DLA The field of complex numbers

Section A.2 The field of complex numbers

In this section.

Now that we have a new number, let's let it out to play in the field with all the other numbers.

Subsection A.2.1 Definition of the field

Example A.2.1.

Suppose we have polynomial \(x^2 - 6x + 13\text{.}\) Is the “number” \(3+2\ci\) a root? To check, substitute \(x = 3+2\ci \text{.}\)

\begin{align*} \amp \phantom{=} (3+2\ci)^2 - 6(3+2\ci) + 13 \\ \amp = (9 + 12\ci + 4(-1)) - 18 - 12\ci + 13 \amp \amp (\ci^2 = -1)\\ \amp = 0. \end{align*}

The answer is yes! By algebraically combining \(\ci\text{,}\) which is a root of the prototypical “unsolvable” polynomial equation \(x^2+1=0\text{,}\) with other real numbers, we obtain roots of other previously “unsolvable” polynomial equations.

It turns out that the result of combining such expressions involving \(\ci\) using the four operations of addition, subtraction, multiplication, and division can always be simplified using \(\ci^2 = -1 \) down to something of the form \(a + b\ci\text{,}\) where \(a,b\) are real numbers.

Definition A.2.2. The field of complex numbers.

The collection \(\C\) of all possible algebraic expressions \(a+b\ci\text{,}\) where \(a,b\) are real numbers. Expressions of this form are called complex numbers. In such an expression, the number \(a\) is called the real part and the number \(b\) is called the imaginary part.

\(\Re(z)\): The real part of the complex number \(z\text{.}\)
\(\Im(z)\): The imaginary part of the complex number \(z\text{.}\)

Example A.2.3. Real and imaginary parts.

The complex number \(-3+\frac{1}{2}\ci\) has real part \(\Re\left(-3+\frac{1}{2}\ci\right)=-3\) and imaginary part \(\Im\left(-3+\frac{1}{2}\ci\right)=\frac{1}{2}\text{.}\)
The complex number \(e - \pi\ci = e + (-\pi)\ci\) has real part \(\Re(e - \pi\ci) = e\) and imaginary part \(\Im(e - \pi\ci) = -\pi\text{.}\)

Remark A.2.4. Real numbers are complex numbers.

Every real number \(a\) can simultaneously be considered to be a complex number, as

\begin{equation*} a = a + 0\ci. \end{equation*}

So the real numbers are precisely those complex numbers that have imaginary part equal to zero. (Complex numbers whose real part is equal to zero are called purely imaginary. For example, \(3\ci\) is purely imaginary.)

Roughly speaking, a field is a mathematical system of “numbers” that carries some versions of the four operations addition, subtraction, multiplication, and division, and in which these four operations behave arithmetically and algebraically the way we would expect them to behave. We explore these four operations for the complex numbers, along with an important new fifth operation, in the following subsections.

Subsection A.2.2 Basic operations with complex numbers

Here are examples of adding, subtracting, multiplying, and exponentiating complex numbers. (We will deal with division afterward.) Notice it mostly involves just familiar algebra: collecting like terms and FOIL.

Example A.2.5. Algebraic operations involving complex numbers.

Addition: \((3+\ci) + (5-3\ci) = 8 - 2\ci \text{.}\)
Subtraction: \((3+\ci) - (5-3\ci) = -2 + 4\ci \text{.}\)
Multiplication: \((3+\ci)(5-3\ci) = 15 - 4\ci - 3(-1) = 18 - 4\ci \text{.}\)
Power: \((3+\ci)^2 = 9 + 6\ci + (-1) = 8 + 6\ci \text{.}\)
A special multiplication example:
\begin{equation*} (5-3\ci)(5+3\ci) = 25 + 0\ci - 9(-1) = 34. \end{equation*}

The last in the list of examples above is particularly important. You may recognize it as related to the factorization of a difference of squares. Since \(\ci^2 = -1\text{,}\) multiplying a complex number against its “difference-of-squares-factorization partner” like this always results in a real number. Because of this, we'll give this partner a name.

Definition A.2.6. Complex conjugate.

For complex number \(z = a + b\ci\text{,}\) we define the complex conjugate of \(z\) to be

\begin{equation*} \cconj{z} = \lcconj{a+b\ci} = a - b\ci. \end{equation*}

Remark A.2.7.

