Complex Numbers

$$\mathbb{C}:=\{a+bi\mid a,b\in \mathbb{R},i^2=-1\}$$

The complex numbers are a field of ordered pairs, in the sense that $(a,b)\ne (c,d)$ if $a\ne c$.

Definition

If $z\in \mathbb{C}$ (and so $c=a+bi$ for $a,b\in \mathbb{R}$) then we denote by:

• Complex Conjugate: $\bar{z}$ and $\bar{z}:=a-bi$
• Modulus: $|z{|}^2$ where $|z{|}^2:=(a+bi)(a-bi)=a^2+b^2\in \mathbb{R}$
  • We note that $z\in \mathbb{R},z=a+0i$ for $a\in \mathbb{R}$. So, $|z|=\sqrt{a^2+0^2}=|a|$.

Notation

If $z=a+bi$ then

$$\begin{array}{rl}\text{Re}(z)& :=a\\ \text{Im}(z)& :=b\end{array}$$

Proposition (Complex Facts)

Let $z,w\in \mathbb{C}$. Then,

• $\overline{z+w}=\bar{z}+\bar{w}$
• $\overline{zw}=\bar{z}\cdot \bar{w}$
• $z+\bar{z}=2\text{Re}(z)$
• $z-\bar{z}=2\text{Im}(z)\cdot i$
• $z\bar{z}=|z{|}^2=\text{Re}(z{)}^2+\text{Im}(z{)}^2\in \mathbb{R}$
• $|z|>0$ if $z\ne 0$ and $|0|=0$
• $|\bar{z}|=|z|$
• $|z\cdot w|=|z|\cdot |w|$
• $|\text{Re}(z)|\le |z|$
• $|\text{Im}(z)|\le |z|$

Proof: (Prop 1)

$$\overline{z+w}=\bar{z}+\bar{w}$$

Let $z=a+bi$ and $w=c+di$. where $a,b,c,d\in \mathbb{R}$.

$$\overline{z+w}=\overline{(a+bi)+(c+di)}=\overline{(a+c)+(b+d)i}$$

On the other hand,

$$\bar{z}+\bar{w}=\overline{a+bi}+\overline{c+di}=(a+c)-(b+d)i$$

Triangle Inequality

$$|z+w|\le |z|+|w|$$

Proof:

Since all terms are non negative, it suffices to show

$$|z+w{|}^2\le (|z|+|w|{)}^2$$

Then,

$$\begin{array}{rl}|z+w{|}^2& =(z+w)\cdot \overline{(z+w)}\\ & =(z+w)\cdot (\bar{z}+\bar{w})\\ & =(z\bar{z})+z\bar{w}+w\bar{z}+w\bar{w}\\ & =|z{|}^2+z\bar{w}+\overline{\stackrel{ˉ}{w}z}+|w{|}^2\\ & =|z{|}^2+2\text{Re}(z\bar{w})+|w{|}^2\end{array}$$

We note $|\text{Re}(z\bar{w})|\le z\bar{w}=|z|\cdot |\bar{w}|=|z|\cdot |w|$.

Continuing the inequality,

$$\begin{array}{rl}|z+w{|}^2& =|z{|}^2+2\text{Re}(z\bar{w})+|w{|}^2\\ & \le |z{|}^2+2|z|\cdot |w|+|w{|}^2\\ & =(|z|+|w|{)}^2\end{array}$$

And so the inequality holds true.

Cauchy-Schwarz Inequality

If $a_1,\cdots a_n$ and $b_1,\cdots b_n$ are complex numbers, then

$$|\sum _{j=1}^n{a}_j\overline{b_j}{|}^2\le \sum _{j=1}^n|a_j{|}^2\sum _{j=1}^n|b_j{|}^2$$

where the proof follows from the idea that $B>0$. and concludes at the inequality.

Also recall that

$$|\vec{u}\cdot \vec{v}|\le ||\vec{u}||\cdot ||\vec{v}||$$

Proof:

$$A=\sum _{j=1}^n|a_j{|}^{2}B=\sum _{j=1}^n|b_j{|}^{2}C=\sum _{j=1}^n{a}_j\overline{b_j}$$

Suppose $B=0$. Then the proof is trivial. Suppose not, in which $B>0$. Then,

$$\begin{array}{rl}\sum _{j=1}^n|B{a}_j-C{b}_j{|}^2& =\sum _{j=1}^n(B{a}_j-C{b}_j)\cdot \overline{(B{a}_j-C{b}_j)}\\ & =\sum _{j=1}^n(B{a}_j-C{b}_j)(B\overline{a_j}-\overline{C}\cdot \overline{b_j})\\ & =b^2\sum _{j=1}^n{a}_j\overline{a_j}-B\overline{C}\sum a_j\overline{b_j}-CB\sum \overline{a_j}{b}_j+C\overline{C}\sum b_j\overline{b_j}\\ & =B^{2}A-B\overline{C}C-CB\overline{C}+C\overline{C}B\\ & =B^2-B|C{|}^2-|C{|}^{2}B+|C{|}^{2}B\\ & =A{B}^2-|C{|}^{2}B\end{array}$$

In whzch $A{B}^2-|C{|}^{2}B\ge 0$ because $B$ is positive. So,

$$\begin{array}{rl}& ⟺A{B}^2\ge |C{|}^{2}B\\ & ⟺AB\ge |C{|}^2\end{array}$$

which is Cauchy-Schwarz.