NoteNextra-origin/content/Math416/Math416_L2.md

Math416 Lecture 2

Review?

z_1=r_1(\cos\theta_1+i\sin\theta_1)=r_1\text{cis}(\theta_1)

z_2=r_2(\cos\theta_2+i\sin\theta_2)=r_2\text{cis}(\theta_2)

z_1z_2=r_1r_2\text{cis}(\theta_1+\theta_2)

\forall n\in \mathbb{Z}, z^n=r^n\text{cis}(n\theta)

De Moivre's Formula

\forall n\in \mathbb{Z}, z^n=r^n\text{cis}(n\theta)

New Fancy stuff

Claim:

\forall n\in \mathbb{Z}, z^{\frac{1}{n}}=\sqrt[n]{r}\text{cis}\left(\frac{1}{n}\theta\right)

Proof:

Take an $n$th power, De Moivre's formula holds \forall rational k\in \mathbb{Q}.

Example:

we calculate 1^{\frac{1}{3}}

1=\text{cis}\left(2k\pi\right)

1^{\frac{1}{3}}=\text{cis}\left(\frac{2k\pi}{3}\right)

When k=0, we get 1

When k=1, we get \text{cis}\left(\frac{2\pi}{3}\right)=-\frac{1}{2}+i\frac{\sqrt{3}}{2}

When k=2, we get \text{cis}\left(\frac{4\pi}{3}\right)=-\frac{1}{2}-i\frac{\sqrt{3}}{2}

Strange example

Let p(x)=a_3x^3+a_2x^2+a_1x+a_0 be a polynomial with real coefficients.

Without loss of generality, Let a_3=1, x=y-\beta

We claim \beta=\frac{a_2}{3}

\begin{aligned}
p(x)&=(y-\beta)^3+a_2(y-\beta)^2+a_1(y-\beta)+a_0\\
&=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)\\
\end{aligned}

It's sufficient to know how to solve real cubic equations.

q(x)=x^3+ax+b

Let x=w+\frac{c}{w}

Solve

\begin{aligned}
(w+\frac{c}{w})^3+a(w+\frac{c}{w})+b=0\\
w^3+3w\frac{c}{w}+3\frac{c^2}{w^2}+aw+\frac{ac}{w}+b=0\\
\end{aligned}

We choose c such that 3c+a=0, c=-\frac{a}{3}

\begin{aligned}
w^3+3\frac{c^2}{w}+b=0\\
w^6+bw^3+c^2=0\\
\end{aligned}

Notice that w^6+bw^3+c^2=0 is a quadratic equation in w^3.

w^3=\frac{-b\pm\sqrt{b^2-4c^3}}{2}

So w is a cube root of \frac{-b\pm\sqrt{b^2-4c^3}}{2}

x=w+\frac{c}{w}=w-\frac{a}{3w}

Example:

p(x)=x^3-3x+1=0

a=-3, b=1, c=-\frac{a}{3}=-\frac{-3}{3}=1

\begin{aligned}
w^3&=\frac{-b\pm\sqrt{b^2-4c^3}}{2}\\
&=\frac{-1\pm\sqrt{1-4}}{2}\\
&=\frac{-1\pm\sqrt{3}i}{2}\\
\end{aligned}

To take cube root of w,

w^3=\text{cis}\left(\frac{2\pi}{3}+2k\pi\right)

So

Case 1:

w=\text{cis}\left(\frac{2\pi}{9}+\frac{2k\pi}{3}\right)

It is sufficient to check k=0,1,2 by nth root of unity.

When k=0, w=\text{cis}\left(\frac{2\pi}{9}\right)

When k=1, w=\text{cis}\left(\frac{8\pi}{9}\right)

When k=2, w=\text{cis}\left(\frac{14\pi}{9}\right)

Case 2:

w=\text{cis}\left(\frac{4\pi}{9}+\frac{2k\pi}{3}\right)

When k=0, w=\text{cis}\left(\frac{4\pi}{9}\right)

When k=1, w=\text{cis}\left(\frac{10\pi}{9}\right)

When k=2, w=\text{cis}\left(\frac{16\pi}{9}\right)

So the final roots are:

w+\frac{c}{w}=w+\frac{1}{w}

\text{cis}(\theta)+\frac{1}{\text{cis}(\theta)}=\text{cis}(\theta)+\text{cis}(-\theta)=2\cos(\theta)

So the final roots are:

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)

Remember \cos(2\pi-\theta)=\cos(\theta)

So the final roots are:

2\cos\left(\frac{2\pi}{9}\right), 2\cos\left(\frac{8\pi}{9}\right), 2\cos\left(\frac{14\pi}{9}\right)

Compact

A set K\in \mathbb{R}^n is compact if and only if it is closed and bounded. Compact Theorem in Math 4111

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

Unfortunately, \mathbb{C} is not compact.

Riemann Sphere and Complex Projective Space

Let \mathbb{C}\sim \mathbb{R}^2\subset \mathbb{R}^3

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.

So all the point on the north pole is mapped to outside of the unit circle in \mathbb{R}^2.

all the point on the south pole is mapped to inside of the unit circle in \mathbb{R}^2.

The line through (0,0,1) and (\xi,\eta,z) intersects the unit sphere at (x,y,0)

Line (tx,ty,1-t) intersects z^2 at t^2x^2+t^2y^2+(1-t)^2=1

So t=\frac{2}{1+x^2+y^2}

z=x+iy\mapsto \frac{1}{1+|z|^2}(2Re(z),2Im(z),|z|^2-1)

(\xi,\eta,z)\mapsto \frac{\xi+i\eta}{1-z}

This is a homeomorphism. \mathbb{C}\setminus\{\infty\}\simeq S^2

Derivative of a function

Suppose \Omega is an open subset of \mathbb{C}.

A function $f:\Omega\to \mathbb{C}$'s derivative is defined as

f'(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}

f=u+iv, u,v:\Omega\to \mathbb{R}

How are f' and derivatives of u and v related?

1. Differentiation and complex linearity applies to f

Chain rule applies

\frac{d}{dz}(f(g(z)))=f'(g(z))g'(z)

Polynomials

\frac{d}{dz}z^n=nz^{n-1}