# Complex Numbers

## Complex Numbers Introduction

The fundamental theorem of algebra states that every non-constant single-variable polynomial

$f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_0x^0$

has at least one root. If you consider $$f(x) = x^2 + 1$$, it is obvious that there is no such root:

If we try to tackle the problem analytically it is obvious that $$x^2=-1$$ has no solution since $$x^2\geq 0$$ and using the quadratic formula, we get a negative square root, which also indicates that the formula does not have a solution:

$\begin{array}{rl}x_{1,2} &= -\frac{p}{2}\pm\sqrt{\left(\frac{p}{2}\right)^2-q}\\&= -\frac{0}{2}\pm\sqrt{\left(\frac{0}{2}\right)^2-1}\\&= \pm\sqrt{-1}\end{array}$

Just as numbers are extended from the natural numbers to integers to get a solution for subtraction into the negative domain, the extension from the integers to the rational numbers to represent fractions, and finally the extension to real numbers by irrational numbers, the numbers are expanded again to be able to solve negative roots. To do this, we introduce a new symbol $$\mathbf{i}$$, called imaginary unit, which is defined as $$\mathbf{i}^2 := -1$$. With this trick, the fundamental theorem of algebra stays true and the quadratic formula stated before has exactly two solutions, $$\mathbf{i}$$ and $$-\mathbf{i}$$, since

$\sqrt{-k}=\sqrt{(-1)\cdot k}=\sqrt{-1}\cdot\sqrt{k} = \mathbf{i}\sqrt{k}$

In general, any polynomial $$f(x)$$ can be factored using complex numbers $$\xi_1,\dots,\xi_n$$ such as

$f(x) = a_n\cdot (x - \xi_1)\cdot\dots\cdot(x-\xi_n)$

Using the imaginary unit, we can finally introduce a complex number to be $$\mathbf{z}:=a+\mathbf{i}b$$ with $$a\in\mathbb{R}$$ being the real part and $$b\in\mathbb{R}$$ being the imaginary part.

Geometrically we use a two-dimensional coordinate-system to express a complex number and keep drawing the real part on the horizontal axis and the imaginary part on the vertical axis. Plotting a complex function would require to draw the graph in the third dimension. This indicates, that numbers contain a hidden dimension, which has immense practical utilty.

By this interpretation, a complex number is a point in two dimensional space. When we draw a line from the origin of the coordinate-system to that point, we can represent a complex number also as the length or the magnitude $$|\mathbf{z}|$$ and an angle $$\theta$$, called the argument or the phase:

The so generated triangle allows us to use trigonometry to calculate the argument $$\arg\mathbf{z}=\theta$$ and the magnitude $$|\mathbf{z}|$$. If we think of the complex number sitting on a circle, we can see the magnitude as the radius $$r = |\mathbf{z}|$$ of that circle. To calculate these properties we can state

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

$\arg\mathbf{z} := \tan^{-1}\left(\frac{b}{a}\right)$

Whereas the argument is not unique, it is defined as a multiple of $$2\pi$$. For example the argument of -1 can be $$\pi, -\pi, 3\pi, ...$$, or more formally $$\arg(-1)=\pi+2k\pi$$ with $$k\in\mathbb{Z}$$.

Another connection can be made to trigonometric functions. A complex number can be expressed with the sinus and co-sinus. The following definition is trivial to find when looking at the sketch above and is called the phasor form:

$\mathbf{z} = |\mathbf{z}|\cos\theta + \mathbf{i}|\mathbf{z}|\sin\theta=|\mathbf{z}|(\cos\theta+\mathbf{i}\sin\theta)=|\mathbf{z}|cis(\theta)$

## Definition of Complex Numbers

A complex number $$\mathbf{z}\in\mathbb{C}$$ is defined as a pair of numbers $$a,b\in\mathbb{R}$$ such that $$\mathbf{z}:=a+\mathbf{i}b = (a, b)$$. The introduced symbol $$\mathbf{i}$$ is specified by $$\mathbf{i}^2:=-1$$ or $$\mathbf{i}:=\sqrt{-1} = (0,1)$$ and is called imaginary unit, $$a=\Re(\mathbf{z})$$ is called the real part of the complex number $$\mathbf{z}$$ and $$b=\Im(\mathbf{z})$$ its imaginary part. By convention we write $$a+\mathbf{i}b$$ for variables $$a$$ and $$b$$, but write $$\mathbf{i}$$ behind $$b$$ if $$b$$ is a constant, for example $$2+3\mathbf{i}$$. We formally define the set of complex numbers $$\mathbb{C}$$ as real linear combinations:

