Houjun Liu

complex number

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


\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 \(\mathbb{R}^{2}\).


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 \(\alpha, \beta,\lambda \in \mathbb{C}\).


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


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

Proven via the same trick from last time


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

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 \(\alpha\).

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

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}