11. Appendices - Introduction to ODEs and Linear Algebra

Complex Numbers

Although often supposedly covered in previous courses such as "pre-calculus", most students I have taught at this level need at least a review of complex numbers. There is also some material (for example the matrix representation) that you may not have seen before.

Complex numbers (often denoted by the "blackboard bold" symbol $\mathbb{C}$ ) are an extension of the real numbers which have all of the same arithmetic properties. We can write any complex number $z \in \mathbb{C}$ as the sum of a real number and another real number multiplied by $i$ , a square root of $-1$ (so $i^2 = -1$ ):

$z = a + i b = Re(z) + i Im(z)$

Note that in electrical engineering literature, current is denoted by $i$ (because the French physicist André-Marie Ampère used it to abbreviate "intensité de courant" in 1820, before the common use of $i$ in mathematics), so $j$ is used instead.

Euler introduced the symbol $i$ to avoid some mistakes which can be made if we use $\sqrt{-1}$ instead. Recall that the square root symbol (or radical) means the principal, or non-negative root. This doesn't make consistent sense once we extend beyond the real numbers, so there are apparent contradictions such as:

$1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{-1}\sqrt{-1} = -1 (!)$

If $z = a + i b$ , the real number $b$ which is the coefficient of $i$ in this representation is called the imaginary part of the complex number $z$ , and $a$ is called the real part of $z$ . A very common source of confusion is that the imaginary part of a complex number is real (!).

When we multiply two complex numbers $z_1 = a_1 + i b_1$ and $z_2 = a_2 + i b_2$ , we can use the fact that $i^2 = -1$ to get

$z_1 z_2 = (a_1 + i b_1)(a_2 + i b_2) = (a_1 a_2 - b_1 b_2) + i (a_1 b_2 + a_2 b_1)$

So the real part of the product is $Re(z_1 z_2) = a_1 a_2 - b_1 b_2$ , and the imaginary part of the product is $Im(z_1 z_2) = a_1 b_2 + a_2 b_1$ .

The complex conjugate $\bar{z}$ of a complex number $z = a + i b$ is defined as $\bar{z} = a - i b$ .

Although it took people a long time to come up with the idea, it is very helpful to consider complex numbers as points on a plane. Usually the real part is drawn horizontally, and the imaginary part vertically. Then the distance from the origin to the complex number $z = a + i b$ is the natural generalization of the absolute value, and is often denoted in the same way:

$|z| = \sqrt{a^2 + b^2} = \sqrt{z \bar{z}}$

This quantity $|z|$ is also called the modulus of $z$ , or the norm of $z$ .

Thinking of complex numbers as points in the plane also suggests using polar coordinates $(r,\theta)$ for them, which turns out to be incredibly useful because of an amazing identity first noticed by Euler. Although Euler came up with hundreds (maybe thousands) of new formulas, this one is enshrined as Euler's Formula because of its extreme importance and elegance:

$e^{i \theta} = \cos(\theta) + i \sin(\theta)$

One way to prove this identity is to use the power series expansions of the functions. The exponential function has the series expansion

$e^x = 1 + x + x^2/2 + x^3/6 + x^4/24 + \ldots = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

which converges for all real $x$ . Substituting the purely imaginary number $i \theta$ in for $x$ we get

$e^{i\theta} = 1 + i \theta - \theta^2/2 - i \theta^3/6 + \theta^4/24 + \ldots$

since $i^2 = -1$ , $i^3 = -i$ , and $i^4 = 1$ . Now if we separate the terms into real and imaginary parts we get

$e^{i\theta} = (1 - \theta^2/2 + \theta^4/24 + \ldots) + i (\theta - \theta^3/6 + \ldots)$

and the real part is the series expansion of $\cos(\theta)$ , while the imaginary part is the series expansion of $\sin(\theta)$ .