$\mathbb{C}:= \{\mathbf{z} = a\cdot 1 + b \cdot\mathbf{i} | a, b\in\mathbb{R}\}$

From which is clear, that complex numbers span a two-dimensional vector space with base $$\{1, \mathbf{i}\}$$. Complex numbers can therefore be identified as points in the Gaussian number plane. We will see that the following interpretations of a complex number are identical, such as a number with two components, a two dimensional vector, a trigonometric expression and also a way to express it as an exponential:

$\begin{array}{rl}\mathbf{z} :=&a+\mathbf{i}b\\:=& a\cdot(1,0)+b\cdot(0,1) = (a, b)\\:=&r\cdot(\cos(\theta)+\mathbf{i}\sin(\theta)) = r\cdot cis(\theta)\\:=&r\cdot e^{\mathbf{i}\theta}\end{array}$

### Complex numbers form a field

The set of complex numbers $$\mathbb{C}$$ form a field. To make a set a field, the following properties are necessary - which are shown within the definition of the complex numbers operators:

• existence of additive identity element
• commutative property of multiplication
• associative property of multiplication
• existence of multiplicative inverse
• existence of multiplicative identity element
• multiplication is distributive over addition

### Complex numbers have no order

The way complex numbers are defined, complex numbers lose the property to get ordered, that means there is no operation $$\leq$$ in $$\mathbb{C}$$. Intuitively this makes sense when complex numbers are seen as points in the complex plane, but more formally if we say $$\mathbf{z}\geq 0$$, we should be able to say $$\mathbf{i}\geq 0$$, from which follows that $$\sqrt{-1}\geq 0$$, or $$-1\geq 0 \text{↯}$$. The same happens the other way round: If $$\mathbf{z}\leq 0$$, it should also be possible to say $$-\mathbf{i}\geq 0$$, from which follows that $$-\sqrt{-1}\geq 0$$, or $$-1\geq 0 \text{↯}$$. It follows that we can not sort complex numbers. Furthermore, if $$\mathbf{z}=a+0\mathbf{i}$$, we simply write $$a$$.

Even if we don’t have an order, we can compare two complex numbers. Two complex numbers are equal if and only if theor real parts are equal and theor imaginary parts are equal.

### Embedding real numbers

The field of real numbers can be embedded into the field of complex numbers with the following injective ring-homomorphism:

$\psi: \mathbb{R}\to\mathbb{C}, x\mapsto (x, 0)$

Using this homomorphism, we are able to define the following two constants: $$\mathbf{0}:=(0,0)$$ and $$\mathbf{1}:=(1, 0)$$

## Operators of complex numbers

The algebra of complex numbers is usually defined by assuming that the algebra of real numbers holds for complex numbers as well, except that $$\mathbf{i}^2=-1$$.

$+:\mathbb{C}\times\mathbb{C}\mapsto\mathbb{C}$

The addition of two complex numbers is defined as vector addition in a two dimensional vector space, therefore $$(a_1, b_1)+(a_2,b_2):= (a_1+ a_2,b_1+ b_2)$$, or

$\begin{array}{rl}\mathbf{z_1} + \mathbf{z_2}:= &(a_1+\mathbf{i}b_1)+(a_2+\mathbf{i}b_2)\\=&a_1+\mathbf{i}b_1+ a_2+ \mathbf{i}b_2\\=&(a_1+ a_2) + \mathbf{i}(b_1+ b_2)\\\end{array}$

Geometrically, the addition of two complex numbers is the ordinary parallelogram law for adding vectors:

Proof: The addition of complex numbers is associative:

