ are the numbers from the previous Example and Exercise, namely $z = \sqrt{3} + i$ and $w = 1 + i \sqrt{3}$ . You can compute the product $z w$ in Cartesian form to check that you get the same result, $4 i$ .

The figure at right shows the graphs of $z$ and $w$ , and their product, $z w$ . You can see that multiplying $z$ by $w$ rotates the graph of $z$ by the argument of $w$ , namely $\frac{π}{3}$ or $60 degrees$ .

Find the polar forms of $z = 3 + 3 i$ and $w = - 2 - 2 i$ .
Compute the product $z w$ .

$z = 3 \sqrt{2} (\cos (\frac{π}{4}) + i \sin (\frac{π}{4}))$ , $w = 2 \sqrt{2} (\cos (\frac{5 π}{4}) + i \sin (\frac{5 π}{4}))$
$minus 12 i$

The quotient of two complex numbers in polar form is computed in a similar fashion. We divide the moduli and subtract the arguments.

Example 4

Find the quotient of $z = 2 (\cos (\frac{π}{6}) + i \sin (\frac{π}{6}))$ and $w = 2 (\cos (\frac{π}{3}) + i \sin (\frac{π}{3}))$ .

We divide the moduli, $\frac{2}{2} = 1$ , and subtract the arguments, $\frac{π}{6} - \frac{π}{3} = \frac{- π}{6}$ .

The polar form of the quotient is thus

$\frac{z}{w} = 1 (\cos (\frac{- π}{6}) + i \sin (\frac{- π}{6})) = \frac{\sqrt{3}}{2} - \frac{1}{2} i$

The figure at right shows the graphs of $z, w$ , and the quotient $\frac{z}{w}$ .

Compute the quotient $\frac{z}{w}$ for $z = 3 + 3 i$ and $w = - 2 - 2 i$ .

$\frac{- 3}{2}$

Powers and Roots of Complex Numbers

Because raising to a power is just repeated multiplication, we can also use the polar form to simplify powers of a complex number. For example, if $z = r (\cos (α) + i \sin (α))$ , then

$z^{2} = z \cdot z = r \cdot r (\cos (α + α) + i \sin (α + α))$

We compute $z^{2}$ by squaring the modulus, $r$ , and doubling the argument, $α$ , so the polar form is

$z^{2} = r^{2} (\cos (2 α) + i \sin (2 α))$

An analogous result holds for all positive integers $n$ , and is known as De Moivre's Theorem.

Example 5

Raise $z = \sqrt{3} + i$ to the fourth power.

We use the polar form, $z = 2 (\cos (\frac{π}{6}) + i \sin (\frac{π}{6}))$ , and apply De Moivre's theorem. Then

$\begin{aligned} z^{4} & = 2^{4} (\cos (4 \cdot \frac{π}{6}) + i \sin (4 \cdot \frac{π}{6})) \\ = 16 (\cos (\frac{2 π}{3}) + i \sin (\frac{2 π}{3})) \end{aligned}$

The Cartesian form of $z to the power 4$ is thus $16 (\frac{- 1}{2} + i \frac{\sqrt{3}}{2})$ , or $- 8 + 8 i \sqrt{3}$ . You can check that you get this same result if you compute ${(\sqrt{3} + i)}^{4}$ by expanding the power.

Compute $w^{3}$ , for $w = - \sqrt{2} + \sqrt{2} i$ .

$4 \sqrt{2} + 4 \sqrt{2} i$

DeMoivre's Theorem also works for rational values of $n$ , so we can compute roots of complex numbers. For example, by applying the theorem with $n = \frac{1}{2}$ , we see that one of the square roots of $z = r (\cos (α) + i \sin (α))$ is

$z^{1 / 2} = r^{1 / 2} (\cos (\frac{α}{2}) + i \sin (\frac{α}{2}))$

Now, every number,whether real or complex, has two square roots. To find the other root, remember that we can add a multiple of $2 π$ to the argument of $z$ , that is, we can also write the polar form of $z$ as