You should think of complex conjugate as “flipping the sign” on the imaginary part of the complex number. In particular, if \(z = a - b\ci\) then \(\cconj{z} = \lcconj{a - b\ci} = a + b\ci\text{.}\)

Example A.2.8. Computing complex conjugates.

\(\lcconj{5+4\ci} = 5 - 4\ci \text{.}\)
\(\lcconj{e-\pi\ci} = e + \pi\ci \text{.}\)

Subsection A.2.3 Division of complex nubers

We can use complex conjugates to help us divide complex numbers.

Example A.2.9. Division of complex numbers.

To simplify a ratio of complex numbers, we can use the conjugate to rationalize the denominator, as if \(\ci\) really were equal to \(\sqrt{-1}\text{.}\)

\begin{align*} \frac{3+4\ci}{2-3\ci} \amp = \frac{3+4\ci}{2-3\ci} \cdot \frac{2+3\ci}{2+3\ci}\\ \amp = \frac{(3+4\ci)(2+3\ci)}{(2-3\ci)(2+3\ci)}\\ \amp = \frac{6 + 17 \ci - 12}{2^2 + 3^2}\\ \amp = \frac{-6 + 17 \ci}{13}\\ \amp = -\frac{6}{13} + \frac{17}{13}\, \ci. \end{align*}

Pattern.

In general, division of complex numbers follows the pattern

\begin{equation*} \frac{a+b\ci}{c+d\ci} = \frac{a+b\ci}{c+d\ci} \cdot \frac{c-d\ci}{c-d\ci} = \left(\frac{ac+bd}{c^2+d^2}\right) + \left(\frac{bc-ad}{c^2+d^2}\right) \; \ci. \end{equation*}

Since we can divide, we can also reciprocate.

Example A.2.10. Reciprocal of a complex number.

Let's compute the reciprocal of \(3+2\ci\text{.}\)

\begin{equation*} \inv{(3+2\ci)} = \frac{1}{3+2\ci} = \frac{1+0\ci}{3+2\ci} \cdot \frac{3-2\ci}{3-2\ci} = \frac{3}{13} - \frac{2}{13}\, \ci. \end{equation*}

If we multiply a number and its reciprocal, we would expect to get \(1\) as the result.

\begin{align*} (3+2\ci)\inv{(3+2\ci)} \amp = (3+2\ci)\left(\frac{3}{13} - \frac{2}{13}\, \ci\right)\\ \amp = \frac{9}{13} - \frac{6}{13}\, \ci + \frac{6}{13}\, \ci - \frac{4}{13} \cdot (-1)\\ \amp = 1 + 0 \ci\\ \amp = 1. \end{align*}

It worked!

There is a pattern to this reciprocal example, but it will have to wait until after Proposition A.2.14.

Subsection A.2.4 Properties of complex numbers

Since the operations of addition, subtraction, multiplication, and division of complex numbers are built on following our usual algebra rules on expressions involving the “variable” \(\ci\) (along with the extra rule \(\ci^2 = -1\)), there is no need to restate all those usual algebra rules. So we will focus on the new operation of complex conjugation.

Proposition A.2.11. Properties of complex conjugates.

The following are valid rules of algebra involving the complex conjugation operation. Let \(w\) and \(z\) represent arbitrary complex numbers, and let \(k\) represent an arbitrary real number.

Conjugating a real number has no effect. That is, \(\cconj{k} = k\text{.}\)
The conjuate of a sum is the sum of the conjugates. That is, \(\lcconj{w+z} = \cconj{w} + \cconj{z}. \)
The conjugate of a product is the product of the conjugates. That is, \(\lcconj{wz} = \cconj{w}\cconj{z}. \)
The conjugate of a “scalar multiple” is the “scalar multiple” of the conjugate. That is, \(\lcconj{kz} = k\cconj{z}. \)
The conjugate of a quotient is the quotient of the conjugates. That is, \(\lcconj{w/z} = \cconj{w} / \cconj{z}\text{.}\)
The conjugate of a reciprocal is the reciprocal of the conjugate. That is, \(\lcconj{1/z} = 1 / \cconj{z}\text{.}\)
For every nonzero, integer exponent, the conjugate of that power is the power of the conjugate. That is, \(\lcconj{z^n} = \cconj{z}^n\text{.}\)

Considering how a complex number interacts with its own conjugate yields several important patterns.

Proposition A.2.12. Relative properties of complex conjugates.

