243 lines
5.0 KiB
Markdown
243 lines
5.0 KiB
Markdown
# Math416 Lecture 2
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## Review?
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$$
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z_1=r_1(\cos\theta_1+i\sin\theta_1)=r_1\text{cis}(\theta_1)
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$$
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$$
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z_2=r_2(\cos\theta_2+i\sin\theta_2)=r_2\text{cis}(\theta_2)
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$$
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$$
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z_1z_2=r_1r_2\text{cis}(\theta_1+\theta_2)
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$$
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$$
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\forall n\in \mathbb{Z}, z^n=r^n\text{cis}(n\theta)
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$$
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### De Moivre's Formula
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$$
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\forall n\in \mathbb{Z}, z^n=r^n\text{cis}(n\theta)
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$$
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## New Fancy stuff
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Claim:
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$$
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\forall n\in \mathbb{Z}, z^{\frac{1}{n}}=\sqrt[n]{r}\text{cis}\left(\frac{1}{n}\theta\right)
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$$
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Proof:
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Take an $n$th power, De Moivre's formula holds $\forall$ rational $k\in \mathbb{Q}$.
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Example:
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we calculate $1^{\frac{1}{3}}$
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$$
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1=\text{cis}\left(2k\pi\right)
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$$
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$$
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1^{\frac{1}{3}}=\text{cis}\left(\frac{2k\pi}{3}\right)
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$$
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When $k=0$, we get $1$
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When $k=1$, we get $\text{cis}\left(\frac{2\pi}{3}\right)=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$
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When $k=2$, we get $\text{cis}\left(\frac{4\pi}{3}\right)=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$
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#### Strange example
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Let $p(x)=a_3x^3+a_2x^2+a_1x+a_0$ be a polynomial with real coefficients.
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Without loss of generality, Let $a_3=1$, $x=y-\beta$
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We claim $\beta=\frac{a_2}{3}$
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$$
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\begin{aligned}
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p(x)&=(y-\beta)^3+a_2(y-\beta)^2+a_1(y-\beta)+a_0\\
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&=y^3+\left(a_2-3\beta\right)y^2+\left(a_1-3\beta^2-2a_2\beta\right)y+\left(a_0-3\beta^3-3a_1\beta-a_2\beta^2\right)\\
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\end{aligned}
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$$
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It's sufficient to know how to solve real cubic equations.
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$$
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q(x)=x^3+ax+b
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$$
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Let $x=w+\frac{c}{w}$
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Solve
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$$
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\begin{aligned}
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(w+\frac{c}{w})^3+a(w+\frac{c}{w})+b=0\\
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w^3+3w\frac{c}{w}+3\frac{c^2}{w^2}+aw+\frac{ac}{w}+b=0\\
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\end{aligned}
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$$
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We choose $c$ such that $3c+a=0$, $c=-\frac{a}{3}$
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$$
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\begin{aligned}
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w^3+3\frac{c^2}{w}+b=0\\
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w^6+bw^3+c^2=0\\
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\end{aligned}
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$$
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Notice that $w^6+bw^3+c^2=0$ is a quadratic equation in $w^3$.
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$$
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w^3=\frac{-b\pm\sqrt{b^2-4c^3}}{2}
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$$
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So $w$ is a cube root of $\frac{-b\pm\sqrt{b^2-4c^3}}{2}$
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$x=w+\frac{c}{w}=w-\frac{a}{3w}$
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Example:
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$$
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p(x)=x^3-3x+1=0
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$$
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$a=-3$, $b=1$, $c=-\frac{a}{3}=-\frac{-3}{3}=1$
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$$
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\begin{aligned}
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w^3&=\frac{-b\pm\sqrt{b^2-4c^3}}{2}\\
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&=\frac{-1\pm\sqrt{1-4}}{2}\\
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&=\frac{-1\pm\sqrt{3}i}{2}\\
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\end{aligned}
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$$
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To take cube root of $w$,
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$$
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w^3=\text{cis}\left(\frac{2\pi}{3}+2k\pi\right)
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$$
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So
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Case 1:
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$$
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w=\text{cis}\left(\frac{2\pi}{9}+\frac{2k\pi}{3}\right)
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$$
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It is sufficient to check $k=0,1,2$ by nth root of unity.
