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Complex Basics.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=0.5cm]{geometry} \begin{document} {\large Complex Numbers} \begin{align*} \text{Definition:}\quad&i\equiv\sqrt{-1}\qquad\text{where $i$ is an \it imaginary \rm number that satisfies $i^2=-1$ .}\\ &\text{(If there is a $j\neq i$ and $j^2=-1$ as well, then it must be that $j=-i=-\sqrt{-1}$, the \it conjugate \rm of $i$.)}\\ \text{Cartesian:}\quad&z=a+ib\\ \text{Polar:}\quad&z=r\:(\cos\theta+i\sin\theta)=r\:\text{cis}\:\theta=r\:e^{i\theta}\:,\quad\text{where }r>0\text{ (if $z\neq 0$) and }0\leq\theta<2\pi\:.\\ &\text{(When $z=0$, it is not presented in polar form, as $\arg(0)$ does not exist.)}\\ \text{Convertions:}\quad &r=|z|\equiv\sqrt{a^2+b^2}\:,\quad\theta=\arg(z)\equiv\tan^{-1}\frac{\:b\:}{a}\:,\quad \text{Re}(z)\equiv a=r\cos\theta\:,\quad\text{Im}(z)\equiv b=r\sin\theta\\ \\ \text{Basics:}\quad&\text{Algebra for $\mathbb{R}$ works mostly but not always }\ldots\quad \text{e.g. }\sqrt{a\times b}\neq\sqrt{a}\times\sqrt{b}\quad\text{when }a<0\text{ and }b<0\:.\\ &\text{Instead }\sqrt{a\times b}=i\sqrt{-a}\times i\sqrt{-b}=i^2\sqrt{-a}\times\sqrt{-b}=-\sqrt{-a}\times\sqrt{-b}\quad\text{(Note: }-a>0\text{ and }-b>0\text{)}\\ &\text{Re}(z_1\pm z_2)=\text{Re}(z_1)\pm\text{Re}(z_2)\:,\quad\text{Im}(z_1\pm z_2)=\text{Im}(z_1)\pm\text{Im}(z_2)\:,\quad |\text{Re}(z)|\leq|z|\:,\quad|\text{Im}(z)|\leq|z|\\ &\text{If $z=x+iy$ and $z^2=a+ib$ where $a,b,x,y\in\mathbb{R}$, then $x^2-y^2=a$ and $2xy=b$\:.}\quad\left(\sqrt{a+ib}=x+iy\right)\\ &\text{If $z_1=z_2$, then Re($z_1$)=Re($z_2$) and Im($z_1$)=Im($z_2$) .}\\ \\ \text{Conjugate:}\quad&\bar{z}\equiv\text{Re}(z)-i\text{Im}(z)=a-ib\quad\text{or}\quad\bar{z}\equiv|z|\left[\cos\big(-\arg(z)\big)+i\sin\big(-\arg(z)\big)]=r\:[\cos(-\theta)+i\sin(-\theta)\right]\\ &\text{For any polynomial equation $\sum_{k=0}^n a_k x^k=0$ where $a_k\in\mathbb{R}$, if $\alpha$ is a complex solution, so is $\overline{\alpha}$.}\\ &\text{Re}(z)=\frac{z+\bar{z}}{2}\:,\quad\text{Im}(z)=\frac{z-\bar{z}}{2i}\:,\quad \overline{\left(z_1\pm z_2\right)}=\overline{z_1}\pm\overline{z_2}\:,\quad \overline{\left(z_1\cdot z_2\right)}=\overline{z_1}\cdot\overline{z_2}\:,\quad \overline{\left(\frac{z_1}{z_2}\right)}=\frac{\:\overline{z_1}\:}{\overline{z_2}}\\ \\ \text{General Form of }&\text{Equation:}\quad\boxed{z^n=a+ib}\\ &\text{Let }z=R\text{ cis}\:\phi\:,\qquad a+ib=r\text{ cis}\:\theta\\ &(R\text{ cis}\:\phi)^n=r\text{ cis}\:\theta\\ &R^n\text{ cis}\:n\phi=r\text{ cis}\:\theta\\ &R^n=r\:,\qquad n\phi=\theta+2k\pi\quad\text{where }k\in\mathbb{Z}\\ &R=r^{\frac{\:1\:}{n}}\:,\qquad\phi=\frac{\theta+2k\pi}{n}=\frac{\:\theta\:}{n}+\frac{2\pi}{n}k\\ &\text{Solutions are }\boxed{\sqrt[n]{r}\cdot\left[\cos\left(\tfrac{\:\theta\:}{n}+\tfrac{2\pi}{n}k\right)+i\cdot\sin\left(\tfrac{\:\theta\:}{n}+\tfrac{2\pi}{n}k\right)\right]\quad\text{where }\theta=\tan^{-1}\tfrac{\:b\:}{a},\:\:k=0,1,\ldots,n-1}\\ \\ \text{Roots of Unity:}\quad&\boxed{z^n=1}\qquad\text{General Form with $a=1$, $b=0$ and $\theta=0$}\\ &\text{Solutions are }\boxed{\cos\left(\tfrac{2\pi}{n}k\right)+i\cdot\sin\left(\tfrac{2\pi}{n}k\right)\quad\text{where }k=0,1,\ldots,n-1}\\ &\text{Let each solution be }w_k=\cos\left(\tfrac{2\pi}{n}k\right)+i\cdot\sin\left(\tfrac{2\pi}{n}k\right)\text{ , and $w=w_1$, then these will follow:}\\ &w_k=w^k=w^{k+mn}\quad\text{where }m\in\mathbb{Z}\:.\qquad(w_0=w^0=w^n=1)\qquad \text{$w^k$ is used hereinafter, instead of $w_k$ .}\\ &\text{All $w^k$ values form a unit circle.}\\ &(w^k)^n=1\quad\because w^k\text{ is a solution.}\\ &\sum_{k=0}^{n-1}w^k=1+w+w^2+\ldots+w^{n-1}=\frac{w^n-1}{w-1}=0\qquad(w^n=1,w\neq 1)\\ &\overline{w_k}=w^{-k}=w^{n-k}\qquad\text{i.e. All non-real roots form pairs of conjugates.}\\ &(w-1)\cdot\sum_{k=0}^{n-1}(k+1)w^k=(w-1)\cdot(1+2w+3w^2+\ldots+nw^{n-1})=n\\ \end{align*} \end{document}