Complex numbers notes

Tomas Koutsky published on

4 min, 762 words

Categories: math random

Imaginary numbers

Quadratic equations as $x^2 = -1$ don't have any real number solution. The solution exists in the complex number system.

Backbone of which is the imaginary unit $i$. For which following is true:

  • $i = \sqrt{-1}$
  • $i^2 = -1$

By taking multiples of $i$, I can create infinitely many pure imaginary numbers, e.g. $3i$, $i\sqrt{5}$, $-12i$.

By squaring an imaginary number I can explore how those relate to the real numbers: $$ (3i)^2 = 3^2i^2 = 9i^2 $$

Now I can apply the fact that $i^2 = -1$:

$$ 9i^2 = 9(-1) = -9 $$

Meaning $3i$ is square root of $-9$

Simplifying pure imaginary numbers

As proved $\sqrt{-9}$ simplifies to $3i$ and following though process can be applied: The square root of $-9$ is an imaginary number. The square root of $9$ is $3$, so the square root of negative $9$ is $3$ imaginary units, or $3i$

More formally: $$ \text{For}~ a > 0, \sqrt{-a} = i\sqrt{a} $$

Powers of $i$

Basic power line can be devised: $$ \begin{aligned} i^0 &= 1 \\ i^1 &= i \\ i^2 &= -1 \\ i^3 &= i^2 \times i = (-1)(i) = -i \\ i^4 &= i^3 \times i = (-i)(i) = (-1)(i)(i) = (-1)(-1) = 1 \\ i^5 &= i^4 \times i = 1 \times i = i \\ i^6 &= i^5 \times i = -1 \\ i^7 &= i^6 \times i = -i \\ i^8 &= i^7 \times i = 1 \\ \end{aligned} $$

Results can be summarised in following table:

$i^1$$i^2$$i^3$$i^4$$i^5$$i^6$$i^7$$i^8$
$i$$-1$$-i$$1$$i$$-1$$-i$1

Hence it appearst that powers of $i$ cycle through the sequence of $i, -1, -i, 1$.

Properties of exponents can be used to devise larger powers of $i$, e.g.: $$ i^{20} = i^{4\times5} = (i^4)^5 = 1^5 = 1 $$ Hence: $$ i^{20} = 1 $$

The fact that $i$ raised to a multiple of $4$ is $1$ (along with properties of exponents) can be used to find large powers of $i$.

Roots of negative numbers

Square roots of negative numbers can be simplified followingly: $$ \sqrt{-20} = \sqrt{-1 \times 4 \times 5} = i\sqrt{4 \times 5} = 2i\sqrt{5} $$

$i$ as the principal root of $-1$

Consider following application of the $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$ rule on the $i$:

GIven $i = \sqrt{-1}$ I can state: $$ -1 = i \times i = \sqrt{-1} \times \sqrt{-1} = \sqrt{-1 \times -1} = \sqrt{1} = 1 $$ Which is obviosly not truthy. In reality the $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$ rule cannot be applied when both $a$ and $b$ are negative.

Complex numbers

Combination of real and imaginary numbers, e.g. $z = 5 + 3i$ which has following parts: Real: $$ Re(z) = 5 $$ Imaginary: $$ Im(z) = 3 $$

The number can be plotted with $Re$ on $x$ axis and $Im$ on $y$ axis: complex_plot.png

More formally a complex number is any number that can be written as $a + bi$.

Classifying complex numbers

An imaginary number is a complex nuimber $a + bi$ where $a = 0$. A real number is a complex number $a + bi$ where $b = 0$.

Hence following hiearchy:

-------------------------
|  Complex numbers      |
|                       |
| ----------------      |
| | Real numbers |      |
| ----------------      |
|                       |
| --------------------- |
| | Imaginary numbers | |
| --------------------- | 
-------------------------

Simplifying complex numbers

Let's consider: $$ 2 + 3i + 7i^2 + 5i^3 + 9i^4 $$

I already know that: $$ i^2 = -1 \\ i^3 = -i \\ i^4 = 1 $$

Hence: $$ 2 + 3i + 7(-1) + 5(-i)+ 9(1) = \\ 2 + 3i - 7 -5i + 9 = \\ 4 - 2i $$

Adding complex numbers

E.g.: $$ (5 + 2i) + (3 - 7i) = \\ 5 + 3 + 2i - 7i = \\ 8 - 5i $$

Subtracting complex numbers

E.g.: $$ (2 - 3i) - (6 - 18i) = \\ 2 - 3i - 6 + 18i = \\ -4 + 15i $$

Distance of complex numbers

E.g. distance between: $$ z = 2 + 3i \\ w = -5 -i $$

Can be calculated using Pythagoras theorem: complex_distance.png

Midpoint of complex numbers

Is the half of real and complex parts separately: $$ z = 2 + 3i \\ w = -5 -i \\ $$

$$ a = (\frac{2 + (-5)}{2}) + (\frac{3 + (- 1)}{2})i \\ a = -\frac{3}{2} + i $$

Multiplying complex numbers

A real number by a complex number

Use a distributive property: $$ -4(13 + 5i) = -4(13) + (-4)(5i) = -52 - 20i $$

A pure imaginary number by a complex number

$$ 2i(3 - 8i) = 2i(3) - 2i(8i) = 6i - 16i^2 $$

Use the fact that $i^2 = -1$: $$ 6i - 16i^2 = 6i - 16(-1) = 6i + 16 = 16 + 6i $$

Two complex numbers

Multiply each term in the first number by each term in the second number:

$$ (1 + 4i)(5 + i) = \\ (1)(5) + (1)(i) + (4i)(5) + (4i)(i) = \\ 5 + i + 20i + 4i^2 = \\ 5 + 21i + 4i^2 = \\ 5 + 21i + 4(-1) = \\ 5 + 21i - 4 = \\ 1 + 21i $$