$\begin{array}{rl}(\mathbf{z}_1 + \mathbf{z}_2) + \mathbf{z}_3 &= ((a_1+\mathbf{i}b_1)+(a_2+\mathbf{i}b_2))+(a_3+\mathbf{i}b_3)\\&= ((a_1+ a_2)+ a_3) + \mathbf{i}((b_1+b_2)+ b_3)\\&= (a_1+(a_2+ a_3)) + \mathbf{i}(b_1+(b_2+ b_3))\\&= (a_1+\mathbf{i}b_1)+((a_2+\mathbf{i}b_2)+(a_3+\mathbf{i}b_3))\\&= \mathbf{z}_1 + (\mathbf{z}_2 + \mathbf{z}_3)\\\end{array}$

Proof: The addition of complex numbers is commutative:

$\begin{array}{rl}\mathbf{z}_1 + \mathbf{z}_2 &= (x_1+\mathbf{i}y_1)+(x_2+\mathbf{i}y_2)\\&= (x_1+x_2)+\mathbf{i}(y_1+y_2)\\&= (x_2+x_1)+\mathbf{i}(y_2+y_1)\\&= (x_2+\mathbf{i}y_2)+(x_1+\mathbf{i}y_1)\\&= \mathbf{z}_2+\mathbf{z}_1\end{array}$

Proof: The additive identity element is $$\mathbf{0}$$:

$\begin{array}{rl}\mathbf{z}+\mathbf{0} &= (a+\mathbf{i}b)+(0+\mathbf{i}0)\\&= (x+0)+\mathbf{i}(b+0)\\&= a+\mathbf{i}b\end{array}$

The opposite direction $$\mathbf{0}+\mathbf{z}$$ follows from the additive commutativity in $$\mathbb{C}$$. Addition is also consistent with the addition of real numbers:

$(a_1 + \mathbf{i}\cdot 0) + (a_2 + \mathbf{i}\cdot 0)=a_1+ a_2$

Summing two complex numbers in polar representation proofs the well-known cosine-law for triangles:

$\begin{array}{rl}\mathbf{z}_1+\mathbf{z}_2 &= \left(r_1e^{\mathbf{i}\theta_1}\right)+ \left(r_2e^{\mathbf{i}\theta_2}\right)\\&= (r_1\cos\theta_1+r_2\cos\theta_2) + \mathbf{i}(r_1\sin\theta_1+r_2\sin\theta_2)\end{array}$

since we can write the magnitude of the vector sum $$|\mathbf{z}_1+\mathbf{z}_2|$$ as:

$\begin{array}{rl}|\mathbf{z}_1+\mathbf{z}_2|^2 &= {(r_1\cos\theta_1+r_2\cos\theta_2)^2 + (r_1\sin\theta_1+r_2\sin\theta_2)^2}\\&= r_1^2+r_2^2+2r_1r_2\cos(\theta_1-\theta_2)\end{array}$

### Complex Negation

$-:\mathbb{C}\mapsto\mathbb{C}$

The negation of a complex number or the additive inverse is defined by the vector negation, which negates componentwise like $$-(a, b) := (-a, -b)$$, or

$\begin{array}{rl}-\mathbf{z}:= &-a-\mathbf{i}b\end{array}$

Geometrically, the negation of a complex number is the reflection across the origin of the coordinate system.

Proof: Complex negation gives additive identity

$\mathbf{z}+(-\mathbf{z}) = (a+\mathbf{i}b)+(-(a+\mathbf{i}b)) = (a+(-a))+\mathbf{i}(b+(-b)) = 0+0\mathbf{i} = \mathbf{0}$

The opposite direction $$(-\mathbf{z})+\mathbf{z}=\mathbf{0}$$ follows from the additive commutativity in $$\mathbb{C}$$.

Apparently, the negation is not multiplicative:

$(-\mathbf{z}_1)\times(-\mathbf{z}_2) = \mathbf{z}_1\times\mathbf{z}_2\text{, not }-(\mathbf{z}_1\times\mathbf{z}_2)$

### Complex Subtraction

$-:\mathbb{C}\times\mathbb{C}\mapsto\mathbb{C}$

The subtraction of complex numbers can now be defined as the addition with the additive inverse, but stays the componentwise subtraction:

$\begin{array}{rl}\mathbf{z_1} - \mathbf{z_2}:=& \mathbf{z}_1+(-\mathbf{z}_2)\\=&a_1+\mathbf{i}b_1+ (-(a_2+ \mathbf{i}b_2))\\=&a_1+\mathbf{i}b_1- a_2- \mathbf{i}b_2\\=&(a_1- a_2) + \mathbf{i}(b_1- b_2)\\\end{array}$