A complex number \(z\) satisfies the following special relationships with its own conjugate.

The conjugate of a conjugate is the original complex number. That is, \(\lcconj{\cconj{z}} = z\text{.}\)
A complex number is actually real (i.e. has imaginary part \(0\)) if and only if it is equal to its own conjugate. That is, \(z\) is real if and only if \(\cconj{z} = z\text{.}\)
A complex number is purely imaginary (i.e. has real part \(0\)) if and only if it is equal to the negative of its own conjugate. That is, \(z\) is purely imaginary if and only if \(\cconj{z} = -z\text{.}\)
Combining a complex number with its conjugate using addition or subtraction returns twice the real part or twice the imaginary part times \(\ci\text{,}\) respectively. That is,
\begin{align*} z + \cconj{z} \amp = 2\cdot\Re(z), \amp z - \cconj{z} = 2\cdot\Im(z)\cdot\ci. \end{align*}
Multiplying a complex number and its complex conjugate always results in a real number, equal to the sum of the squares of the real and imaginary parts. That is, if \(z = a + b\ci\text{,}\) then \(z\cconj{z} = a^2 + b^2\text{.}\)

The sum-of-squares pattern in Statement 5 of Proposition A.2.12 has a connection to geometry via Pythagoras that we will explore in Section A.3. (And by the time you are reading this appendix, you may already have encountered the sum-of-squares formula's connection to geometry in a linear algebra context in Chapter 13.) For now, we will give a version of this formula a name.

Definition A.2.13. Complex modulus.

For complex number \(z = a + b\ci\text{,}\) the real number \(\cmodulus{z} = \sqrt{a^2 + b^2}\) is called the modulus of \(z\text{.}\)

Proposition A.2.14. Properties of the complex modulus.

The following are valid rules of algebra involving the complex modulus. Let \(w\) and \(z\) represent arbitrary complex numbers, and let \(k\) represent an arbitrary real number.

The square of the modulus is equal to the product of the complex number with its own conjugate. That is, \(\cmodulus{z}^2 = z\cconj{z}\text{.}\)
A modulus is never negative, and only zero has modulus zero. That is, \(\cmodulus{z} \ge 0\text{,}\) with \(\cmodulus{z} = 0\) if and only if \(z=0\text{.}\)
A complex number and its conjugate have the same modulus. That is, \(\cmodulus{\cconj{z}} = \cmodulus{z}\text{.}\)
The modulus of a real number is equal to the absolute value of that number.
The modulus of a product is the product of the moduli. That is, \(\cmodulus{wz} = \cmodulus{w}\cmodulus{z}\text{.}\)
The modulus of a “scalar multiple” is the absolute “scalar multiple” of the modulus. That is, \(\cmodulus{kz} = \abs{k}\cmodulus{z}\text{,}\) where \(\abs{k}\) is the regular absolute value of the real number \(k\text{.}\)
The modulus of a quotient is the quotient of the moduli. That is, \(\cmodulus{w/z} = \cmodulus{w}/\cmodulus{z}\text{.}\)
The modulus of a reciprocal is the reciprocal of the modulus. That is, \(\cmodulus{1/z} = 1/\cmodulus{z}\text{.}\)
For every nonzero, integer exponent, the modulus of that power is the power of the modulus. That is, \(\cmodulus{z^n} = \cmodulus{z}^n\text{.}\)
The modulus satisfies the Triangle Inequality: \(\cmodulus{w + z} \le \cmodulus{w} + \cmodulus{z}\text{.}\)

Pattern.

We can use Statement 1 of Proposition A.2.14 to help us reciprocate:

\begin{align*} \cmodulus{z}^2 \amp = z\cconj{z} \amp \amp \implies \amp \inv{z} \amp = \frac{\cconj{z}}{\cmodulus{z}^2}. \end{align*}

Example A.2.15. Reciprocal of a complex number, revisited.

Let's recompute the reciprocal of \(3+2\ci\) (from Example A.2.10).

We have

\begin{equation*} \cmodulus{3+2\ci}^2 = 3^2 + 2^2 = 13, \end{equation*}

and so

\begin{equation*} \inv{(3+2\ci)} = \frac{\lcconj{3+2\ci}}{\cmodulus{3+2\ci}^2} = \frac{3-2\ci}{13} = \frac{3}{13} - \frac{2}{13}\, \ci, \end{equation*}

which is the same reciprocal as previously calculated in Example A.2.10.