$z = r (\cos (α + 2 π) + i \sin (α + 2 π))$

The second square root of $z$ is thus

$r^{1 / 2} (\cos (\frac{α}{2} + π) + i \sin (\frac{α}{2} + π))$

As an example consider $w = - 8 + 8 i \sqrt{3},$ whose polar form is

$16 (\cos (\frac{2 π}{3}) + i \sin (\frac{2 π}{3}))$

or by adding $2 π$ to the argument,

$16 (\cos (\frac{8 π}{3}) + i \sin (\frac{8 π}{3}))$

The two square roots of $w$ are

$\begin{aligned} z_{1} & = 16^{1 / 2} (\cos (\frac{π}{3}) + i \sin (\frac{π}{3})) = 4 (\frac{1}{2} + i \frac{\sqrt{3}}{2}) = 2 + 2 i \sqrt{3} \\ z_{2} & = 16^{1 / 2} (\cos (\frac{4 π}{3}) + i \sin (\frac{4 π}{3})) \\ = 4 (\frac{- 1}{2} + i \frac{- \sqrt{3}}{2}) = - 2 - 2 i \sqrt{3} \end{aligned}$

You can verify that both of these numbers are square roots of $w = - 8 + 8 i \sqrt{3}$ , for instance,

$\begin{aligned} {(2 + 2 i \sqrt{3})}^{2} & = 4 + 2 (2) (2 i \sqrt{3}) + 12 i^{2} \\ = - 8 + 8 i \sqrt{3} \end{aligned}$

The graphs of $w$ and its two square roots are shown at right.

It is not hard to show that every number has three complex cube roots, four complex fourth roots, and so on.

Example 6

Find three complex cube roots of $8$ .

One polar form of $z equals 8$ is

$z = 8 (\cos 0 + i \sin 0)$

so one of its cube roots is

$z_{1} = 8^{1 / 3} (\cos (\frac{0}{3}) + i \sin (\frac{0}{3})) = 2 (\cos 0 + i \sin 0) = 2$

But by adding multiples of $2 π$ to the argument, we can also write the polar form of as

$z = 8 (\cos 2 π + i \sin 2 π) and z = 8 (\cos 4 π + i \sin 4 π)$

Applying De Moivre's theorem to each of these expressions gives us two more cube roots

$\begin{aligned} z_{2} & = 2 (\cos (\frac{2 π}{3}) + i \sin (\frac{2 π}{3})) = - 1 + i \sqrt{3} \\ z_{3} & = 2 (\cos (\frac{4 π}{3}) + i \sin (\frac{4 π}{3})) = - 1 - i \sqrt{3} \end{aligned}$

The three roots are shown at right. Note that they are evenly spaced around a circle of radius 2. You can check that adding further muliples of $2 π$ to the argument does not generate any new cube roots.

Note

Roots of a Complex Number

A complex number $z = r (\cos α + i \sin α)$ in polar form has $n$ complex $n$ th roots, given by

$z_{k} = r^{1 / n} (\cos (\frac{α + 2 π k}{n}) + i \sin (\frac{α + 2 π k}{n}))$

for $k = 0, 1, 2, \dots, n - 1$ .

Write $i$ in polar form.
Find four complex fourth roots of $i$ .

$\cos (\frac{π}{2}) + i \sin (\frac{π}{2})$
$\cos (\frac{π}{8}) + i \sin (\frac{π}{8})$ , $\cos (\frac{5 π}{8}) + i \sin (\frac{5 π}{8})$ , $\cos (\frac{9 π}{8}) + i \sin (\frac{9 π}{8})$ , $\cos (\frac{13 π}{8}) + i \sin (\frac{13 π}{8})$

Review the following skills you will need for this section.

Algebra Refresher

Write with fractional exponents.