### Complex Multiplication

$\times:\mathbb{C}\times\mathbb{C}\mapsto\mathbb{C}$

The multiplication is designed in such a way that $$\mathbf{i}^2=-1$$, associativity, commutativity and als distributivity holds. To derive a multiplication that holds true for these properties, we can simply multiply out $$(a_1,b_1)\times(a_2,b_2)=(a_1a_2-b_1b_2,a_1b_2+a_2b_1)$$, or

$\begin{array}{rl}\mathbf{z_1} \times \mathbf{z_2}:= & (a_1+\mathbf{i}b_1)\cdot(a_2+\mathbf{i}b_2)\\=& a_1a_2 +\mathbf{i}a_1b_2+\mathbf{i}a_2b_1+\mathbf{i}^2b_1b_2\\=& \underbrace{(a_1a_2-b_1b_2)}_{\Re(\mathbf{z_1} \times \mathbf{z_2})}+\mathbf{i}\underbrace{(a_1b_2+a_2b_1)}_{\Im(\mathbf{z_1} \times \mathbf{z_2})}\end{array}$

Geometrically the multiplication of a complex number with $$\mathbf{i}$$ is a rotation by 90° and looks like this:

In fact multiplication with a complex number of length 1 always rotates the other complex number, as we will see later. A multiplication of two arbitrary complex numbers can be seen as a rotation and scaling in one step. Lets say we want to multiply $$\mathbf{z}_1$$ and $$\mathbf{z}_2$$:

Each of the vectors formed by the complex numbers span a triangle with $$1$$ on the real axis like this:

We can now rotate the red triangle and align it to the green like this:

And finally we scale the red triangle such that the site which had length $$1$$ now has length $$|\mathbf{z}_1|$$. The point where the former $$\mathbf{z}_2'$$ was pointing to is the product of the two complex numbers:

This means, complex multiplication is just a rotation and scaling. More explicitly, we can show that the multiplication also holds using polar notation (goniometric form):

$\begin{array}{rl}\mathbf{z_1} \times \mathbf{z_2}:= & (r_1 e^{\mathbf{i}\theta_1})\cdot (r_2 e^{\mathbf{i}\theta_2})\\=& r_1 r_2\cdot e^{\mathbf{i}(\theta_1 +\theta_2)}\\=& r_1 r_2 \cdot((\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2) + \mathbf{i}(\cos\theta_1\sin\theta_2+\sin\theta_1\cos\theta_2))\\=& r_1 r_2\cdot (\cos(\theta_1+\theta_2) + \mathbf{i}\sin(\theta_1+\theta_2))\\=& r_1 r_2\cdot cis(\theta_1+\theta_2)\end{array}$

Multiplying complex numbers in polar notation is thus just the product of their magnitudes and the sum of their arguments:

$|\mathbf{z_1}\mathbf{z_2}| = |\mathbf{z_1}|\cdot|\mathbf{z_2}|$

$\arg(\mathbf{z_1}\mathbf{z_2}) = \arg\mathbf{z_1}+\arg\mathbf{z_2}$

If the magnitude of both numbers were 1, then the multiplication amounts to a rotation in the plane.

Proof: The multiplication of complex numbers is associative:

$\begin{array}{rl}(\mathbf{z_1}\times\mathbf{z_2})\times\mathbf{z_3} &= ((a_1+\mathbf{i}b_1)\cdot(a_2+\mathbf{i}b_2))\cdot(a_3+\mathbf{i}b_3)\\&= (a_1a_2-b_1b_2)\cdot(a_3+\mathbf{i}b_3)\\&= (a_1a_2a_3 - b_1b_2a_3 - a_1b_2b_3 - b_1a_2b_3) + \mathbf{i}(a_1a_2b_3-b_1b_2b_3+a_1b_2a_3+b_1a_2a_3)\\&= (a_1+\mathbf{i}b_1)\cdot(a_2a_3-b_2b_3)\\&= (a_1+\mathbf{i}b_1)\cdot((a_2+\mathbf{i}b_2)\cdot(a_3+\mathbf{i}b_3))\\&= \mathbf{z_1}\times(\mathbf{z_2}\times\mathbf{z_3})\end{array}$