Subsection A.2.5 Complex polynomials

The most important property of the field of complex numbers is that by merely inventing a solution to one previously unsolvable polynomial equation (i.e. the imaginary solution \(\ci\) to \(x^2 + 1 = 0\)), we have actually created solutions to all polynomial equations.

Now that we know how to perform arithmetic with complex numbers, we can expand to doing some algebra. A complex polynomial is just a polynomial where we allow the coefficients to be complex numbers, and assume the variable to represent an unknown complex number.

Example A.2.16. Checking a solution to a complex polynomial.

Is \(z = 4-\ci\) a solution to the complex polynomial equation \(z^2 + \ci z + (-16+4\ci)=0\text{?}\)

\begin{align*} \amp \phantom{=} (4-\ci)^2 + \ci (4-\ci) + (-16+4\ci) \\ \amp = \bigl[16 - 8\ci + (-1)\bigr] + \bigl[4\ci - (-1)\bigr] + (-16+4\ci) \\ \amp = 0. \end{align*}

Yes, \(z = 4-\ci\) is a solution to \(z^2 + \ci z + (-16+4\ci)\text{.}\)

Theorem A.2.17. The Fundamental Theorem of Algebra (Complex Version).

If \(p(z)\) is a complex polynomial of degree \(n \ge 1\text{,}\) then the polynomial equation \(p(z) = 0\) has exactly \(n\) roots (where repeated roots are counted multiple times, according to the number of times that root is repeated).

Because determining roots of a polynomial is equivalent to factoring that polynomial, we can reinterpret this theorem in terms of factoring.

Corollary A.2.18.

Every complex polynomial \(p(z)\) can be completely factored as

\begin{equation*} p(z) = w_0 (z-w_1)(z-w_2) \dotsm (z-w_n), \end{equation*}

for some constant complex numbers \(w_0,w_1,w_2,\dotsc,w_n\text{,}\) where \(n\) is the degree of \(p(z)\text{.}\)

We can also use this complex version of the Fundamental Theorem to refine the real version (Theorem A.1.3), since a real polynomial can be considered to be a complex polynomial that just happens to have purely real numbers as coefficients.

Corollary A.2.19.

If \(p(x)\) is a real polynomial of degree \(n \ge 1\text{,}\) then the polynomial equation \(p(x) = 0\) has exactly \(n\) roots when we allow complex numbers as roots (and where again, repeated roots are counted multiple times).

Example A.2.20. Roots of a cubic polynomial.

Consider the cubic polynomial equation

\begin{gather} x^3 + 3x^2 + 2x - 6 = 0.\label{equation-complex-field-cubic-fully-factored}\tag{\(\star\)} \end{gather}

We can guess one root: \(x=1\) solves this equation. This means that \(x-1\) is a factor of this polynomial. Using long division of polynomials, we could calculate

\begin{equation*} \frac{x^3 + 3x^2 + 2x - 6}{x-1} = x^2+4x+6, \end{equation*}

so that

\begin{equation*} x^3 + 3x^2 + 2x - 6 = (x-1)(x^2+4x+6). \end{equation*}

So to find the other roots of (\(\star\)), we only need to determine the roots of \(x^2+4x+6 = 0\text{.}\) As a real polynomial, this quadratic factor is irreducible (i.e. has no roots). However, as a complex polynomial, we can use the quadratic formula:

\begin{align*} x \amp = \frac{-4 \pm \sqrt{16 - 4 \cdot 1 \cdot 6}}{2} \\ \amp = \frac{-4 \pm \sqrt{-8}}{2} \\ \amp = \frac{-4 \pm \sqrt{-1}\sqrt{2}\sqrt{4}}{2} \\ \amp = -2 \pm \ci\sqrt{2}. \end{align*}

Therefore, the degree-three polynomial equation in (\(\star\)) does indeed have precisely three roots:

\begin{equation*} x = -1, -2+\ci\sqrt{2}, -2-\ci\sqrt{2}. \end{equation*}

And the cubic polynomial in (\(\star\)) can be fully factored as

\begin{equation*} x^3 + 3x^2 + 2x - 6 = (x-1)\bbrac{x - (-2+\ci\sqrt{2})}\bbrac{x - (-2-\ci\sqrt{2})}. \end{equation*}