Euler's formula lets us express trigonometric functions in terms of exponentials, which are often much easier to deal with. One of the many consequences of this is an easy proof of the trigonometric identity known as De Moivre's formula, which is otherwise not so easy to prove. Since $(e^{i \theta})^n = e^{i n \theta}$ , Euler's formula tells us that

$(\cos(\theta) + i \sin(\theta))^n = \cos(n \theta) + i \sin(n \theta)$

The connections between exponentials, trigonometric functions, and complex numbers are crucial to understanding linear differential equations. Oscillatory behavior of solutions to linear ODEs can be thought of as rotation in the "phase space" of position and velocity, and rotations are naturally associated with complex numbers through Euler's formula.

From rotation matrices to complex numbers

Another way to represent complex numbers is with two by two matrices:

$z = a + ib = r e^{i \theta} \ \ \longleftrightarrow \ \ \left ( \begin{array}{cc} a & -b \\ b & a \end{array} \right ) = r \left ( \begin{array}{cc} \cos(\theta) & -\sin{(\theta)} \\ \sin{(\theta)} & \cos(\theta) \end{array} \right )$

and we can see that exponential of purely complex numbers $e^{i \theta}$ correspond to planar rotation matrices.

Example: complex root calculation

Problem: Compute all of the sixth roots of $4i$ .

For computing roots or powers of complex numbers it is most convenient to use the polar form. In this case, to find the sixth roots of $i$ we first put $i$ into its polar form. For a complex number $z = a + ib$ , the polar radius is the norm $r = |z| = \sqrt{a^2 + b^2}$ , so in this case $r = \sqrt{0^2 + 4^2} = 4$ . The angle of $i$ from the real axis is $\pi/2$ , so

$4i = 4 e^{i \pi/2}$

In taking roots of complex numbers it is important to consider the ambiguity of the angle: any additional multiple of $2 \pi$ gives the same number. In this case,

$4i = 4 e^{i \pi/2} = 4 e^{5 i \pi/2} = 4 e^{9 i \pi/2} = \ldots = 4 e^{21 i \pi/2}$

The sixth roots $w_i$ will all have the same radius $4^{1/6} = 2^{1/3}$ , but with angles that differ by $2 \pi/ 6 = \pi/3$ . From the representation $4i = 4 e^{i \pi/2}$ , we get the sixth root

$w_1 = 2^{1/3} e^{i \pi/12}$

and then we can add multiples of $\pi/3$ to obtain the other five:

$w_2 = 2^{1/3} e^{5 i \pi/12}, \ \ \ w_3 = 2^{1/3} e^{i 9 \pi/12}, \ \ w_4 = 2^{1/3} e^{i 13 \pi/12} = 2^{1/3} e^{- i \pi/12}$

For the $w_4$ expression above we have switched to using negative angles; we can also subtract multiples of $\pi/3$ to get the remaining roots

$w_5 = 2^{1/3} e^{- i 5 \pi/12}, \ \ \ w_6 = 2^{1/3} e^{- i 9 \pi/12}$

These roots $w_i$ are the blue points in the illustration below.

Exercises:

Compute the real and imaginary parts of $z = \frac{-2 + i}{3 - 2 i}$ .
Find the polar form $(r (\cos(\theta) + i \sin(\theta)))$ of $z = \left [2(\cos(\frac{\pi}{3})- i \sin(\frac{\pi}{3}))\right ]^7$ . The angle $\theta$ should be in the interval $[0, 2 \pi)$ .
Find the cube roots of $z=-27$ .
Use Euler's formula ( $e^{i \theta} = \cos(\theta) + i \sin(\theta)$ ) to write both sides of the following equation in terms of $\cos(\theta_1)$ and $\sin(\theta_2)$

$e^{i \theta_1} e^{i \theta_2} = e^{i (\theta_1 + \theta_2)}$

and then write the resulting identity between the imaginary parts of both sides.

License

This work (text, mathematical images, and javascript applets) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. Biographical images are from Wikipedia and have their own (similar) licenses.