Proof: The multiplication of complex numbers is commutative:

$\begin{array}{rl}\mathbf{z}_1\times\mathbf{z}_2 &= (a_1+\mathbf{i}b_1)(a_2+\mathbf{i}b_2)\\&= a_1a_2-b_1b_2+\mathbf{i}(a_1b_2+b_1a_2)\\&= a_2a_1-b_2b_1+\mathbf{i}(a_2b_1+b_2a_1)\\&= (a_2+\mathbf{i}b_2)(a_1+\mathbf{i}b_1)\\&= \mathbf{z}_2\times\mathbf{z}_1\end{array}$

Proof: The distributive law of complex numbers:

From left: $\begin{array}{rl}\mathbf{z}_1(\mathbf{z}_2 + \mathbf{z}_3) &= (a_1 + \mathbf{i}b_1)(a_2 + a_3 + \mathbf{i}(b_2 + b_3))\\&= a_1 a_2 + a_1 a_3 - b_1 b_2 - b_1 b_3 + \mathbf{i}(a_1 b_2 + a_2 b_1 + a_3 b_1 + a_1 b_3)\end{array}$

From right: $\begin{array}{rl}(\mathbf{z}_1 + \mathbf{z}_2)\mathbf{z}_3 &= (a_1 + a_2 + \mathbf{i}(b_1 + b_2))(a_3 + \mathbf{i}b_3)\\&= a_1 a_3 + a_2 a_3 - b_1 b_3 - b_2 b_3 + \mathbf{i}(a_1 b_3 + a_2 b_3 + a_3 b_1 + a_3 b_2)\end{array}$

Proof: The multiplicative identity element is $$\mathbf{1}$$:

$\mathbf{1}\times\mathbf{z} = (1+\mathbf{i}0)(a+\mathbf{i}b) = (a-0)+\mathbf{i}(b+0)=a+\mathbf{i}b=\mathbf{z}$

The opposite direction $$\mathbf{z}\times\mathbf{1}$$ follows from the multiplicative commutativity in $$\mathbb{C}$$. Multiplication is also consistent with the multiplication of real numbers:

$(a_1, 0)\times(a_2,0)=(a_1\cdot a_2,0)$

### Complex Scalar Multiplication

$\cdot:\mathbb{R}\times\mathbb{C}\mapsto\mathbb{C}$

The complex scalar multiplication (the multiplication of a complex number and a real number or a complex number and a complex number with $$\Im=0$$ - is defined as a vector scalar multiplication where the scalar is multiplied component-wise. Let $$s\in\mathbb{R}$$ then

$s\cdot\mathbf{z}:=sa+\mathbf{i}sb$

Geometrically, this can be seen as moving on a straight line:

### Complex Dot product

If we consider complex numbers as a two-dimensional vector, we can simply expand the dot product to it and see the same geometric interpretation:

$\begin{array}{rl}\mathbf{z_1}\cdot\mathbf{z_2}&:= (a_1, b_1)\cdot(a_2, b_2)\\&= a_1a_2+b_1b_2\\\end{array}$

### Complex Conjugate

$\overline{\cdot}:\mathbb{C}\mapsto\mathbb{C}$

The complex conjugate of a complex number $$\mathbf{z}$$ is the reflection across the real axis and is defined as $$\overline{(a,b)}=(a,-b)$$, or

$\begin{array}{rl}\overline{\mathbf{z}} :=& \overline{a+\mathbf{i}b}\\=& a-\mathbf{i}b\end{array}$

Or in polar form:

$\begin{array}{rl}\overline{\mathbf{z}} :=& r\cdot(\cos(-\theta) + \mathbf{i}\sin(-\theta))\\=& r\cdot cis(-\theta)\end{array}$

The complex conjugate is a field automorphism and thus compatible with addition and multiplication:

$\overline{\mathbf{z_1}}\pm\overline{\mathbf{z_2}} =(a_1-\mathbf{i}b_1)\pm (a_2 - \mathbf{i}b_2) = (a_1 \pm a_2) - \mathbf{i}(b_1\pm b_2) =\overline{\mathbf{z_1}\pm\mathbf{z_2}}$