$\sqrt[3]{a}$
$\frac{1}{\sqrt[4]{n}}$
$the 4th root of t cubed$
$\sqrt[3]{z^{5}}$
$\frac{1}{\sqrt{x + 2}}$
$\sqrt[3]{(w - 1)^{2}}$
$z^{3} \sqrt[5]{w^{4}}$
$\frac{a^{2}}{\sqrt[4]{b^{3}}}$

$\underset{―}{}$

Algebra Refresher Answers

$a to the power 1 / 3$
$n^{- 1 / 4}$
$t to the power 3 / 4$
$z to the power 5 / 3$
$(x + 2)^{- 1 / 2}$
$(w - 1)^{2 / 3}$
$z cubed w to the power 4 / 5$
$a^{2} b^{- 3 / 4}$

Section 10.4 Summary

Vocabulary

Argument
Modulus

Concepts

Multiplying a complex number by $i$ rotates its graph by $90 degrees$ around the origin.
Note
Polar Form for a Complex Number
The complex number $z = a + b i$ can be written in the polar form
$z = r (\cos (θ) + i \sin (θ))$
where
$r = \sqrt{a^{2} + b^{2}}$
and $θ$ is defined by
$a = r \cos (θ), b = r \sin (θ), 0 \leq θ \leq 2 π$
The angle $θ$ is called the argument of the complex number, and $r$ is its length, or modulus.
Note
Product and Quotient in Polar Form
If $z_{1} = r (\cos (α) + i \sin (α))$ and $z_{2} = R (\cos (β) + i \sin (β))$ , then
$z_{1} z_{2} = r R (\cos (α + β) + i \sin (α + β))$
and
$\frac{z_{1}}{z_{2}} = \frac{r}{R} (\cos (α - β) + i \sin (α - β))$
Note
De Moivre's Theorem
If $z = r (\cos (α) + i \sin (α))$ is a complex number in polar form, and $n$ is a positive integer, then
$z^{n} = r^{n} (\cos (n α) + i \sin (n α))$
Note
Roots of a Complex Number
A complex number $z = r (\cos (α) + i \sin (α))$ in polar form has $n$ complex $n$ th roots, given by
$z_{k} = r^{1 / n} (\cos (\frac{α + 2 π k}{n}) + i \sin (\frac{α + 2 π k}{n}))$
for $k = 0, 1, 2, \dots, n - 1$ .

Study Questions

What happens to the vector representing a complex number when we multiply the number by $- i$ ?
If $z = p + q i$ , what do these expressions represent? $p^{2} + q^{2}, p - q i, \frac{q}{p}$
If $z sub 1$ and $z sub 2$ lie on the unit circle, explain why $z sub 1 z sub 2$ and $\frac{z_{1}}{z_{2}}$ also lie on the unit circle.
Explain how you could use DeMoivre's theorem to compute $z to the power 2 / 3$ .

Skills

Convert from polar form to standard form #5–12
Write a complex number in polar form #13-22
Find the product or quotient of two complex numbers in polar form #25–32
Find a power of a complex number #33–42
Find the complex roots of a number #43–48, 51–52, 55–60

Homework 10-4

For Problems 1–4, simplify and plot each complex number as a point on the complex plane.

$1, i, i squared, i cubed$ and $i to the power 4$

$1$ , $i$ , $minus 1$ , $- i$ , $1$

$- 1, - i, - i^{2}, - i^{3}$ and $- i^{4}$

$1 plus 2 i$ and $i open parenthesis 1 plus 2 i close parenthesis$

$1 plus 2 i$ , $- 2 + i$

$3 minus 4 i$ and $i open parenthesis 3 minus 4 i close parenthesis$

For Problems 5–8, write the complex numbers in standard form. Give exact values for your answers.

$6 (\cos (\frac{2 π}{3}) + i \sin (\frac{2 π}{3}))$

$- 3 + 3 i \sqrt{3}$

$4 (\cos (\frac{7 π}{4}) + i \sin (\frac{7 π}{4}))$

$\sqrt{2} (\cos (\frac{3 π}{4}) + i \sin (\frac{3 π}{4}))$

$- 1 + i$

$\frac{3}{2} (\cos (\frac{5 π}{6}) + i \sin (\frac{5 π}{6}))$

For Problems 9–12, write the complex numbers in standard form. Round your answers to hundredths.

