## Imaginary numbers

• Invented to solve problems where an equation has no real roots e.g. $x^2 + 16 = 0$, the idea of declaring the existence of a quantity $i$ such that $i^2 = -1$ allows us to express the solution as
$$x = \sqrt{-16} = \sqrt{16i^2} = \pm4i$$

The set represented by $\mathbb{I}$ defines an imaginary number as

$$i^2 = -1$$

### Powers of i

If $i^2 = -1$ then $i^4 = i^2i^2 = -1 * -1 = 1$

Therefore we have the sequence

$$\begin{array}{ccccc} \hline i & i^2 & i^3 & i^4 & i^5 & \ldots \\ \hline i & -1 & -i & 1 & i & \ldots \\ \hline \end{array}$$

## Complex numbers

A complex number is just the sum of a real and an imaginary number

$$z = a + bi \quad a,b \in \mathbb{R}, \quad i^2 = -1$$

### Operations on complex numbers

Given two complex numbers

$$z_1 = a_1 + b_1i \\ z_2 = a_2 + b_2i$$

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

#### Product

\begin{align*} z_1z_2 &= a_1a_2 + a_1b_2i + a_2b_1i + b_1b_2i^2 \quad \text{given that i^2 = -1} \\ &= (a_1a_2 - b_1b_2) + (a_1b_2 + b_1a_2)i \end{align*}

Given the complex number

$$z = a + bi$$

#### Norm (modulus or absolute value)

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

#### Complex conjugate

The product of two complex numbers where the only difference between them is the sign of the imaginary part is

$$(a + bi)(a - bi) = a^2 - abi + abi - b^2i^2 = a^2 + b^2$$

This quantity $a - bi$ is called the complex conjugate of $z$ (denoted as $z^*$), it implies that

$$zz^* = |z|^2$$

#### Inverse

$$z^{-1} = \frac{1}{z}$$

Multiplying the numerator and denominator with the conjugate of $z$ (so that we have a real part on the denominator)

$$z^{-1} = \frac{1}{z} \frac{z^*}{z^*} = \frac{z^*}{zz^*} = \frac{z^*}{|z|^2}$$

#### Square root of $i$

We’re trying to find a complex number $z$ such that

$$\sqrt{i} = z \\ i = z^2$$

Assuming that $z$ is the complex number $z = a + bi$

\begin{align} i &= (a + bi)^2 \nonumber \\ &= (a + bi)(a + bi) \nonumber \\ &= a^2 - b^2 + 2abi \label{square-imaginary} \end{align}

Therefore

$$(a^2 - b^2) + (2ab)i = 0 + 1i$$

Equaling real and imaginary parts

\begin{align*} a^2 - b^2 &= 0 \\ 2ab = 1 \end{align*}

Therefore $a = \pm b$, replacing $a = -b$ in the second equation we obtain $-2b^2 = 1$ which is not satisfied by any real number $b$ therefore the case $a = -b$ is impossible, replacing $a = b$ in the second equation we obtain $2a^2 = 1$ so

$$2a^2 = 1 \\ a^2 = \frac{1}{2} \\ a = b = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}}$$

Finally the value of $\sqrt{i}$ is

$$\sqrt{i} = (a + bi) = \pm{\frac{1}{\sqrt{2}}} (1 + i)$$

The value of $\sqrt{-i}$ is found in the same way (by replacing $b = -a$ in the equation $-2ab = 1$ found from multiplying \eqref{square-imaginary} by $-1$)

$$\sqrt{-i} = (a + bi) = \pm{\frac{1}{\sqrt{2}}} (1 - i)$$

### Matrix representation of a complex number

The matrix $C$ for a complex number is the sum of two other matrices representing the real $R$ and imaginary $I$ parts:

$$C = R + I$$

which can be written as

$$C = a \hat{R} + b \hat{I} \quad\quad a, b \in \mathbb{R}$$

Where $R = 1$ and $I = i$

The matrix representation of $R = 1$ in 2d is the identity matrix

$$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

To find the matrix representation of $i$ we have to analyze the definition of $i$ which is a quantity which squares to $-1$, given that we already know the value of $1$ in matrix form

\begin{align*} i^2 &= -1 * \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \end{align*}

Squaring the following matrix gives the matrix above, then the value of $i$ expressed in matrix form is

$$I = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$$

Finally the value of $C$ is

$$C = a \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + b \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$$

