10.14 Complex Numbers

Complex numbers

\(z\cdot \bar z=|z|^{2}\), this implies \(|z\cdot w|^2=|z|^2\cdot |w|^2\)

Polar version

Let \(a=r\cos(\alpha),b=r\sin(\alpha)\), then \(a+ib=r\cdot \text{cis}(\alpha)\) where \(r=|z|=\sqrt{a^2+b^2}\)

For argument, \(\arg:\mathbb{C}\setminus \{0\}\to (-\pi,\pi]\) where \(\alpha =\arg(a+ib)\) which is principal branch

\(z_1\cdot z_2=(r_1\text{cis}\theta_1)(r_2\text{cis}\theta_2)=r_1r_2\text{cis}(\theta_1+\theta_2)\)

---

Distance

Distance\((z,w):=|z-w|=\sqrt{|z-w|^2}\)

If we define, \(\forall m\in\N\), consider \(z_m=x_m+iy_m\in\R^2\). Then \(z_m\to z=x+iy\iff x_m\to x,y_m\to y\)

Important example of convergent sequence of complex number

Fix \(z\in\mathbb{C},m\in \N,E_m=\sum^m_{k=0}\frac{z^k}{k!}=1+z+\frac{z^2}{2!}+...+\frac{z^m}{m!}\)

\(|E_{m+N}-E_{m}|=\left|\sum^{m+N}_{k=m+1}\frac{z^{k}}{k!}\right|\leq \sum^{m+N}_{k=m+1} \left|\frac{z^{k}}{k!}\right|=\sum^{m+N}_{k=m+1}\frac{|z|^{k}}{k!}\leq \sum^{\infty} _{k=m+1}\frac{|z|^{k}}{k!}\to 0\) when \(m\to \infty\)

Thus \(E_m\) is Cauchy sequence in \(\R^2=\mathbb{C}\Rightarrow\) convergent

Property

\(|z+w|\leq |z|+|w|\)

Proof

\(|z+w|\leq |z|+|w|\iff |z+w|^2\leq |z|^2+|w|^2+2|z||w|\iff (z+w)(\bar z+\bar w)\leq |z|^2+|w|^2+2|z||w|\)

\(\iff \text{Re}(z\bar w)\leq |z|\cdot |w|\iff 2abcd\leq a^2d^2+b^2c^2\)

Definition

\(\exp(z)=e^z:=\sum^\infty_{m=0}\frac{z^m}{m!}\)

Euler formula: \(e^{ix}=\cos(x)+i\sin(x)\)

Proof

\(\sum^{N}_{m=0}\frac{(ix)^{m}}{m!}=\sum^{N}_{m=0}\frac{i^{m}x^{m}}{m!}=\sum_{0\leq m\leq N,m\text{ is even}}\frac{i^{m}x^{m}}{m!}+\sum_{0\leq m\leq N,m\text{ is odd}} \frac{i^{m}x^{m}}{m!}\)

\(=\sum_{0\leq k\leq N/2}\frac{i^{2k}x^{2k}}{2k!}+\sum_{0\leq k\leq N/2} \frac{i^{2k+1}x^{2k+1}}{2k+1!}\)

Thus \(\lim_{N\to\infty}\sum^{N}_{m=0}\frac{(ix)^{m}}{m!}=\sum_{k=0}^\infty\frac{(-1)^kx^{2k}}{2k!}+\sum_{k=0}^{\infty}\frac{(-1)^kx^{2k+1}}{2k+1!}i=\cos(x)+i\sin(x)\)

Non-trivial but not so hard exercise
​\(\exp(z+w) = \sum_{n=0}^{\infty} \frac{(z+w)^n}{n!} = \sum_{n=0}^{\infty} \frac{1}{n!} \left(\sum_{k=0}^{n} \binom{n}{k} z^k w^{n-k}\right)\)
exercise! \(= \left(\sum_{k=0}^{\infty} \frac{z^k}{k!}\right) \left(\sum_{j=0}^{\infty} \frac{w^j}{j!}\right)\) \(= \exp(z) \cdot \exp(w)\)

---

\((re^{i\alpha})(r'e^{i\beta})=rr'e^{i(\alpha+\beta)}=rr'(\cos(\alpha+\beta)+\sin( \alpha+\beta))\), then \(rr'(\cos\alpha+i\sin\alpha)(\cos\beta+i\sin\beta)=rr'(\cos(\alpha+\beta)+\sin( \alpha+\beta))\)

Complex numbers and \(2\times 2\) real matrices

\((a+ib)(x+iy)=(ax-by)+i(ay+bx)\)

\(\cdot:\R^{2}\to \R^{2}\) where \((a,b)\cdot \begin{pmatrix} x \\ y \end{pmatrix} \mapsto \begin{pmatrix} ax - by \\ ay + bx \end{pmatrix} = \begin{pmatrix} a & -b \\ b & a \end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix}\)

Then \(a+ib\longleftrightarrow\begin{pmatrix} a&-b\\ b&a \end{pmatrix}\)

\(\mathbb{C}\longleftrightarrow \text{multiply by a fixed complex number}\) is a linear map

Multiply by \(i\) is equivalent to multiply \(J=\begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix}\) (ratation by 90 degree)

\(J^2=-Id\), then \(z_1\cdot z_2=\begin{pmatrix} a&-b\\ b&a \end{pmatrix}\cdot \begin{pmatrix} c&-d\\ d&c \end{pmatrix}=(aId+bJ)(cId+dJ)=\begin{pmatrix} ac-bd&-(bc+ad)\\ bc+ad&ac-bd \end{pmatrix}\)

\(\begin{pmatrix} a & c \\ b & d \end{pmatrix}^{-1}=\frac{1}{ad-bc} \begin{pmatrix} d & -c \\ -b & a \end{pmatrix}\Rightarrow \begin{pmatrix} a & -b \\ b & a \end{pmatrix}^{-1}=\frac{1}{a^2+b^2} \begin{pmatrix} a & b \\ -b & a \end{pmatrix}\)

Then \(z^{-1}=(a+bi)^{-1}=\frac{a-ib}{a^2+b^2}=\frac{\bar z}{|z|^2}\)

\(\text{Tr}\begin{pmatrix} a&-b\\ b&a \end{pmatrix}=2a=2\text{Re}(a+ib)\) and \(\det\begin{pmatrix} a&-b\\b&a \end{pmatrix}=a^2+b^2=|a+ib|^2\)

‍