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Math416Complex Variables (Lecture 1)

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,⋅)(G,\cdot) such that: for all a,b,c∈Ga,b,c\in G:

  1. a⋅b∈Ga\cdot b\in G
  2. a⋅(b⋅c)=(a⋅b)⋅ca\cdot (b\cdot c)=(a\cdot b)\cdot c
  3. a⋅1=aa\cdot 1=a
  4. a⋅a−1=1a\cdot a^{-1}=1

Ring

A group with two operations: addition and multiplication. (R,+,⋅)(R,+,\cdot) such that: for all a,b,c∈Ra,b,c\in R:

  1. Commutative under addition: a+b=b+aa+b=b+a
  2. Associative under multiplication: (a⋅b)⋅c=a⋅(b⋅c)(a\cdot b)\cdot c=a\cdot (b\cdot c)
  3. Distributive under addition: a⋅(b+c)=a⋅b+a⋅ca\cdot (b+c)=a\cdot b+a\cdot c

Example:

{a+6b∣a,b∈Z}\{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 C\mathbb{C} of ordered pairs (x,y)(x,y) with x,y∈Rx,y\in \mathbb{R} and the operations:

  • Addition: (x1,y1)+(x2,y2)=(x1+x2,y1+y2)(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2)
  • Multiplication: (x1,y1)(x2,y2)=(x1x2−y1y2,x1y2+x2y1)(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 C\mathbb{C} satisfies the following conditions (The field axioms):

For all z1,z2,z3∈Cz_1,z_2,z_3\in \mathbb{C}:

  1. z1+z2=z2+z1z_1+z_2=z_2+z_1 (commutative law of addition)
  2. (z1+z2)+z3=z1+(z2+z3)(z_1+z_2)+z_3=z_1+(z_2+z_3) (associative law of addition)
  3. z1⋅z2=z2⋅z1z_1\cdot z_2=z_2\cdot z_1 (commutative law of multiplication)
  4. (z1⋅z2)⋅z3=z1⋅(z2⋅z3)(z_1\cdot z_2)\cdot z_3=z_1\cdot (z_2\cdot z_3) (associative law of multiplication)
  5. z1⋅(z2+z3)=z1⋅z2+z1⋅z3z_1\cdot (z_2+z_3)=z_1\cdot z_2+z_1\cdot z_3 (distributive law)
  6. There exists an additive identity element 0=(0,0)0=(0,0) such that z+0=zz+0=z for all z∈Cz\in \mathbb{C}.
  7. There exists a multiplicative identity element 1=(1,0)1=(1,0) such that z⋅1=zz\cdot 1=z for all z∈Cz\in \mathbb{C}.
  8. There exists an additive inverse −z=(−x,−y)-z=(-x,-y) for all z=(x,y)∈Cz=(x,y)\in \mathbb{C} such that z+(−z)=0z+(-z)=0.
  9. There exists a multiplicative inverse z−1=(xx2+y2,−yx2+y2)z^{-1}=\left(\frac{x}{x^2+y^2},-\frac{y}{x^2+y^2}\right) for all z=(x,y)∈Cz=(x,y)\in \mathbb{C} such that z⋅z−1=1z\cdot z^{-1}=1.
(a,b)−1=(aa2+b2,−ba2+b2)(a,b)^{-1}=\left(\frac{a}{a^2+b^2},-\frac{b}{a^2+b^2}\right)

Embedding of R\mathbb{R} in C\mathbb{C} 1.3

Let z=x+iy∈Cz=x+iy\in \mathbb{C} where a,b∈Ra,b\in \mathbb{R}.

  • xx is called the real part of zz and
  • yy is called the imaginary part of zz.
  • ∣z∣=x2+y2|z|=\sqrt{x^2+y^2} is called the absolute value or modulus of zz.
  • The angle between the positive real axis and the line segment from 00 to zz is called the argument of zz and is denoted by θ\theta (argument of zz).
  • z‾=x−iy\overline{z}=x-iy is called the conjugate of zz. (z⋅z‾=∣z∣2z\cdot \overline{z}=|z|^2)
  • z1+z2=(x1+x2,y1+y2)z_1+z_2=(x_1+x_2,y_1+y_2) (vector addition)

Lemma 1.3

∣z1z2∣=∣z1∣∣z2∣|z_1z_2|=|z_1||z_2|

Theorem 1.5 (Triangle Inequality)

∣z1+z2∣≤∣z1∣+∣z2∣|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,

(∣z1+z2∣)2−∣z1+z2∣2=∣z1+z2∣2−2∣z1+z2∣−(z1+z2)(z1‾+z2‾)=∣z1∣2+∣z2∣2+2∣z1∣∣z2∣−(∣z1∣2+∣z2∣2+z1‾z2+z2‾z1)=2∣z1∣∣z2∣−2Re(z1‾z2)=2(∣z1∣∣z2∣−∣z1z2∣)≥0\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(∣z1∣∣z2∣−∣z1z2∣)=02(|z_1||z_2|-|z_1z_2|)=0, and z1‾z2\overline{z_1}z_2 is a non-negative real number cc, then ∣z1∣∣z2∣=∣z1z2∣|z_1||z_2|=|z_1z_2|…

What is the use of this?

Let arg⁡(z)=θ∈(−π,π]\arg(z)=\theta\in (-\pi,\pi], z1=r1(cos⁡θ1+isin⁡θ1)z_1=r_1(\cos\theta_1+i\sin\theta_1), z2=r2(cos⁡θ2+isin⁡θ2)z_2=r_2(\cos\theta_2+i\sin\theta_2).

z1z2=r1r2[cos(θ1+θ2)+isin⁡(θ1+θ2)]z_1z_2=r_1r_2[cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)]

(Define cis(θ)=cos⁡θ+isin⁡θ\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 z1,z2z_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 ∣z1−z2∣2+∣z1+z2∣2|z_1-z_2|^2+|z_1+z_2|^2, and the sum of the squares of the lengths of the sides is 2∣z1∣2+2∣z2∣22|z_1|^2+2|z_2|^2.

∣z1−z2∣2+∣z1+z2∣2=(x1−x2)2+(y1−y2)2+(x1+x2)2+(y1+y2)2=2x12+2x22+2y12+2y22=2(∣z1∣2+∣z2∣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 zz is defined as the angle θ\theta between the positive real axis and the ray from the origin through zz.

De Moivre’s Formula

Theorem 1.10 De Moivre’s Formula

Let z=rcis(θ)z=r\text{cis}(\theta), then

∀n∈Z\forall n\in \mathbb{Z}:

zn=rncis(nθ)z^n=r^n\text{cis}(n\theta)

Proof

For n=0n=0, z0=1=1cis(0)z^0=1=1\text{cis}(0).

For n=−1n=-1, z−1=1z=1rcis(−θ)=1r(cos(−θ)+isin⁡(−θ))z^{-1}=\frac{1}{z}=\frac{1}{r}\text{cis}(-\theta)=\frac{1}{r}(cos(-\theta)+i\sin(-\theta)).

Application:

(cis(θ))3=cis(3θ)=cos⁡(3θ)+isin⁡(3θ)=cos3(θ)−3cos(θ)sin2(θ)+i(3cos2(θ)sin(θ)−sin3(θ))\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}
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