$5 open parenthesis cos ⁡ open parenthesis 5.2 close parenthesis plus i sin ⁡ open parenthesis 5.2 close parenthesis close parenthesis$

$2.34 minus 4.21 i$

$3 open parenthesis cos ⁡ open parenthesis 3.5 close parenthesis plus i sin ⁡ open parenthesis 3.5 close parenthesis close parenthesis$

$12 open parenthesis cos ⁡ open parenthesis 115 degrees close parenthesis plus i sin ⁡ open parenthesis 115 degrees close parenthesis close parenthesis$

$- 5.07 + 10.88 i$

)

$20 open parenthesis cos ⁡ open parenthesis 250 degrees close parenthesis plus i sin ⁡ open parenthesis 250 degrees close parenthesis close parenthesis$

For Problems 13–16, write the complex numbers in polar form. Give exact values for your answers.

$3 i$ and $minus 3 i$

$3 (\cos (\frac{π}{2}) + i \sin (\frac{π}{2}))$ , $3 (\cos (\frac{3 π}{2}) + i \sin (\frac{3 π}{2}))$

$2 plus 2 i$ and $2 minus 2 i$

$- 3 - \sqrt{3} i$ and $3 - \sqrt{3} i$

$2 \sqrt{3} (\cos (\frac{7 π}{6}) + i \sin (\frac{7 π}{6}))$ , $2 \sqrt{3} (\cos (\frac{11 π}{6}) + i \sin (\frac{11 π}{6}))$

$2 \sqrt{3} + 2 i$ and $- 2 \sqrt{3} + 2 i$

For Problems 17–22, write the complex numbers in polar form. Round your answers to hundredths.

$- 4 + 2 i$ and $4 minus 2 i$

$4.47 open parenthesis cos ⁡ open parenthesis 2.68 close parenthesis plus i sin ⁡ open parenthesis 2.68 close parenthesis close parenthesis,$ $4.47 open parenthesis cos ⁡ open parenthesis 5.82 close parenthesis plus i sin ⁡ open parenthesis 5.82 close parenthesis close parenthesis$

$- 3 - 8 i$ and $3 plus 8 i$

$9 minus 5 i$ and $9 plus 5 i$

$8.60 open parenthesis cos ⁡ open parenthesis 5.78 close parenthesis plus i sin ⁡ open parenthesis 5.78 close parenthesis close parenthesis,$ $8.60 open parenthesis cos ⁡ open parenthesis 0.51 close parenthesis plus i sin ⁡ open parenthesis 0.51 close parenthesis close parenthesis$

$2 plus 6 i$ and $2 minus 6 i$

$3 + 4 i, 3 - 4 i, - 3 + 4 i,$ and $- 3 - 4 i$

$5 open parenthesis cos ⁡ open parenthesis 0.93 close parenthesis plus i sin ⁡ open parenthesis 0.93 close parenthesis close parenthesis,$ $5 open parenthesis cos ⁡ open parenthesis 5.36 close parenthesis plus i sin ⁡ open parenthesis 5.36 close parenthesis close parenthesis,$ $5 open parenthesis cos ⁡ open parenthesis 2.21 close parenthesis plus i sin ⁡ open parenthesis 2.21 close parenthesis close parenthesis,$ $5 open parenthesis cos ⁡ open parenthesis 4.07 close parenthesis plus i sin ⁡ open parenthesis 4.07 close parenthesis close parenthesis$

$1 + 3 i, 1 - 3 i, - 1 + 3 i,$ and $- 1 - 3 i$

What can you conclude about the polar forms of $z$ and $\bar{z}$ ?

If $z = r (\cos (θ) + i \sin (θ)),$ then $\bar{z} = r (\cos (2 π - θ) + i \sin (2 π - θ)$

What can you conclude about the polar forms of $z$ and $- z$ ?

For Problems 25–28, find the product $z sub 1 z sub 2$ and the quotient $\frac{z_{1}}{z_{2}}$ .

$z_{1} = 4 (\cos (\frac{4 π}{3}) + i \sin (\frac{4 π}{3}))$ $z_{2} = \frac{1}{2} (\cos (\frac{5 π}{6}) + i \sin (\frac{5 π}{6}))$

$z_{1} z_{2} = 2 (\cos (\frac{π}{6}) + i \sin (\frac{π}{6})) = \sqrt{3} + i$ ; $\frac{z_{1}}{z_{2}} = 8 (\cos (\frac{π}{2}) + i \sin (\frac{π}{2})) = 8 i$

