AATA Multiplicative Group of Complex Numbers

The polar representation of a complex number makes it easy to find products and powers of complex numbers. The proof of the following proposition is straightforward and is left as an exercise.

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Proposition 4.20.

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Let

z = r cis θ

and

w = s cis ϕ

be two nonzero complex numbers. Then

z w = r s cis (θ + ϕ) .

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Example 4.21.

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z = 3 cis (π / 3)

and

w = 2 cis (π / 6),

then

z w = 6 cis (π / 2) = 6 i .

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Theorem 4.22. DeMoivre.

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Let

z = r cis θ

be a nonzero complex number. Then

[r cis θ]^{n} = r^{n} cis (n θ)

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for

n = 1, 2, \dots .

Proof.

We will use induction on

n .

For

n = 1

the theorem is trivial. Assume that the theorem is true for all

k

such that

1 \leq k \leq n .

Then

\begin{aligned} z^{n + 1} & = z^{n} z \\ = r^{n} (\cos n θ + i \sin n θ) r (\cos θ + i \sin θ) \\ = r^{n + 1} [(\cos n θ \cos θ - \sin n θ \sin θ) + i (\sin n θ \cos θ + \cos n θ \sin θ)] \\ = r^{n + 1} [\cos (n θ + θ) + i \sin (n θ + θ)] \\ = r^{n + 1} [\cos (n + 1) θ + i \sin (n + 1) θ] . \end{aligned}

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Example 4.23.

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Suppose that

z = 1 + i

and we wish to compute

z^{10} .

Rather than computing

(1 + i)^{10}

directly, it is much easier to switch to polar coordinates and calculate

z^{10}

using DeMoivre’s Theorem:

\begin{aligned} z^{10} & = (1 + i)^{10} \\ = {(\sqrt{2} cis (\frac{π}{4}))}^{10} \\ = (\sqrt{2})^{10} cis (\frac{5 π}{2}) \\ = 32 cis (\frac{π}{2}) \\ = 32 i . \end{aligned}

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Subsection The Circle Group and the Roots of Unity

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The multiplicative group of the complex numbers,

C^{*},

possesses some interesting subgroups. Whereas

Q^{*}

and

R^{*}

have no interesting subgroups of finite order,

C^{*}

has many. We first consider the circle group,

T = {z \in C : | z | = 1} .

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The following proposition is a direct result of Proposition 4.20.

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Proposition 4.24.

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The circle group is a subgroup of

C^{*} .

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Although the circle group has infinite order, it has many interesting finite subgroups. Suppose that

H = {1, - 1, i, - i} .

Then

H

is a subgroup of the circle group. Also,

1,

- 1,

i,

and

- i

are exactly those complex numbers that satisfy the equation

z^{4} = 1 .

The complex numbers satisfying the equation

z^{n} = 1

are called the

n

th roots of unity.

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Theorem 4.25.

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z^{n} = 1,

then the

n

th roots of unity are

z = cis (\frac{2 k π}{n}),

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where

k = 0, 1, \dots, n - 1 .

Furthermore, the

n

th roots of unity form a cyclic subgroup of

T

of order

n

Proof.

By DeMoivre’s Theorem,

z^{n} = cis (n \frac{2 k π}{n}) = cis (2 k π) = 1 .

The

z

’s are distinct since the numbers

2 k π / n

are all distinct and are greater than or equal to 0 but less than

2 π .

The fact that these are all of the roots of the equation

z^{n} = 1

follows from from Corollary 17.9, which states that a polynomial of degree

n

can have at most

n

roots. We will leave the proof that the

n

th roots of unity form a cyclic subgroup of

T

as an exercise.

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A generator for the group of the

n

th roots of unity is called a primitive

n

th root of unity.

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Example 4.26.

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The 8th roots of unity can be represented as eight equally spaced points on the unit circle (Figure 4.27). The primitive 8th roots of unity are

\begin{aligned} ω & = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \\ ω^{3} & = - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \\ ω^{5} & = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i \\ ω^{7} & = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i . \end{aligned}

The 8 roots of unity are spaced evenly around the unit circle beginning with 1 on the positive horizontal axis and followed by omega, i, the cube of omega, -1, omega to the fifth power, -i, and omega to the seventh power. — Figure 4.27. 8th roots of unity

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