# Complex Numbers

**A nonempty bounded subset of ZZ contains a maximal
element**

This assertion is used several times in these notes, here is its proof.

Let m be an arbitrary element of the set M in question, there is at least one,
by assumption.

Further, let c be the bound. Then the algorithm

**for **n=c:-1:m** do: if **n ∈ M**, break, fi, od**

is guaranteed to halt after finitely many steps, and the current value of n is the
maximal element

of M.

Also, note the corollary that a bounded function into the integers takes on its
maximal value:

its range then contains a maximal element and any preimage of that maximal
element will do.

**Complex numbers**

A complex number is of the form

z = a + ib.

with a and b real numbers, called, respectively, the **real part of **z and
the** imaginary part of** z,

and i the** imaginary unit**, i.e.,

Actually, there are two complex numbers whose square is −1. We denote the other
one by −i. Be

aware that, in parts of Engineering, the symbol j is used instead of i.

MATLAB works internally with (double precision) complex numbers. Both variables
i

and j in MATLAB are initialized to the value i.

One adds complex numbers by adding separately their real and imaginary parts.
One multiplies

two complex numbers by multiplying out and rearranging, mindful of the fact that
i^{2} = −1. Thus,

(a + ib)(c + id) = ac + aid + bic − bd = (ac − bd) + i(ad + bc).

Note that both addition and multiplication of complex numbers is commutative.
Further, the

product of z = a + ib with its **complex conjugate**

is the nonnegative number

and its (nonnegative) squareroot is called the **absolute value** or **modulus** of z
and denoted by

For z ≠ 0, we have l z l ≠ 0, hence
is a well-defined complex number. It

is the **reciprocal** of z since
,
of use for division by z. Note that, for any two complex

numbers z and
,

It is very useful to visualize complex numbers as points in the so called **
complex plane**, i.e.,

to identify the complex number a + ib with the point (a, b) in IR^{2}. With this identification, its

absolute value corresponds to the (Euclidean) distance of the corresponding
point from the origin.

The sum of two complex numbers corresponds to the vector
sum of their corresponding points. The

product of two complex numbers is most easily visualized in terms of the** polar
form
**

with r ≥ 0, hence r = l z l its modulus, and ∈ IR is called its

**argument**. Indeed, for any real ,

has absolute value 1, and is the angle (in radians) that the vector

(a, b) makes with the positive real axis. Note that, for z ≠ 0, the argument, , is only defined up to

a multiple of 2π , while, for z = 0, the argument is arbitrary. If now also ,

then, by the law of exponents,

Thus, as already noted, the absolute value of the product is the product of the absolute values of

the factors, while the argument of a product is the sum of the arguments of the factors.

For example, in as much as the argument of is the negative of the argument of z, the argument

of the product is necessarily 0. As another example, if z = a + ib is of modulus 1, then z lies

on the unit circle in the complex plane, and so does any power z

^{k}of z. In fact, then

for some real number , and therefore . Hence, the sequence , appears

as a sequence of points on the unit circle, equally spaced around that circle, never accumulating

anywhere unless = 0, i.e., unless z = 1.

(17.1) Lemma: Let z be a complex number of
modulus 1. Then the sequence of powers of z lies on the unit circle, but fails to converge except when z = 1. |

**Convergence of a scalar sequence**

A subset Z of C is said to be **bounded** if it lies in some ball

of ( finite) radius r. Equivalently, Z is bounded if, for some r,
< r for all
∈ Z. In either case,

the number r is called a** bound for Z**.

In particular, we say that the scalar sequence is
**bounded** if the set

is bounded. For example, the sequence (1, 2, 3, ...) is not bounded.

(17.2) Lemma: The sequence
is
bounded if and only if . Here,
denotes the kth power of the scalar z. |

**Poof: **Assume that
> 1. I claim that, for all m,

(17.3)

This is certainly true for m = 1. Assume it correct for m = k. Then

The first term on the right-hand side gives

since > 1, while, for
the second term, by induction
hypothesis. Consequently,

showing that (17.3) also holds for m = k + 1.

In particular, for any given c, choosing m to be any natural number bigger than
,

we have . We conclude that the sequence
is unbounded when
> 1.

Assume that . Then, for any m,
, hence the sequence

is not only bounded, it lies entirely in the
**unit disk**

A sequence of (real or complex) scalars is said to
**converge to the
scalar** , in

symbols:

if, for all ε > 0, there is some so
that, for all .

Assuming without loss the scalars to be complex, we can profitably visualize this
definition as

saying the following: Whatever small circle of radius ε we
draw around the

point , all the terms of the sequence except the
first few are inside that circle.

(17.4) Lemma: A convergent sequence is bounded. |

**Proof: **If , then
there is some so that, for all
.

Therefore, for all m,

Note that r is indeed a well-defined nonnegative number, since a finite set of real
numbers always

has a largest element.

(17.5) Lemma: The sequence
is
convergent if and only if either
< 1 orelse = 1. In the former case, , while in the latter case . |

**Proof:** Since the sequence is not even bounded when
> 1, it cannot be convergent in

that case. We already noted that it cannot be convergent when
= 1 unless
=
1, and in that

case for all m, hence also
.

This leaves the case < 1. Then either
= 0, in which case
= 0 for all
m, hence also

. Else, 0 <
< 1, therefore
is a well-defined complex number of
modulus

greater than one, hence, as we showed earlier,
grows
monotonely to infinity as

m→∞. But this says that decreases monotonely to 0. In other words,
.

**Horner, or: How to divide a polynomial by a linear factor**

Recall that, given the polynomial p and one of its roots, μ, the polynomial
can

be constructed by **synthetic division**. This process is also known as **nested
multiplication** or

**Horner's scheme** as it is used, more generally, to evaluate a polynomial efficiently. Here are the

details, for a polynomial of degree ≤ 3.

If , and z is any scalar, then

In other words, we write such a polynomial in **nested form
**and then evaluate from the inside out.

Each step is of the form

"horner (17.6)

it involves one multiplication and one addition. The last number calculated is
, it is the value

of p at z. There are 3 such steps for our cubic polynomial (the definition
requires no

calculation!). So, for a polynomial of degree n, we would use n multiplications
and n additions.

Now, not only is of interest, since it equals p(z), the other
are also
useful since

We verify this by multiplying out and rearranging terms according to powers of
t. This gives

The last equality holds since, by (17.6),

for j < 3 while by definition.

(17.7) Nested Multiplication (aka Horner): To
evaluate the polynomial at the point z, compute the sequence by the prescription
Then , with q(t) = p(t)/(t − z). |

Since p(t) = (t − z)q(t), it follows that degq < deg p.
This provides another proof (see (3.21))

for the easy part of the Fundamental Theorem of Algebra, namely that a
polynomial of degree k

has at most k roots.