$z_{1} = 6 (\cos (\frac{5 π}{8}) + i \sin (\frac{5 π}{8}))$ $z_{2} = \frac{3}{2} (\cos (\frac{π}{8}) + i \sin (\frac{π}{8}))$

$z_{1} = 3 (\cos (\frac{3 π}{5}) + i \sin (\frac{3 π}{5}))$ $z_{2} = 2 (\cos (\frac{3 π}{10}) + i \sin (\frac{3 π}{10}))$

$z_{1} z_{2} = 6 (\cos (\frac{9 π}{10}) + i \sin (\frac{9 π}{10}))$ ; $\frac{z_{1}}{z_{2}} = \frac{3}{2} (\cos (\frac{3 π}{10}) + i \sin (\frac{3 π}{10}))$

$z_{1} = 4 (\cos (\frac{5 π}{12}) + i \sin (\frac{5 π}{12}))$ $z_{2} = 6 (\cos (\frac{3 π}{4}) + i \sin (\frac{3 π}{4}))$

For Problems 29–32, convert the complex numbers to polar form, then find the product $z sub 1 z sub 2$ and the quotient $\frac{z_{1}}{z_{2}}$ .

$z_{1} = 2 i, z_{2} = 4 i$

$z sub 1 z sub 2 equals 8$ ; $\frac{z_{1}}{z_{2}} = \frac{1}{2}$

$z_{1} = - 2 i, z_{2} = 3 i$

$z_{1} = 2 \sqrt{3} - 2 i$ , $z_{2} = - 1 + i$

$z_{1} z_{2} = 4 \sqrt{2} (\cos \frac{7 π}{12} + i \sin \frac{7 π}{12})$ ; $\frac{z_{1}}{z_{2}} = 2 \sqrt{2} (\cos \frac{13 π}{12} + i \sin \frac{13 π}{12})$

$z_{1} = \sqrt{3} + i$ , $z_{2} = - 1 + i \sqrt{3}$

For Problems 33–38, find the power.

$(2 + 2 i)^{5}$

$- 128 - 128 i$

$(\sqrt{2} - \sqrt{2} i)^{6}$

$(- 1 + \sqrt{3} i)^{8}$

$- 128 - 128 \sqrt{3} i$

${(\frac{1}{2} + \frac{\sqrt{3}}{2} i)}^{12}$

$(\sqrt{3} - i)^{10}$

$512 + 512 \sqrt{3} i$

$(1 - i)^{20}$

For Problems 39–42, use De Moivre's theorem to find the reciprocal.

$2 minus 2 i$

$\frac{1}{4} + \frac{1}{4} i$

$3 + \sqrt{3} i$

$- \sqrt{2} + \sqrt{6} i$

$\frac{- \sqrt{2}}{8} - \frac{\sqrt{6}}{8} i$

$- 1 - i$

For Problems 43–48,

Find the roots and plot them in the complex plane.
Write the roots in standard form.

The square roots of $9 i$

$3 (\cos \frac{π}{4} + i \sin \frac{π}{4})$ , $3 (\cos \frac{3 π}{4} + i \sin \frac{3 π}{4})$
$\frac{3}{\sqrt{2}} + \frac{3}{\sqrt{2}} i$ , $\frac{- 3}{\sqrt{2}} - \frac{3}{\sqrt{2}} i$

The fourth roots of $minus 81$

The fifth roots of $32$

$2, 2 (\cos \frac{2 π}{5} + i \sin \frac{2 π}{5})$ , $2 (\cos \frac{4 π}{5} + i \sin \frac{4 π}{5})$ , $2 (\cos \frac{6 π}{5} + i \sin \frac{6 π}{5})$ , $2 (\cos \frac{8 π}{5} + i \sin \frac{8 π}{5})$
$2, 0.618 plus 1.9 i$ , $- 1.618 + 1.176 i$ , $- 1.618 - 1.176 i$ , $0.618 minus 1.902 i$

The cube roots of $i$

The cube roots of $4 \sqrt{3} + 4 i$

$4 (\cos \frac{π}{18} + i \sin \frac{π}{18})$ , $4 (\cos \frac{13 π}{18} + i \sin \frac{13 π}{18})$ , $4 (\cos \frac{25 π}{18} + i \sin \frac{25 π}{18})$
$1.97 plus 0.347 i$ , $- 1.286 + 1.532 i$ , $- 0.684 - 1.879 i$

The square roots of $- 2 + 2 \sqrt{3} i$

Show that any complex number of the form $z = \cos (θ) + i \sin (θ)$ lies on the unit circle in the complex plane.