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When $k=0$, $w=\text{cis}\left(\frac{2\pi}{9}\right)$
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When $k=1$, $w=\text{cis}\left(\frac{8\pi}{9}\right)$
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When $k=2$, $w=\text{cis}\left(\frac{14\pi}{9}\right)$
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Case 2:
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$$
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w=\text{cis}\left(\frac{4\pi}{9}+\frac{2k\pi}{3}\right)
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$$
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When $k=0$, $w=\text{cis}\left(\frac{4\pi}{9}\right)$
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When $k=1$, $w=\text{cis}\left(\frac{10\pi}{9}\right)$
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When $k=2$, $w=\text{cis}\left(\frac{16\pi}{9}\right)$
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So the final roots are:
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$$
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w+\frac{c}{w}=w+\frac{1}{w}
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$$
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$$
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\text{cis}(\theta)+\frac{1}{\text{cis}(\theta)}=\text{cis}(\theta)+\text{cis}(-\theta)=2\cos(\theta)
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$$
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So the final roots are:
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$$
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2\cos\left(\frac{2\pi}{9}\right), 2\cos\left(\frac{8\pi}{9}\right), 2\cos\left(\frac{14\pi}{9}\right), 2\cos\left(\frac{4\pi}{9}\right), 2\cos\left(\frac{10\pi}{9}\right), 2\cos\left(\frac{16\pi}{9}\right)
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$$
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Remember $\cos(2\pi-\theta)=\cos(\theta)$
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So the final roots are:
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$$
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2\cos\left(\frac{2\pi}{9}\right), 2\cos\left(\frac{8\pi}{9}\right), 2\cos\left(\frac{14\pi}{9}\right)
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$$
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### Compact
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A set $K\in \mathbb{R}^n$ is compact if and only if it is closed and bounded. [Compact Theorem in Math 4111](https://notenextra.trance-0.com/Math4111/Math4111_L12#theorem-241)
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If $\{x_n\}\in K$, then there must be some point $w$ such that every disk $D(w,\epsilon)$ contains infinitely many points of $K$. [Infinite Point Theorem about Compact Set in Math 4111](https://notenextra.trance-0.com/Math4111/Math4111_L11#theorem-237)
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Unfortunately, $\mathbb{C}$ is not compact.
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### Riemann Sphere and Complex Projective Space
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Let $\mathbb{C}\sim \mathbb{R}^2\subset \mathbb{R}^3$
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We put a unit sphere on the origin, and project the point on sphere to $\mathbb{R}^2$ by drawing a line through the north pole and the point on the sphere.
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So all the point on the north pole is mapped to outside of the unit circle in $\mathbb{R}^2$.
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all the point on the south pole is mapped to inside of the unit circle in $\mathbb{R}^2$.
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The line through $(0,0,1)$ and $(\xi,\eta,\zeta)$ intersects the unit sphere at $(x,y,0)$
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Line $(tx,ty,1-t)$ intersects $\zeta^2$ at $t^2x^2+t^2y^2+(1-t)^2=1$
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So $t=\frac{2}{1+x^2+y^2}$
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$$
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\zeta=x+iy\mapsto \frac{1}{1+|\zeta|^2}(2Re(\zeta),2Im(\zeta),|\zeta|^2-1)
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$$
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$$
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(\xi,\eta,\zeta)\mapsto \frac{\xi+i\eta}{1-\zeta}
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$$
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This is a homeomorphism. $\mathbb{C}\setminus\{\infty\}\simeq S^2$
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#### Derivative of a function
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Suppose $\Omega$ is an open subset of $\mathbb{C}$.
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A function $f:\Omega\to \mathbb{C}$'s derivative is defined as
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$$
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f'(\zeta_0)=\lim_{\zeta\to \zeta_0}\frac{f(\zeta)-f(\zeta_0)}{\zeta-\zeta_0}
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$$
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$f=u+iv$, $u,v:\Omega\to \mathbb{R}$
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How are $f'$ and derivatives of $u$ and $v$ related?
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1. Differentiation and complex linearity applies to $f$
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Chain rule applies
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$$
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\frac{d}{d\zeta}(f(g(\zeta)))=f'(g(\zeta))g'(\zeta)
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$$
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Polynomials
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$$
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\frac{d}{d\zeta}\zeta^n=n\zeta^{n-1}
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$$
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