The Complex Numbers

Adding a number A to itself n times gives
us the process of multiplication with product nA. Now let's multiply a number A
by itself n times and let's call the result the n-th power of A (represented as
A^{n}). In this case I don't want to represent the set in the metaphor
of gathering together countable objects, such as pebbles, but rather in the
metaphor of the branches of a bush. We begin at a node from which A branches
emanate. From the tip of each branch another A branches emanate and so on. The
n-th generation of branches numbers A^{n}.

In that process of generating powers of A
we have a new application of numbers. The number n becomes the exponent of A.
But we can also write B = A^{n}, so we can also call n the logarithm of
B to the base A.

Because the exponents are themselves
numbers, we can apply arithmetic procedures to them. Thus, if we have B = A^{n}
and C = A^{m}, then we have as true to mathematics

A^{n+m} = A^{n}A^{m}
= BC

(Eq'n 1)

A^{n-m} = A^{n}/A^{m}
= B/C

(Eq'n 2)

and

A^{nm} = (A^{n})^{m}
= B^{m}

(Eq'n 3)

We reverse that process by specifying a
number of branches B and a number of generations n and seeking the number A
(which we call the n-th root of B) such that B = A^{n} or A =.
With this process we determine the number of branches in a generation. We can
use Equation 3 to show that we have another way of describing the root of a
number. We have A = B^{x} and B = A^{n}, so we have B = B^{xn},
which means that x = 1/n, which means in turn that the root of a number is also
the fractional power of that number.

All of the numbers that we have discovered so far work with powers and roots with the sole exception of the even-numbered roots of negative numbers: we seem to have no possibility of finding such roots among the numbers that we have in hand. Again, we must invent a new kind of number, one based on i =. The numbers multiplied by that new number comprise the imaginary numbers and those combine with the real numbers to create complex numbers. In general we describe a complex number as Z = a+ib, in which a represents the real part and b represents the imaginary part of the number. Just as we have described the real numbers as lying on an infinitely long line, we describe the complex numbers as lying in an infinitely wide plane defined by two mutually perpendicular lines, one carrying the real numbers and the other carrying the purely imaginary numbers. Our complex number Z, then, represents a point in that plane with the coordinates (a,b).

Now we can look at how we can carry out our arithmetic processes with complex numbers. We have:

1. Addition,

(a+ib) + (c+id) = (a+c) + i(b+d);

(Eq'n 4)

2. Subtraction,

(a+ib) - (c+id) = (a-c) +i(b-d);

(Eq'n 5)

3. Multiplication,

(a+ib)(c+id) = (ac-bd) +i(ad+bc);

(Eq'n 6)

4. Division,

(a+ib)/(c+id) = [(ac+bd) +
i(bc-ad)]/(c^{2}+d^{2}),

(Eq'n 7)

in which we have carried out the calculation by multiplying and dividing the original quotient by (c-id), the complex conjugate of the divisor, converting the division into a division by a real number to reflect the fact that we truly do not know how to divide by a complex number;

5. Powers,

(a+ib)^{c+id} = (a^{2}+b^{2})^{(c+id)/2}
exp[i(c+id)arg(a+ib)],

(Eq'n 8)

in which

arg(a+ib) = arctan(b/a);

(Eq'n 9)

and

6. Roots,

(Eq'n 10)

The first four of those processes come easily enough that we don't need to prove them. But how do I know that the powers and roots of complex numbers go as I wrote them? How can I put that statement to the test?

Let me begin by referring to the
representation of a complex number as a point in a plane and then convert the
pseudo-Cartesian coordinates (a,b) into their polar counterparts (r, ). In this
case a represents a distance along the x-axis and b represents a distance along
the y-axis, while r represents a distance from the origin to the point (a,b) and
represents the angle that the line r makes with the x-axis. We then have Euler's
relation

(Eq'n 11)

in which we also have

(Eq'n 12)

and

(Eq'n 13)

which is equivalent to Equation 9.

