Introduction#

Mathematics Methods 2

Complex numbers#

In complex analysis we are dealing with the set of complex numbers \(\mathbb{C}\) which are ordered pairs of the form \((x, y)\), where \(x\) and \(y\) are real numbers. If we introduce the imaginary unit \(i = \sqrt{-1}\) we can write it in its rectangular form:

(109)#\[ z = x + iy \]

Every complex number \(z = x + iy\) has a real part Re\(z = x\) and an imaginary part Im\(z = y\). Sometimes \(\mathfrak{R}(z)\) and \(\mathfrak{I}(z)\) are used instead of Re and Im. Two complex numbers \(z_1\) and \(z_2\) are equal iff their real and imaginary parts are equal,

\[ z_1 = z_2 \iff \text{Re}(z_1) = \text{Re}(z_2) \wedge \text{Im}(z_1) = \text{Im}(z_2). \]

If Re\((z) = 0\) and Im\((z) \neq 0\) we say that \(z\) is pure imaginary. On the other hand, if Im\((z) = 0\) and Re\((z) \neq 0\) we recover the pure real number \(x\). Therefore, we identify real numbers as complex numbers whose imaginary part is equal to 0, so \(\mathbb{R}\) is a subset of \(\mathbb{C}\), i.e. \(\mathbb{R} \subset \mathbb{C}\).

Complex algebra#

Mathematics Methods 2

Addition and multiplication#

Adding complex numbers is performed by adding real and imaginary parts separately:

\[ z_1 \pm z_2 = (x_1 + i y_1) \pm (x_2 + iy_2) = (x_1 \pm x_2) + i(y_1 \pm y_2) \]

Multiplying two complex numbers is done by expanding the brackets and grouping real and imaginary parts:

\[\begin{split} \begin{align} z_1 \cdot z_2 & = (x_1 + i y_1) \cdot (x_2 + iy_2) = x_1 x_2 + i^2 y_1 y_2 + i x_1 y_2 + i x_2 y_1 \\ & = (x_1 x_2 - y_1 y_2) + i(x_1 y_2 + x_2 y_1). \end{align} \end{split}\]

Complex conjugate#

For every complex number \(z = x + iy\) we define its complex conjugate \(z^*\) or \(\overline{z}\) by changing the sign of its imaginary part:

\[ z^* \equiv x - iy \]

This is an involutory operation since \( (z^)^ = z \) with the following properties:

\[\begin{split} (z_1 \pm z_2)^* = z_1^* \pm z_2^* \\ (z_1 \cdot z_2)^* = z_1^* \cdot z_2^* \\ (z^{-1})^* = (z^*)^{-1} \end{split}\]

We can write the real and imaginary parts of \(z\) using the complex conjugate:

\[ \text{Re}(z) = \frac{1}{2} (z + z^*), \quad \text{Im}(z) = \frac{1}{2i} (z - z^*) \]

Modulus#

To every complex number \(z\) we assign a modulus (absolute value) \(| z |\), a non-negative real number defined as:

(110)#\[ | z | \equiv \sqrt{x^2 + y^2} \]

Zero is the only complex number whose modulus is 0.

Properties

The modulus of a complex number satisfies the following properties:

\[ |z_1| \cdot |z_2| = |z_1 z_2|, \quad | z^{-1} | = |z|^{-1}, \]
\[|z^*| = |z|, \quad z^* \cdot z = |z|^2 \]

Division#

We divide a complex number \(z_1\) by another complex number \(z_2 \neq 0\) by expanding the fraction such that the denominator becomes a real number. We do this by multiplying both the numerator and the denominator by \(z_2^*\) (see the last property of a modulus above).

\[\begin{split} \begin{align} \frac{z_1}{z_2} & = \frac{x_1 + iy_1}{x_2 + iy_2} = \frac{x_1 + iy_1}{x_2 + iy_2} \cdot \frac{x_2 - iy_2}{x_2 - iy_2} \\ & = \left ( \frac{x_1x_2 + y_1y_2}{x_2^2 + y_2^2} \right ) + i \left ( \frac{x_2y_1- x_1y_2}{x_2^2 + y_2^2} \right ) \end{align} \end{split}\]

Multiplicative inverse#

Let us find the reciprocal of a complex number \(z = x + iy\):

\[\begin{split} \begin{align} z^{-1} & = \frac{1}{z} = \frac{1}{x + iy} \cdot \frac{x - iy}{x -iy} = \frac{x - iy}{x^2 + y^2} \\ & = \frac{x}{x^2 + y^2} -i \frac{y}{x^2 + y^2} \end{align} \end{split}\]

Powers and roots#

Some operations are quite tedious and much more difficult to perform on complex numbers in their rectangular form \(z = x + iy\). Raising a complex number to some power and taking a root are one of such operations. We will therefore show the formula here for these operations for a complex number in its Trigonometric form (we introduce this in the next notebook).

\[\begin{split} z = r( cos \varphi + i \sin \varphi) \\ z^n = r^n ( cos n \varphi + i \sin n \varphi) \end{split}\]

Let \(w = z^{1/n} \), then

\[ w = r^{1/n} \left ( \cos \left ( \frac{ \varphi}{n} + \frac{2k \pi}{n} \right ) + i \sin \left ( \frac{ \varphi}{n} + \frac{2k \pi}{n} \right ) \right ), \]

where \(k \in \mathbb{Z}\).