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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,\zeta) intersects the unit sphere at (x,y,0)
Line (tx,ty,1-t) intersects \zeta^2 at t^2x^2+t^2y^2+(1-t)^2=1
So t=\frac{2}{1+x^2+y^2}
\zeta=x+iy\mapsto \frac{1}{1+|\zeta|^2}(2Re(\zeta),2Im(\zeta),|\zeta|^2-1)
(\xi,\eta,\zeta)\mapsto \frac{\xi+i\eta}{1-\zeta}
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'(\zeta_0)=\lim_{\zeta\to \zeta_0}\frac{f(\zeta)-f(\zeta_0)}{\zeta-\zeta_0}
f=u+iv, u,v:\Omega\to \mathbb{R}
How are f' and derivatives of u and v related?
- Differentiation and complex linearity applies to
f
Chain rule applies
\frac{d}{d\zeta}(f(g(\zeta)))=f'(g(\zeta))g'(\zeta)
Polynomials
\frac{d}{d\zeta}\zeta^n=n\zeta^{n-1}