### The complex plane

The powers of $i$ give rise to the sequence $(1, i, -1, -i, 1, \ldots)$ which is quite similar to the pattern $(x, y, -x, -y, x, \ldots)$, the resemblance is no coincidence as complex number belong to a 2-dimensional plane, this complex plane allows us to visualize complex numbers using the horizontal axis for the real part and the vertical axis for the imaginary part

$1, i, -1, -i$

We can see that the positions of $i^0 = 1, i^1 = i, i^2 = -1, i^3 = -i, \ldots$ suggest that the multiplication of a complex number by $i$ is equivalent to rotating through 90 degrees

e.g.

\begin{align*} z_1 &= 2 + i \\ z_2 &= (2 + i)(i) = -1 + 2i \\ z_3 &= (-1 + 2i)(i) = -2 - i \\ z_4 &= (-2 - i)(i) = 1 - 2i \\ z_5 &= (1 - 2i)(i) = 2 + i = z_1 \end{align*}

A complex number is rotated $\pm 90^{\circ}$ by multiplying it by $\pm i$

Let’s graph the roots of $\sqrt{i} = \pm \frac{1}{\sqrt{2}} (1 + i)$

We can see that $\tfrac{1}{\sqrt{2}} (1 + i)$ is exactly at $45^{\circ}$ and $- \tfrac{1}{\sqrt{2}} (1 + i)$ is exactly at $225^{\circ}$

Let’s multiply the complex number $2 + i$ by $\sqrt{i}$ (it should rotate it by $45^{\circ}$)

\begin{align*} z_1 &= 2 + i \\ z_2 &= (2 + i)(\sqrti + \sqrti i) = \sqrti + 3 \sqrti i \end{align*}

Multiplying $z_2$ by $\sqrt{i}$ again should be equal to multiplying $z_1$ by $i$ (because $z_2$ is already rotated by $45^{\circ}$)

\begin{align*} z_2 &= \sqrti + 3 \sqrti i \\ z_3 &= (\sqrti + 3 \sqrti i)(\sqrti + \sqrti i) \\ &= (\frac{1}{2} - \frac{3}{2}) + (\frac{1}{2} + \frac{3}{2})i \\ &= -1 + 2i \end{align*}

Which is exactly what we find if we multiply $z_i$ by $i$, these observations suggest that we can build a complex number which can rotate another complex number by any angle

A complex number is rotated $45^{\circ}$ by multiplying it by $\sqrti + \sqrti i$

A complex number is rotated $225^{\circ}$ by multiplying it by $-\sqrti + \sqrti i$

### Polar representation

Instead of using coordinates in the complex plane we can represent a polar number with the length of the vector from the origin to the complex coordinate and the angle between the complex vector and the positive real axis

$$r = |z| = \sqrt{a^2 + b^2} \\ \theta = arctan(\frac{b}{a})$$

The horizontal component of $z$ is then $r * cos(\theta)$ and the vectical component is $r * sin(\theta)$, expressing the complex number using these quantities

\begin{align*} z &= a + bi \\ &= r * \cos \theta + ri\; \sin \theta \\ &= r \; (\cos \theta + i \sin \theta) \end{align*}

Euler provided the identity

$$$$\label{rotor} e^{i\theta} = \cos \theta + i \sin \theta$$$$

Which allows us to represent any complex number as

$$z = r\,e^{i\theta}$$

Given two polar numbers

$$z = r\,e^{i\theta} \\ w = s\,e^{i\phi} \\$$

Their product is

$$zw = rs\, e^{i(\theta + \phi)} = rs [ \cos (\theta + \phi) + i \sin (\theta + \phi)]$$

Which effectively rotated the complex number $z$ by $\phi$ angles! However the quantity $zw$ was scaled $s$ units, to avoid scalling we can normalize $w$ (i.e. making $r = 1$ which is equal to \eqref{rotor})

A rotor is a complex number that rotates another complex number by an angle $\theta$ (through multiplication) and has the form

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

Rotating a complex number $x + yi$ by an angle $\theta$

\begin{align*} x' + y'i &= (x + yi)(\cos \theta + i \sin \theta) \\ &= (x \cos \theta - y \sin \theta) + (x \sin \theta + y \cos \theta)i \end{align*}

Which in matrix form is

$$\begin{bmatrix} x' & -y' \\ y' & x' \end{bmatrix} = \begin{bmatrix} x & -y \\ y & x \end{bmatrix} \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$$

Note that because of the way the complex product is defined, the multiplication between two complex numbers commutes

\begin{align*} x' + y'i &= (\cos \theta + i \sin \theta)(x + yi)\\ &= (x \cos \theta - y \sin \theta) + (x \sin \theta + y \cos \theta)i \end{align*}