<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/9a222675-0be5-4eca-aa9e-5f7f5df84120/Complex_number_z.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/9a222675-0be5-4eca-aa9e-5f7f5df84120/Complex_number_z.png" width="40px" /> Complex number $z$:
$$ z=x+iy \qquad x,y\in \R $$
where $i^2=-1$ and $z\in\mathbb C$, $\operatorname {Re}(z)=x$ and $\operatorname{Im}(z)=y$
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operations:
$$ \begin{aligned} ➕\quad (x+iy)+(a+ib)&=(x+a)+i(b+y) \\ ❌\, \qquad (x+iy)(a+ib)&=(xa-yb)+i(ya+xb) \\ \end{aligned} $$
Complex conjugate: $\overline z=x-iy$ which allows us to write
$$ |z|^2=\overline zz=x^2+y^2 \ge 0 $$
argand diagram
$$ z=re^{i\theta} \quad \text{where:} \quad r=\sqrt{x^2+y^2} \qquad \theta=\arg(z)=\arctan\left ( \frac yx \right) $$
multiplication:
$$ (re^{i\theta})(\rho e^{i\phi})=r\rho e^{i(\theta+\phi)} $$
Euler's identity
$$ re^{i\theta}=r[\cos(\theta)+i\sin(\theta)] $$
<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/e7067d54-74b5-44d8-90c0-92d0578f52b0/De_moivres.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/e7067d54-74b5-44d8-90c0-92d0578f52b0/De_moivres.png" width="40px" /> De Moivres theorem:
$$ z^n=r^ne^{in\theta}=r^n[\cos(n\theta)+i\sin(n\theta)] $$
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🗒️ Note: solutions are not unique $e^{i(\theta+2\pi)}=e^{2\pi i}e^{i\theta}=e^{i\theta}$
Principle argument, since $\arg(z)=\theta+2m\pi$ where $m\in \Z$ we define the principle value
$$ -\pi<\operatorname{Arg}(z)\le \pi $$
🧔♀️ Theorem: Fundamental theorem of algebra, given any positive integer $n\ge 1$ and any choice of complex numbers $a_0, a_1,\ldots ,a_n$ such that $a_n\ne 0$ the polynomial equation
$$ P(z)=a_nz^n+\cdots +a_1 z+a_0=0 $$
has at least one solution $z\in \mathbb C$
💃 Example: consider the polynomial $z^6-1=0$
$$ z^6=r^6e^{i6\theta}=r^6(\cos(6\theta)+i(\sin(6\theta))=1 $$
Thus we have the imaginary part $r^6\sin(6\theta)=0$ which means $\theta=m\pi/6$
$$ \theta=\left \{ 0,\frac{\pi}{3},\frac{2\pi}{3},\pi,\frac{4\pi}{3},\frac{5\pi}{3} \right \} $$
Thus we have the following solutions
$$ \begin{aligned} z_p=e^\frac{ip\pi}{3}=\left \{ 1,e^\frac{i\pi}{3},e^\frac{i2\pi}{3},-1,e^\frac{i4\pi}{3},e^\frac{i5\pi}{3} \right \} \end{aligned} $$
💼 Case: Consider the function $w=z^{1/2}$, in polar this gives $w=r^{1/2}e^{i\theta/2}$
Plot of $w$
