Complex numbers

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{16 i^{2}} = \pm 4 i

The set represented by $I$ defines an imaginary number as

i^{2} = - 1

Powers of i

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

Therefore we have the sequence

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

Complex numbers

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

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

Operations on complex numbers

Given two complex numbers

z_{1} = a_{1} + b_{1} i z_{2} = a_{2} + b_{2} i

Addition and subtraction

z_{1} \pm z_{2} = a_{1} \pm a_{2} + (b_{1} \pm b_{2}) i

Product

\begin{aligned} z_{1} z_{2} & = a_{1} a_{2} + a_{1} b_{2} i + a_{2} b_{1} i + b_{1} b_{2} i^{2} given that i^{2} = - 1 \\ = (a_{1} a_{2} - b_{1} b_{2}) + (a_{1} b_{2} + b_{1} a_{2}) i \end{aligned}

Given the complex number

z = a + b i

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 + b i) (a - b i) = a^{2} - a b i + a b i - b^{2} i^{2} = a^{2} + b^{2}

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

z z^{*} = | 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^{*}}{z z^{*}} = \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 $z = a + b i$

\begin{aligned} i & = (a + b i)^{2} \\ = (a + b i) (a + b i) \\ (1) & = a^{2} - b^{2} + 2 a b i \end{aligned}

Therefore

(a^{2} - b^{2}) + (2 a b) i = 0 + 1 i

Equaling real and imaginary parts

\begin{aligned} a^{2} - b^{2} & = 0 \\ 2 a b = 1 \end{aligned}

Therefore $a = \pm b$ , replacing $a = - b$ in the second equation we obtain $- 2 b^{2} = 1$ 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 $2 a^{2} = 1$ so

2 a^{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 + b i) = \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 $- 2 a b = 1$ found from multiplying $(1)$ by $- 1$ )

\sqrt{- i} = (a + b i) = \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} a, b \in R

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

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

[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]

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{aligned} i^{2} & = - 1 * [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] \\ = [\begin{array}{c} - 1 & 0 \\ 0 & - 1 \end{array}] \end{aligned}

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

I = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}]

Finally the value of $C$ is

C = a [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + b [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}] = [\begin{matrix} a & - b \\ b & a \end{matrix}]

The complex plane

The powers of $i$ give rise to the sequence $(1, i, - 1, - i, 1, \dots)$ which is quite similar to the pattern $(x, y, - x, - y, x, \dots)$ , 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 $i^{0} = 1, i^{1} = i, i^{2} = - 1, i^{3} = - i, \dots$ suggest that the multiplication of a complex number by $i$ is equivalent to rotating through 90 degrees

e.g.

\begin{aligned} z_{1} & = 2 + i \\ z_{2} & = (2 + i) (i) = - 1 + 2 i \\ z_{3} & = (- 1 + 2 i) (i) = - 2 - i \\ z_{4} & = (- 2 - i) (i) = 1 - 2 i \\ z_{5} & = (1 - 2 i) (i) = 2 + i = z_{1} \end{aligned}

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 $\frac{1}{\sqrt{2}} (1 + i)$ is exactly at $45^{\circ}$ and $- \frac{1}{\sqrt{2}} (1 + i)$ is exactly at $225^{\circ}$

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

\begin{aligned} z_{1} & = 2 + i \\ z_{2} & = (2 + i) (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i) = \frac{1}{\sqrt{2}} + 3 \frac{1}{\sqrt{2}} i \end{aligned}

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{aligned} z_{2} & = \frac{1}{\sqrt{2}} + 3 \frac{1}{\sqrt{2}} i \\ z_{3} & = (\frac{1}{\sqrt{2}} + 3 \frac{1}{\sqrt{2}} i) (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i) \\ = (\frac{1}{2} - \frac{3}{2}) + (\frac{1}{2} + \frac{3}{2}) i \\ = - 1 + 2 i \end{aligned}

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 $\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i$

A complex number is rotated $225^{\circ}$ by multiplying it by $- \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i$

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}} θ = a r c t a n (\frac{b}{a})

The horizontal component of $z$ is then $r * c o s (θ)$ and the vectical component is $r * s i n (θ)$ , expressing the complex number using these quantities

\begin{aligned} z & = a + b i \\ = r * \cos θ + r i \sin θ \\ = r (\cos θ + i \sin θ) \end{aligned}

Euler provided the identity

\begin{matrix} (2) & e^{i θ} = \cos θ + i \sin θ \end{matrix}

Which allows us to represent any complex number as

z = r e^{i θ}

Given two polar numbers

z = r e^{i θ} w = s e^{i ϕ}

Their product is

z w = r s e^{i (θ + ϕ)} = r s [\cos (θ + ϕ) + i \sin (θ + ϕ)]

Which effectively rotated the complex number $z$ by $ϕ$ angles! However the quantity $z w$ was scaled $s$ units, to avoid scalling we can normalize $w$ (i.e. making $r = 1$ which is equal to $(2)$ )

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

e^{i θ} = \cos θ + i \sin θ

Rotating a complex number $x + y i$ by an angle $θ$

\begin{aligned} x^{'} + y^{'} i & = (x + y i) (\cos θ + i \sin θ) \\ = (x \cos θ - y \sin θ) + (x \sin θ + y \cos θ) i \end{aligned}

Which in matrix form is

[\begin{matrix} x^{'} & - y^{'} \\ y^{'} & x^{'} \end{matrix}] = [\begin{matrix} x & - y \\ y & x \end{matrix}] [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}]

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

\begin{aligned} x^{'} + y^{'} i & = (\cos θ + i \sin θ) (x + y i) \\ = (x \cos θ - y \sin θ) + (x \sin θ + y \cos θ) i \end{aligned}

Imaginary numbers

Powers of i

Complex numbers

Operations on complex numbers

Addition and subtraction

Product

Norm (modulus or absolute value)

Complex conjugate

Inverse

Square root of i

Matrix representation of a complex number

The complex plane

Polar representation

See Also

Quaternions

Integer Factorization

Divisor Function

Square root of $i$