$\overline{\mathbf{z_1}}\times\overline{\mathbf{z_2}} = (a_1-\mathbf{i}b_1)\times (a_2 - \mathbf{i}b_2) = (a_1a_2 - b_1b_2) - \mathbf{i}(a_1b_2+a_2b_1) = \overline{\mathbf{z_1}\times\mathbf{z_2}}$

Addition and subtraction with a complex numbers conjugate reveals the following properties:

$\mathbf{z} + \overline{\mathbf{z}} = (a+a, b-b) = 2\Re(\mathbf{z})\leq 2|\mathbf{z}|$

$\mathbf{z} - \overline{\mathbf{z}} = (a-a, b+b) = 2\mathbf{i}\Im(\mathbf{z})\leq 2|\mathbf{z}|$

These findings can be used to extract the real part and imaginary part of a complex number using the complex conjugate:

$a = \Re(\mathbf{z}) = \frac{1}{2}(\mathbf{z}+\overline{\mathbf{z}})$

$b = \Im(\mathbf{z})= \frac{1}{2\mathbf{i}}(\mathbf{z}-\overline{\mathbf{z}}) = \frac{1}{2}\mathbf{i}(\overline{\mathbf{z}}-\mathbf{z})$

Multiplication with a complex numbers conjugate shows an interesting connection to the norm of a complex number, which is always a positive real number:

$\mathbf{z}\times\overline{\mathbf{z}} = (a+\mathbf{i}b)(a-\mathbf{i}b) = a^2+b^2 = \Re(\mathbf{z})^2+\Im(\mathbf{z})^2 = |\mathbf{\mathbf{z}}|^2$

The opposite direction $$\overline{\mathbf{z}}\times\mathbf{z}$$ follows from the multiplicative commutativity in $$\mathbb{C}$$. Additionally the following properties hold for all complex numbers:

$\overline{\overline{\mathbf{z}}} = \mathbf{z}$

$|\overline{\mathbf{z}}| = |\mathbf{z}|$

$(\overline{\mathbf{z}})^{-1} = \overline{\mathbf{z}^{-1}}$

$\overline{\left(\frac{\mathbf{z}_1}{\mathbf{z}_2}\right)} = \frac{\overline{\mathbf{z}_1}}{\overline{\mathbf{z}_2}}$

$\exp(\overline{\mathbf{z}}) = \overline{\exp(\mathbf{z})}$

$\log(\overline{\mathbf{z}}) = \overline{\log(\mathbf{z})} \;\forall \mathbf{z}\neq 0$

$\overline{\mathbf{z}} = \mathbf{z} \Leftrightarrow \Im(\mathbf{z})=0$

$\overline{\mathbf{z}} = -\mathbf{z} \Leftrightarrow \Re(\mathbf{z})=0$

### Complex multiplicative inverse

The multiplicative inverse or reciprocal of a complex number $$\mathbf{z}\neq 0$$ is the normalized conjugate:

$\begin{array}{rrl}& \mathbf{z}^{-1} &= \frac{\mathbf{1}}{\mathbf{z}}\\\Leftrightarrow & \mathbf{z}^{-1}\mathbf{z} &= \mathbf{1}\\\Leftrightarrow & \mathbf{z}^{-1}\mathbf{z}\overline{\mathbf{z}} &= \overline{\mathbf{z}}\\\Leftrightarrow & \mathbf{z}^{-1}|\mathbf{z}|^2 &= \overline{\mathbf{z}}\\\Leftrightarrow & \mathbf{z}^{-1} &= \frac{\overline{\mathbf{z}}}{|\mathbf{z}|^2}\\& &= \frac{a-\mathbf{i}b}{a^2 + b^2}\end{array}$

Proof: Multiplicative inverse

$\begin{array}{rl}\mathbf{z}\times\mathbf{z}^{-1} &= (a+\mathbf{i}b)\times \left(\frac{a-\mathbf{i}b}{a^2+b^2}\right)\\&= \left(\frac{a^2-\mathbf{i}^2b^2}{a^2+b^2}\right)\\&= \left(\frac{a^2+b^2}{a^2+b^2}\right)\\& = \mathbf{1}\end{array}$

The opposite direction $$\mathbf{z}^{-1}\times\mathbf{z}=\mathbf{1}$$ follows from the multiplicative commutativity in $$\mathbb{C}$$. From the definition of the multiplicative inverse also follows that $$\mathbf{i}^{-1} = - \mathbf{i}$$ and $$\mathbf{1}^{-1} = \mathbf{1}$$.

