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

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 $$k \in \mathbb{Z}$$.