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Multiplicative Group of Complex Numbers

The complex numbers are defined as \[\begin{aligned}\end{aligned}\], where i^2 = -1. If z = a + bi, then a is the real part of z and b is the imaginary part of z. To add two complex numbers z=a+bi and w= c+di,

Multiplicative Group of Complex Numbers

The complex numbers are defined as \[\begin{aligned}\end{aligned}\], where \(i^2 = -1\). If \(z = a + bi\), then \(a\) is the real part of \(z\) and \(b\) is the imaginary part of \(z\).

To add two complex numbers \(z=a+bi\) and \(w= c+di\), we just add the corresponding real and imaginary parts: \[\begin{aligned}\end{aligned}\]. Remembering that \(i^2 = -1\), we multiply complex numbers just like polynomials. The product of \(z\) and \(w\) is \[\begin{aligned}\end{aligned}\].

Every nonzero complex number \(z = a +bi\) has a multiplicative inverse; that is, there exists a \(z^{-1} \in {\mathbb C}^\ast\) such that \(z z^{-1} = z^{-1} z = 1\). If \(z = a + bi\), then \[\begin{aligned}\end{aligned}\]. The complex conjugate of a complex number \(z = a + bi\) is defined to be \(\overline{z} = a- bi\). The absolute value or modulus of \(z = a + bi\) is \(|z| = \sqrt{a^2 + b^2}\).

Example

Let \(z = 2 + 3i\) and \(w = 1-2i\). Then \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\]. Also, \[\begin{aligned}z^{-1} & = \frac{2}{13} - \frac{3}{13}i \\ |z| & = \sqrt{13} \\ \overline{z} & = 2-3i\end{aligned}\].

There are several ways of graphically representing complex numbers. We can represent a complex number \(z = a +bi\) as an ordered pair on the \(xy\) plane where \(a\) is the \(x\) (or real) coordinate and \(b\) is the \(y\) (or imaginary) coordinate. This is called the rectangular or Cartesian representation. The rectangular representations of \(z_1 = 2 + 3i\), \(z_2 = 1 - 2i\), and \(z_3 = - 3 + 2i\) are depicted in .

Nonzero complex numbers can also be represented using polar coordinates. To specify any nonzero point on the plane, it suffices to give an angle \(\theta\) from the positive \(x\) axis in the counterclockwise direction and a distance \(r\) from the origin, as in . We can see that \[\begin{aligned}\end{aligned}\]. Hence, \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}a & = r \cos \theta \\ b & = r \sin \theta\end{aligned}\]. We sometimes abbreviate \(r( \cos \theta + i \sin \theta)\) as \(r \cis \theta\). \(\cis \theta\) \(\cos \theta + i \sin \theta\) To assure that the representation of \(z\) is well-defined, we also require that \(0^{\circ} \leq \theta \lt 360^{\circ}\). If the measurement is in radians, then \(0 \leq \theta \lt2 \pi\).

Example

Suppose that \(z = 2 \cis 60^{\circ}\). Then \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\]. Hence, the rectangular representation is \(z = 1+\sqrt{3}\, i\).

Conversely, if we are given a rectangular representation of a complex number, it is often useful to know the number's polar representation. If \(z = 3 \sqrt{2} - 3 \sqrt{2}\, i\), then \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\], so \(3 \sqrt{2} - 3 \sqrt{2}\, i=6 \cis 315^{\circ}\).

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.

Example

If \(z = 3 \cis( \pi / 3 )\) and \(w = 2 \cis(\pi / 6 )\), then \(zw = 6 \cis( \pi / 2 ) = 6i\).

Theorem: DeMoivre

Let \(z = r \cis \theta\) be a nonzero complex number. Then \[\begin{aligned}\end{aligned}\] for \(n = 1, 2, \ldots\).

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

Example

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: \[\begin{aligned}z^{10} & = (1+i)^{10} \\ & = \left( \sqrt{2} \cis \left( \frac{\pi }{4} \right) \right)^{10} \\ & = ( \sqrt{2}\, )^{10} \cis \left( \frac{5\pi }{2} \right) \\ & = 32 \cis \left( \frac{\pi }{2} \right) \\ & = 32i\end{aligned}\].

The Circle Group and the Roots of Unity

The multiplicative group of the complex numbers, \({\mathbb C}^*\), possesses some interesting subgroups. Whereas \({\mathbb Q}^*\) and \({\mathbb R}^*\) have no interesting subgroups of finite order, \({\mathbb C}^*\) has many. We first consider the circle group, \(\mathbb T\) the circle group \[\begin{aligned}\end{aligned}\]. The following proposition is a direct result of .

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.

Theorem

If \(z^n = 1\), then the \(n\)th roots of unity are \[\begin{aligned}\end{aligned}\], where \(k = 0, 1, \ldots, n-1\). Furthermore, the \(n\)th roots of unity form a cyclic subgroup of \({\mathbb T}\) of order \(n\)

Proof

By DeMoivre's Theorem, \[\begin{aligned}\end{aligned}\]. The \(z\)'s are distinct since the numbers \(2 k \pi /n\) are all distinct and are greater than or equal to 0 but less than \(2 \pi\). The fact that these are all of the roots of the equation \(z^n=1\) follows from from , 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 \({\mathbb T}\) as an exercise.

A generator for the group of the \(n\)th roots of unity is called a primitive \(n\)th root of unity.

Example

The 8th roots of unity can be represented as eight equally spaced points on the unit circle (). The primitive 8th roots of unity are \[\begin{aligned}\omega & = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \\ \omega^3 & = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \\ \omega^5 & = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i \\ \omega^7 & = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\end{aligned}\].

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\theta

theta
The usual name for an angle.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

imaginary unit
i² = −1.

^\circ

degrees
1/360 of a full turn. 180° = π radians.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

\blacksquare\ \text{or}\ \square

end of proof (halmos)
Marks the point where the statement has been established.

a \equiv b \pmod n

congruent modulo n
n divides a − b; a and b have the same remainder.

a \mid b,\ \gcd(a,b)

divides, greatest common divisor
b is a multiple of a; the largest number dividing both.

(G, \cdot),\ e,\ g^{-1}

group, identity, inverse
A set with an operation; the do-nothing element; the element that undoes g.

G \cong H,\ G / N

isomorphic, quotient group
Same structure; the group of cosets of a normal subgroup N.

\mathbb{Z}/n\mathbb{Z},\ \mathbb{Z}_n

integers modulo n
The remainders 0…n−1 with clock arithmetic.

\operatorname{Hom}(A, B),\ f \circ g

arrows from A to B, composition
The set of morphisms; do g then f.

Questions people ask

What is a group, in plain words?

A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group — that is where the idea came from.

What is the difference between a ring and a field?

A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.

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Parts of this page are adapted from Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Condensed and re-explained here; errors are ours.

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