Lecture 2 — Complex Numbers

Lecture 2 — Complex Numbers

\renewcommand\Re{\operatorname{Re}} \renewcommand\Im{\operatorname{Im}}

Complex numbers have been around for a while.

\mathbb C as a field

B.C section 1-3

\begin{aligned} \mathbb N &= \{1, 2, 3, \ldots \} \\ \mathbb N_0 &= \{0, 1, 2, 3, \ldots \} \\ \mathbb Z &= \{0, \pm1, \pm2, \pm3, \ldots \} \\ \mathbb Q &= \{p/q : p, q \in \mathbb Z, q \ne 0 \} \\ \mathbb R &= \text{real numbers} \\ \mathbb C &= \text{complex numbers} \end{aligned}

Note that \mathbb Q is actually equivalence classes of “quotients” of integers because certain expressions are equivalent (see MATH2401). \mathbb R can be defined in several technical ways, such as Dedekind cuts or limits of sequences.

\mathbb C can be represented in various (equivalent) ways:

i is the complex number represented by (0,1). We say \mathbb R \subset \mathbb C by identifying the complex number x + 0i with the real number x.

Addition in \mathbb C

\begin{aligned} (x_1, y_1) + (x_2, y_2) &= (x_1 + x_2, y_1 + y_2) \\ (x_1 + iy_1) + (x_2 + iy_2) &= (x_1 + x_2) + i(y_1 + y_2) \end{aligned}

Multiplication in \mathbb C

Denoted by \times or \cdot or juxtaposition (that is, putting things next to each other). \begin{aligned} (x_1, y_1)\cdot(x_2, y_2) &= (x_1x_2 - y_1y_2, y_1x_2 + x_1y_2) \\ (x_1 + iy_1) \cdot (x_2 + iy_2) &= (x_1 x_2 - y_1y_2) + i(y_1x_2 + x_1y_2) \end{aligned} The definition of multiplication formally applies if we use the usual rules for algebra in \mathbb R and set i^2 = -1.

Note: Multiplication of two complex numbers sums their angles (where positive is CCW) and multiples their radius.

\mathbb C is a field

With this addition and multiplication, \mathbb C is a field. Check: \mathbb C must be closed under the binary operations + and \cdot.

F2: + has identity 0+0i and inverse (-x) + i(-y). F5: \cdot has identity 1 + 0i and inverse z^{-1} = 1/(x+iy) \cdot (x-iy)/(x-iy) = \frac x{x^2+y^2} - i\frac{y}{x^2+y^2}

Since \mathbb C is a field, it holds: z_1, z_2 = 0 \implies z_1 = 0 \text{ or }z_2 = 0. This is the null-factor law and holds because on all fields. Also, we have (z_1z_2)^{-1} = z_1^{-1}z_2^{-1}.

Note: i^2 = -1 and (-i)^2 = -1. These are the only two solutions of z^2 = -1 in the complex numbers (we cannot check this yet). This is due to the Fundamental Theorem of Algebra.

Remark: \mathbb C is not ordered and, in fact, cannot be ordered. Thus, i is no more special then -i.

B.C. 4, 5

Given z = x + iy \in \mathbb C, there are a few useful functions to have: - modulus: |\cdot| : \mathbb C \to \mathbb [0, \infty), where |z| = \sqrt{x^2 + y^2}, - real part: \operatorname{Re}(z) = x, imaginary part: \operatorname{Im}(z) = y (both \mathbb C \to \mathbb R),