A complex number is a type of number. They are usually written as \(a+bi\).

Formally—

\begin{equation} \mathbb{C} = \left\{a+bi\ \middle |\ a,b \in \mathbb{R} \right\} \end{equation}

This set generates solutions to every single polynomial with unique solutions. Its plane looks like \(\mathbb{R}^{2}\).

## constituents

an order pair of two elements \((a,b)\) where \(a,b\in \mathbb{R}\).

## properties of complex arithmetic

there are 6. For all statements below, we assume \(\alpha = a+bi\) and \(\beta=c+di\), \(\lambda = e+fi\), where \(a,b,c,d,e,f \in \mathbb{R}\) and therefore \(\alpha, \beta,\lambda \in \mathbb{C}\).

### commutativity

\(\alpha + \beta = \beta + \alpha\) and \(\alpha\beta = \beta\alpha\) for all \(\alpha,\beta \in \mathbb{C}\).

#### Proof of complex number commutativity

We desire \(\alpha + \beta = \beta + \alpha\).

\begin{align} \alpha + \beta &= (a+bi)+(c+di) \\ &=(a+c)+(b+d)i \\ &=(c+a)+(d+b)i \\ &=(c+di) + (a+bi) \\ &=\beta+\alpha\ \blacksquare \end{align}

leveraging the commutativity inside real numbers.

Insights: combining and splitting

This proof has the feature of combining, operating (commuting, here), the splitting.

### associativity

\((\alpha +\beta) + \lambda = \alpha + (\beta +\lambda)\) and \((\alpha\beta) \lambda = (\alpha \beta) \lambda\)

Proven via the same trick from last time

### identities

\(\lambda + 0 = \lambda\), \(\lambda 1 = \lambda\)

#### Proof of complex number additive identity

We desire that \(\lambda + 0 = 0\).

\begin{align} \lambda + 0 &= (e+fi) + (0+0i) \\ &= (e+0) + (f+0)i \\ &= e+fi\ \blacksquare \end{align}

multiplicative identity is proven in the same way

### additive inverse

\(\forall \alpha \in \mathbb{C}, \exists !\ \beta \in \mathbb{C}: \alpha + \beta = 0\)

#### Proof of complex number additive inverse

We desire to claim that \(\forall \alpha \in \mathbb{C}, \exists !\ \beta \in \mathbb{C}: \alpha + \beta = 0\), specifically that there *is* a *unique* \(\beta\) which is the additive inverse of every \(\alpha\).

Take a number \(\alpha \in \mathbb{C}\). We have that \(\alpha\) would then by definition be some \((a+bi)\) where \(a,b \in \mathbb{R}\).

Take some \(\beta\) for which \(\alpha + \beta = 0\); by definition we again have \(\beta\) equals some \((c+di)\) where \(c,d \in \mathbb{R}\).

- \(\because \alpha + \beta =0\), \(\therefore (a+bi) + (c+di) = 0\).
- \(\therefore (a+c) + (b+d)i = 0\)
- \(\therefore a+c = 0, b+d = 0\)
- \(\therefore c = -a, d = -b\)

We have created a unique definition of \(c,d\) and therefore \(\beta\) given any \(\alpha\), implying both uniqueness and existence.

Insights: construct then generalize

In this case, the cool insight is the construct and generalize pattern. We are taking a single case \(\alpha\), manipulating it, and wrote the result we want in terms of the constituents of \(\alpha\). This creates both an existence and uniqueness proof.

### multiplicative inverse

\(\forall \alpha \in \mathbb{C}, \alpha \neq 0, \exists!\ \beta \in \mathbb{C} : \alpha\beta =1\)

This is proven exactly in the same way as before.

### distributive property

\(\lambda(\alpha+\beta) = \lambda \alpha + \lambda \beta\ \forall\ \lambda, \alpha, \beta \in \mathbb{C}\)

#### Proof of complex number distributive property

We desire to claim that \(\lambda(\alpha+\beta) = \lambda \alpha + \lambda \beta\).

\begin{align} \lambda(\alpha+\beta) &= (e+fi)((a+bi)+(c+di))\\ &=(e+fi)((a+c)+(b+d)i)\\ &=((ea+ec)-(fb+fd))+((eb+ed)+(fa+fc))i\\ &=ea+ec-fb-fd+(eb+ed+fa+fc)i\\ &=ea-fb+ec-fd+(eb+fa+ed+fc)i\\ &=(ea-fb)+(ec-fd)+((eb+fa)+(ed+fc))i\\ &=((ea-fb)+(eb+fa)i) + ((ec-fd)+(ed+fc)i)\\ &=(e+fi)(a+bi) + (e+fi)(c+di)\\ &=\lambda \alpha + \lambda \beta\ \blacksquare \end{align}

Insights: try to remember to go backwards

At some point in this proof I had to reverse complex addition then multiplication, which actually tripped me up for a bit (“how does

`i`

distribute!!!”, etc.) Turns out, there was already a definition for addition and multiplication of complex numbers so we just needed to use that.

## additional information

### addition and multiplication of complex numbers

\begin{align} (a+bi) + (c+di) &= (a+c)+(b+d)i \\ (a+bi)(c+di) &= (ac-bd)+(ad+bc)i \end{align}

where, \(a,b,c,d\in\mathbb{R}\).

### subtraction and division of complex numbers

Let \(\alpha, \beta \in \mathbb{C}\), and \(-a\) be the additive inverse of \(\alpha\) and \(\frac{1}{\alpha}\) be the multiplicative inverse of \(\alpha\).

: \(\beta-\alpha = \beta + (-\alpha)\)**subtraction**: \(\frac{\beta}{\alpha} = \beta\frac{1}{\alpha}\)**division**

Simple enough, subtraction and division of complex numbers is just defined by applying the inverses of a number to a different number.

### complex numbers form a field

See properties of complex arithmetic, how we proved that it satisfies a field.

### complex conjugate

The complex conjugate of a complex number is defined as

\begin{equation} \bar{z} = \text{Re}\ z - (\text{Im}\ z)i \end{equation}

i.e. taking the complex part to be negative. Say, \(z = 3+2i\), then \(\bar{z}=3-2i\).

### absolute value (complex numbers)

The absolute value (complex numbers) of a complex number is:

\begin{equation} |z| = \sqrt{{(\text{Re}\ z)^{2} + (\text{Im}\ z)^{2}}} \end{equation}