$| z | = | \cos (θ) + i \sin (θ) | = \sqrt{\cos^{2} (θ) + \sin^{2} (θ)} = 1$

Show that if $| z | equals 1$ , then $\frac{1}{z} = \bar{z}$ .

Find three distinct cube roots of 1.
Find four distinct fourth roots of 1.
Find five distinct fifth roots of 1.
Find six distinct sixth roots of 1.

$1, (\cos \frac{2 π}{3} + i \sin \frac{2 π}{3}), (\cos \frac{4 π}{3} + i \sin \frac{4 π}{3})$
$1, i, - 1, - i$
$1, (\cos \frac{2 π}{5} + i \sin \frac{2 π}{5}), (\cos \frac{4 π}{5} + i \sin \frac{4 π}{5}),$ $(\cos \frac{6 π}{5} + i \sin \frac{6 π}{5}), (\cos \frac{8 π}{5} + i \sin \frac{8 π}{5})$
$1, (\cos \frac{π}{3} + i \sin \frac{π}{3}), (\cos \frac{2 π}{3} + i \sin \frac{2 π}{3}), - 1,$ $(\cos \frac{4 π}{3} + i \sin \frac{4 π}{3}), (\cos \frac{5 π}{3} + i \sin \frac{5 π}{3})$

Find the sum of the three distinct cube roots of 1. (Hint: Plot the roots.)
Find the sum of the four distinct fourth roots of 1.
Find the sum of the five distinct fifth roots of 1.
Find the sum of the six distinct sixth roots of 1.

If $n$ is a positive integer, define

$ω_{k} = \cos (\frac{2 π k}{n}) + i \sin (\frac{2 π k}{n})$

for $k = 0, 1, 2, \dots, n - 1.$ Show that $(ω_{k})^{n} = 1$ . (We call $ω_{k}$ an $n^{t h}$ root of unity.)

$(ω_{k})^{n} = 1^{n} (\cos (n \cdot \frac{2 π k}{n}) + i \sin (n \cdot \frac{2 π k}{n})) = 1 (\cos 2 π k + i \sin 2 π k) = 1$

Let $ω = \cos (\frac{2 π}{n}) + i \sin (\frac{2 π}{n})$ , where $n$ is a positive integer. Show that the $n$ distinct $n^{t h}$ roots of unity are $ω, ω^{2}, ω^{3}, \dots, ω^{n - 1}$ .

For Problems 55-60, solve the equation.

$z^{4} + 4 z^{2} + 8 = 0$

$8^{1 / 4} (\cos (\frac{3 π}{8}) + i \sin (\frac{3 π}{8}))$ , $8^{1 / 4} (\cos (\frac{5 π}{8}) + i \sin (\frac{5 π}{8}))$ , $8^{1 / 4} (\cos (\frac{11 π}{8}) + i \sin (\frac{11 π}{8}))$ , $8^{1 / 4} (\cos (\frac{13 π}{8}) + i \sin (\frac{13 π}{8}))$

$z^{6} + 4 z^{3} + 8 = 0$

$z^{6} - 8 = 0$

$the square root of 2$ , $\sqrt{2} (\cos (\frac{π}{3}) + i \sin (\frac{π}{3}))$ , $\sqrt{2} (\cos (\frac{2 π}{3}) + i \sin (\frac{2 π}{3}))$ , $- \sqrt{2}$ , $\sqrt{2} (\cos (\frac{4 π}{3}) + i \sin (\frac{4 π}{3}))$ , $\sqrt{2} (\cos (\frac{5 π}{3}) + i \sin (\frac{5 π}{3}))$