### Complex division

$/:\mathbb{C}\times\mathbb{C}\mapsto\mathbb{C}$

Using the multiplicative inverse we can now define the quotient of $$\mathbf{z_1}$$ and $$\mathbf{z_2}$$:

$\begin{array}{rl}\frac{\mathbf{z_1}}{\mathbf{z_2}}:=& \mathbf{z_1} \times \mathbf{z^{-1}_2}\\=& \frac{\mathbf{z_1}\overline{\mathbf{z_2}}}{|\mathbf{z_2}|^2}\end{array}$

We can introduce the complex division also in a more unintuitive way (schools seem to prefer this method) like:

$\begin{array}{rl}\frac{\mathbf{z_1}}{\mathbf{z_2}}:=& \frac{a_1+\mathbf{i}b_1}{a_2+\mathbf{i}b_2}\\=& \frac{a_1+\mathbf{i}b_1}{a_2+\mathbf{i}b_2} \cdot \frac{a_2-\mathbf{i}b_2}{a_2-\mathbf{i}b_2}\\=& \frac{(a_1+\mathbf{i}b_1)(a_2-\mathbf{i}b_2)}{(a_2+\mathbf{i}b_2)(a_2-\mathbf{i}b_2)}\\=& \frac{a_1a_2-\mathbf{i}b_2a_1+\mathbf{i}b_1a_2-\mathbf{i}^2b_1b_2}{a_2^2-\mathbf{i}b_2a_2+\mathbf{i}b_2a_2-\mathbf{i}^2b_2^2}\\=& \frac{(a_1a_2+b_1b_2)+\mathbf{i}(b_1a_2-b_2a_1)}{a_2^2+b_2^2}\\=& \underbrace{\frac{(a_1a_2+b_1b_2)}{a_2^2+b_2^2}}_{\Re(\mathbf{z_1}/\mathbf{z_2})}+\mathbf{i}\underbrace{\frac{(b_1a_2-a_1b_2)}{a_2^2+b_2^2}}_{\Im(\mathbf{z_1}/\mathbf{z_2})}\\=& \frac{(a_1a_2+b_1b_2)}{a_2^2+b_2^2}-\mathbf{i}\frac{(a_1b_2-b_1a_2)}{a_2^2+b_2^2}\\=& \frac{\mathbf{z_1}\overline{\mathbf{z_2}}}{\mathbf{z_2}\overline{\mathbf{z_2}}}\\=& \frac{\mathbf{z_1}\overline{\mathbf{z_2}}}{|\mathbf{z_2}|^2}\\\end{array}$

For the complex multiplication we have seen that multiplying by $$\mathbf{i}$$ is a rotation by 90°. Dividing by $$\mathbf{i}$$ correlates to a rotation in the opposite direction by 90°.

### Absolute value of a Complex Number

$|\cdot|:\mathbb{C}\mapsto\mathbb{R}$

For real numbers the absolute value $$|s|$$ is the positive distance to zero. For complex numbers we can use Pythagorean theorem to define a similar distance measure in two dimensions, which is called absolute value or modulus. It can easily be shown that the absolute value of a complex number fulfills the properties of a norm. It turns out that the field $$(\mathbb{C}, +, \times)$$ is complete under the norm. We start with the derivation of the squared absolute value:

$|\mathbf{z}|^2 = |\overline{\mathbf{z}}|^2 = \mathbf{z}\times\overline{\mathbf{z}}=\overline{\mathbf{z}}\times\mathbf{z}=a^2+b^2=\Re(\mathbf{z})^2+\Im(\mathbf{z})^2$

It follows that the squared norm is always a positive real number. Taking the square root keeps it a real positive number and we can find the complex norm:

$|\mathbf{z}| = \sqrt{|\mathbf{z}|^2}$

From which follows that the absolute value of each component is always less or equal than the norm of the complex number:

$|\Re(\mathbf{z})|\leq|\mathbf{z}|$

$|\Im(\mathbf{z})|\leq|\mathbf{z}|$

We can also see that definiteness holds for complex norms:

$|\mathbf{z}| = 0\Leftrightarrow \mathbf{z}=0$

And also absolute homogeneity holds true for complex norms:

$|\mathbf{z_1} \times \mathbf{z_2}| = |\mathbf{z_1}| \cdot |\mathbf{z_2}|$

Proof: Absolute homogeneity for complex norms:

$\begin{array}{rl}|\mathbf{z_1} \times \mathbf{z_2}|^2 &= (\mathbf{z_1}\mathbf{z_2})(\overline{\mathbf{z_1}\mathbf{z_2}})\\&= \mathbf{z_1}\mathbf{z_2}\overline{\mathbf{z_1}}\overline{\mathbf{z_2}}\\&= (\mathbf{z_1}\overline{\mathbf{z_1}})(\mathbf{z_2}\overline{\mathbf{z_2}})\\&= |\mathbf{z_1}|^2 \cdot |\mathbf{z_2}|^2\\&= (|\mathbf{z_1}| \cdot |\mathbf{z_2}|)^2\end{array}$

The norm of the inverse of a complex number has the following property:

$|\mathbf{z}^{-1}| = |\mathbf{z}|^{-1}$

The following Triangle inequality holds:

$|\mathbf{z_1}+\mathbf{z_2}|\leq|\mathbf{z_1}|+|\mathbf{z_2}|$

Proof: Triangle inequality

$\begin{array}{rl}|\mathbf{z_1}+\mathbf{z_2}|^2 &= (\overline{\mathbf{z_1}+\mathbf{z_2}})(\mathbf{z_1}+\mathbf{z_2}) \\&= (\overline{\mathbf{z_1}}+\overline{\mathbf{z_2}})(\mathbf{z_1}+\mathbf{z_2}) \\&= \overline{\mathbf{z}_1}\mathbf{z}_1 + \overline{\mathbf{z}_2}\mathbf{z}_2 +\underbrace{\overline{\mathbf{z}_1}\mathbf{z}_2 + \overline{\mathbf{z}_2}\mathbf{z}_1}_{=2\Re{(\mathbf{z}_2}\overline{\mathbf{z}_1})}\\&= |\mathbf{z}_1|^2+|\mathbf{z}_2|^2 + 2\Re{(\mathbf{z}_2}\overline{\mathbf{z}_1})\\&\leq |\mathbf{z}_1|^2+|\mathbf{z}_2|^2 + 2|\mathbf{z}_2\overline{\mathbf{z}_1}|\\&= |\mathbf{z}_1|^2+|\mathbf{z}_2|^2 + 2|\mathbf{z}_2| |\overline{\mathbf{z}_1}|\\&= (|\mathbf{z}_1|+|\mathbf{z}_2|)^2\end{array}$

### Generalized Triangle inequality

I was wondering if it is possible to generalize the inequality formula. The longest inequality chains I was able to generate and automatically proof are the following:

$|\mathbf{z_1}| - |\mathbf{z_2}| \leq ||\mathbf{z_1}| - |\mathbf{z_2}|| \leq |\mathbf{z_1} - \mathbf{z_2}| \leq |\mathbf{z_1}| + |\mathbf{z_2}|$

$|\mathbf{z_1}| - |\mathbf{z_2}| \leq ||\mathbf{z_1}| - |\mathbf{z_2}|| \leq |\mathbf{z_1} + \mathbf{z_2}| \leq |\mathbf{z_1}| + |\mathbf{z_2}|$

$|\mathbf{z_1}-\mathbf{z_2}|\leq|\mathbf{z_1}-\mathbf{u}|+|\mathbf{z_2}-\mathbf{u}|\forall\mathbf{u}\in\mathbb{C}$

### Distance measure

The distance $$d(\mathbf{z_1}, \mathbf{z_2})$$ between two complex numbers can be calculated using the norm:

$d(\mathbf{z_1}, \mathbf{z_2}) = |\mathbf{z_1}- \mathbf{z_2}|$

The norm itself is therefore a special case of distance measure: $$|\mathbf{z}|=d(\mathbf{0}, \mathbf{z})$$