5.0 KiB
Math416 Lecture 1
Chapter 1: Complex Numbers
Preface
I don't know what happened to the first class. I will try to rewrite the notes from my classmates here.
Rigidity
Integral is preserved for any closed path.
Group
A set with a multiplication operator. (G,\cdot) such that: for all a,b,c\in G:
a\cdot b\in Ga\cdot (b\cdot c)=(a\cdot b)\cdot ca\cdot 1=aa\cdot a^{-1}=1
Ring
A group with two operations: addition and multiplication. (R,+,\cdot) such that: for all a,b,c\in R:
- Commutative under addition:
a+b=b+a - Associative under multiplication:
(a\cdot b)\cdot c=a\cdot (b\cdot c) - Distributive under addition:
a\cdot (b+c)=a\cdot b+a\cdot c
Example:
\{a+\sqrt{6}b\mid a,b\in \mathbb{Z}\} is a ring
Definition 1.1
the complex number is defined to be the set \mathbb{C} of ordered pairs (x,y) with x,y\in \mathbb{R} and the operations:
- Addition:
(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2) - Multiplication:
(x_1,y_1)(x_2,y_2)=(x_1x_2-y_1y_2,x_1y_2+x_2y_1)
Axiom 1.2
The operation of addition and multiplication on \mathbb{C} satisfies the following conditions (The field axioms):
For all z_1,z_2,z_3\in \mathbb{C}:
z_1+z_2=z_2+z_1(commutative law of addition)(z_1+z_2)+z_3=z_1+(z_2+z_3)(associative law of addition)z_1\cdot z_2=z_2\cdot z_1(commutative law of multiplication)(z_1\cdot z_2)\cdot z_3=z_1\cdot (z_2\cdot z_3)(associative law of multiplication)z_1\cdot (z_2+z_3)=z_1\cdot z_2+z_1\cdot z_3(distributive law)- There exists an additive identity element
0=(0,0)such thatz+0=zfor allz\in \mathbb{C}. - There exists a multiplicative identity element
1=(1,0)such thatz\cdot 1=zfor allz\in \mathbb{C}. - There exists an additive inverse
-z=(-x,-y)for allz=(x,y)\in \mathbb{C}such thatz+(-z)=0. - There exists a multiplicative inverse
z^{-1}=\left(\frac{x}{x^2+y^2},-\frac{y}{x^2+y^2}\right)for allz=(x,y)\in \mathbb{C}such thatz\cdot z^{-1}=1.
(a,b)^{-1}=\left(\frac{a}{a^2+b^2},-\frac{b}{a^2+b^2}\right)
Embedding of \mathbb{R} in \mathbb{C} 1.3
Let z=x+iy\in \mathbb{C} where a,b\in \mathbb{R}.
xis called the real part ofzandyis called the imaginary part ofz.|z|=\sqrt{x^2+y^2}is called the absolute value or modulus ofz.- The angle between the positive real axis and the line segment from
0tozis called the argument ofzand is denoted by\theta(argument ofz). \overline{z}=x-iyis called the conjugate ofz. (z\cdot \overline{z}=|z|^2)z_1+z_2=(x_1+x_2,y_1+y_2)(vector addition)
Lemma 1.3
|z_1z_2|=|z_1||z_2|
Theorem 1.5 (Triangle Inequality)
|z_1+z_2|\leq |z_1|+|z_2|
Proof
Geometrically, the triangle inequality states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side.
Algebraically,
\begin{aligned}
(|z_1+z_2|)^2-|z_1+z_2|^2&=|z_1+z_2|^2-2|z_1+z_2|-(z_1+z_2)(\overline{z_1}+\overline{z_2})\\
&=|z_1|^2+|z_2|^2+2|z_1||z_2|-(|z_1|^2+|z_2|^2+\overline{z_1}z_2+\overline{z_2}z_1)\\
&=2|z_1||z_2|-2Re(\overline{z_1}z_2)\\
&=2(|z_1||z_2|-|z_1z_2|)\\
&\geq 0
\end{aligned}
Suppose 2(|z_1||z_2|-|z_1z_2|)=0, and \overline{z_1}z_2 is a non-negative real number c, then |z_1||z_2|=|z_1z_2|...
What is the use of this?
Let \arg(z)=\theta\in (-\pi,\pi], z_1=r_1(\cos\theta_1+i\sin\theta_1), z_2=r_2(\cos\theta_2+i\sin\theta_2).
z_1z_2=r_1r_2[cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)]
(Define \text{cis}(\theta)=\cos\theta+i\sin\theta)
Theorem 1.6 Parallelogram Equality
The sum of the squares of the lengths of the diagonals of a parallelogram equals the sum of the squares of the lengths of the sides.
Proof
Let z_1,z_2 be two complex numbers representing the two sides of the parallelogram, then the sum of the squares of the lengths of the diagonals of the parallelogram is |z_1-z_2|^2+|z_1+z_2|^2, and the sum of the squares of the lengths of the sides is 2|z_1|^2+2|z_2|^2.
\begin{aligned}
|z_1-z_2|^2+|z_1+z_2|^2 &= (x_1-x_2)^2+(y_1-y_2)^2+(x_1+x_2)^2+(y_1+y_2)^2 \\
&= 2x_1^2+2x_2^2+2y_1^2+2y_2^2 \\
&= 2(|z_1|^2+|z_2|^2)
\end{aligned}
Definition 1.9
The argument of a complex number z is defined as the angle \theta between the positive real axis and the ray from the origin through z.
De Moivre's Formula
Theorem 1.10 De Moivre's Formula
Let z=r\text{cis}(\theta), then
\forall n\in \mathbb{Z}:
z^n=r^n\text{cis}(n\theta)
Proof
For n=0, z^0=1=1\text{cis}(0).
For n=-1, z^{-1}=\frac{1}{z}=\frac{1}{r}\text{cis}(-\theta)=\frac{1}{r}(cos(-\theta)+i\sin(-\theta)).
Application:
\begin{aligned}
(\text{cis}(\theta))^3&=\text{cis}(3\theta)\\
&=\cos(3\theta)+i\sin(3\theta)\\
&=cos^3(\theta)-3cos(\theta)sin^2(\theta)+i(3cos^2(\theta)sin(\theta)-sin^3(\theta))\\
\end{aligned}