$z^{4} - 9 i = 0$

$z^{4} + 2 z^{2} + 4 = 0$

$\sqrt{2} (\cos (\frac{π}{3}) + i \sin (\frac{π}{3}))$ , $\sqrt{2} (\cos (\frac{2 π}{3}) + i \sin (\frac{2 π}{3}))$ , $\sqrt{2} (\cos (\frac{4 π}{3}) + i \sin (\frac{4 π}{3}))$ , $\sqrt{2} (\cos (\frac{5 π}{3}) + i \sin (\frac{5 π}{3}))$

$z^{4} - 2 z^{2} + 4 = 0$

Let $z = \cos (θ) + i \sin (θ)$ . Compute $z^{2}$ by expanding the product.
Use De Moivre's theorem to compute $z^{2}$ .
Compare your answers to (a) and (b) to write identities for $sin ⁡ open parenthesis 2 θ close parenthesis$ and $cos ⁡ open parenthesis 2 θ close parenthesis$ .

$\cos^{2} (θ) - \sin^{2} (θ) + (2 \sin (θ) \cos (θ)) i$
$cos ⁡ open parenthesis 2 θ close parenthesis plus i sin ⁡ open parenthesis 2 θ close parenthesis$
$\sin (2 θ) = 2 \sin (θ) \cos (θ); \cos (2 θ) = \cos^{2} (θ) - \sin^{2} (θ)$

Let $z = \cos (θ) + i \sin (θ)$ . Compute $z^{3}$ by expanding the product.
Use De Moivre's theorem to compute $z^{3}$ .
Compare your answers to (a) and (b) to write identities for $sin ⁡ open parenthesis 3 θ close parenthesis$ and $cos ⁡ open parenthesis 3 θ close parenthesis$ .

Problems 63 and 64 show that multiplication by $i$ results in a rotation of $90 degrees$ .

Suppose that $z = a + b i$ and that the real numbers $a$ and $b$ are both nonzero.

What is the slope of the segment in the complex plane joining the origin to $z$ ?
What is the slope of the segment in the complex plane joining the origin to $z i$ ?
What is the product of the slopes of the two segments from parts (a) and (b)? What can you conclude about the angle between the two segments?

$\frac{b}{a}$
$\frac{- a}{b}$
$minus 1$ , $\frac{π}{2}$

Suppose that $z = a + b i$ and that $a$ and $b$ are both real numbers.

If $a not equal to 0$ and $b equals 0$ , then what is the slope of the segment in the complex plane joining the origin to $z$ ? What is the slope of the segment joining the origin to $z i$ ?
If $a equals 0$ and $b not equal to 0$ , then what is the slope of the segment in the complex plane joining the origin to $z$ ? What is the slope of the segment joining the origin to $z i$ ?
What can you conclude about the angle between the two segments from parts (a) and (b)?

Prove the product rule by following the steps.

Suppose $z_{1} = a + b i$ and $z_{2} = c + d i$ . Compute $z sub 1 z sub 2$ .
Now suppose that $z_{1} = r (\cos (α) + i \sin (α))$ and $z_{2} = R (\cos (β) + i \sin (β))$ . Write $a, b, c$ and $d$ in terms of $r, R, α$ and $β$ .
Substitute your expressions for $a, b, c$ and $d$ into your formula for $z sub 1 z sub 2$ .
Use the laws of sines and cosines to simplify your answer to part (c).

$z_{1} z_{2} = (a c - b d) + (a d + b c) i$
$a = r \cos (α), b = r \sin (α), c = R \cos (β), d = R \sin (β)$
$(a c - b d) + (a d + b c) i = (r R \cos (α) \cos (β) - r R \sin (α) \sin (β))$ $+ (r R \cos (α) \sin (β) + r R \sin (α) \cos (β)) i$
$r R (\cos (α + β) + i \sin (α + β))$

Let $z_{1} = r (\cos (α) + i \sin (α))$ and $z_{2} = R (\cos (β) + i \sin (β))$ . Prove the quotient rule as follows: Set $w = \frac{r}{R} (\cos (α - β) + i \sin (α - β)),$ and show that $z sub 1 equals w z sub 2$ .

Trigonometry by Katherine Yoshiwara (yoshiwarabooks.org), GNU Free Documentation License 1.2 or later. Adapted for the XYZ HTML edition with the authors' permission (recorded 2026-07-04). License: GFDL-1.2-or-later.