In light of that representation of
complex numbers, let's now look at the powers of i:

Putting those values into Equation 11
gives us readily

(Eq'n 14)

Because we know that i^{n-n} = i^{0}
= 1, we also know that

(Eq'n 15)

In accordance with Equation 12 we must
have as true to mathematics the statement that the square of a complex number
equals that number multiplied by its complex conjugate (the version of the
number in which we have reversed the algebraic sign of the imaginary part). With
that fact in mind we now want to look at the square of -1 or, more to the point,
the logarithm of that square. We know that we must have ln((-1)^{2}) = 0
because ln(1) = 0. Thus we know that ln(-1)+ln(-1)* = 0, which tells us that ln(-1)
is a purely imaginary number, because if ln(-1) had a real part, it would
reinforce in the addition rather than cancel, as the imaginary part does. We
thus have ln(-1) = ig (or -1 = e^{ig}). We now want to determine the
value of g.

As we saw in the matrix above, the powers
of i track the cardinal points of the unit circle in the complex plane. That
fact gives us the clue we need to begin to solve the problem. By Equation 14 we
have

(Eq'n 16)

Next we turn our attention to the relationship between the exponential function and the basic trigonometric functions.

We have an algebraic description of the
exponential function based on the definition

(Eq'n 17)

The binomial theorem tells us that

(Eq'n 18)

But we know that

(Eq'n 19)

so we have as true to mathematics

(Eq'n 20)

If we then substitute x=iθ,
we can separate the real and imaginary parts of the series and thereby obtain
the algebraic equivalents of the fundamental trigonometric functions; that is,
we have

(Eqn 21)

and

(Eq'n 22)

So now we can rewrite Equation 11 in the
form

(Eq'n 23)

Now we go back to -1=e^{ig} and
Equation 16 and determine that g=π.
So now we know that we can express powers, roots, and logarithms of complex
numbers with other complex numbers.

With the relationship in Equation 23 we
can now put Equations 8 and 10 to the proof. We start with

(a+ib)^{c+id} = (a^{2}+b^{2})^{(c+id)/2}
exp[i(c+id)arg(a+ib)],

(Eq'n 8)

In exponential form we have

(Eq'n 24)

But we know that
θ=arcTan(b/a).
And we know that the exponential factor has a magnitude of one, so r must
represent the magnitude of the complex number a+ib. But by Equation 23 we have

(Eq'n 25)

Thus we can rewrite Equation 24 as Equation 8.

As for Equation 10, we have

(Eq'n 10)

To prove and verify that equation we need only note that the exponent representing a root comprises a fraction, in this case one whose denominator is a complex number. As we did in the case of dividing by a complex number, we multiply and divide the exponent by the complex conjugate of the denominator and obtain Equation 10 directly.

Quod Erat Demonstrandum

Will we find any more new kinds of numbers? So far we have devised six arithmetic processes - addition, multiplication, exponentiation, and their inverses, subtraction, division, and the extraction of roots. Those inverses have obliged us to invent/discover three new kinds of numbers (negative, broken (fractions), and imaginary). If we want to discover a new kind of number, we must develop a new arithmetic process, in particular an analytic process. To get to that point we must first define the synthetic version of that analytic process.

We started with addition, then defined multiple addition, and then multiple multiple addition. By analogy we have as the next synthetic process the power of a power of a number; that is, we want to look at the m-th power of the n-th power of some number A. But that merely gives us the mn-th power of A. Our new process doesn't differ from the exponentiation of A in a way that gives us enough intellectual traction to require a new process. We seem to have reached the end of the line.

We see that fact reflected in the
Fundamental Theorem of Algebra, in which the definitive polynomial employs all
three of the synthetic processes. We know now that we have enough numbers of the
right kind to ensure that we will always find a number Z that solves the
equation

(Eq'n 26)

for any and all possible values of n and
the complex coefficients a_{i}.

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