@Chen Siyi

October 26, 2020

Note5 Complex Analysis II

Note5 Complex Analysis IIMore About Holomorphic FunctionsHolomorphic Functions are AnalyticUniqueness of Holomorphic FunctionsAnalytic ContinuationSingularities and PolesSingularitiesMultiplicity of PolesHow do you judge singularities?Representation Near PolesThe Principle PartThe ResidueHow do you find the residue?An Important Property of the ResidueResidue CalculusThe Residue TheoremThe Complex Logarithm, Power, and RootComplex LogarithmComplex PowersComplex RootsResidue Calculus Overview: Evaluation of Real IntegralsReview: Jordan’s LemmaSemicircle and Indented SemicircleCircle with KeyholesMultiple KeholeSquareResidue Calculus for Functions with Branch pointsThe Heaviside Operator MethodThe Heaviside FunctionThe Heaviside Operator Method(Basic)The Heaviside OperatorApply to Solve the Initial Value ProblemThe Heaviside Operator Method(More General)The General IdeaThe Explicit Transformation

More About Holomorphic Functions

Holomorphic Functions are Analytic

$f$ holomorphic $Ω$ $D$ $z_0$ $Ω$ $z_0$

f(z)=\sum_{n=0}^{\infty} a_{n}\left(z-z_{0}\right)^{n}

$z ∈ D$ and the coefficients are given by

a_{n}=\frac{f^{(n)}\left(z_{0}\right)}{n !}, \quad n \in \mathbb{N}

Uniqueness of Holomorphic Functions

$Ω ⊂ \mathbb{C}$ $f,g : Ω → \mathbb{C}$ $S ⊂ Ω$ $Ω$ and that

f(z)=g(z) \quad \quad \text { for all } z \in S

$f(z)=g(z)$ $z ∈Ω$ .

Analytic Continuation

$M ⊂ \mathbb{C}$ $f : M → \mathbb{C}$ $Ω$ $M ⊂ Ω$ $g : Ω → \mathbb{C}$ holomorphic $g(z) = f (z)$ $z ∈ M$ analytic continuation $f$ $Ω$ .

Singularities and Poles

Singularities

$Ω ⊂ \mathbb{C}$ $z_0 ∈ Ω$ $f : Ω$ ${z_0} → \mathbb{C}$ $f$ isolated singularity $z_0$ ).

removable $\tilde{f} : Ω → \mathbb{C}$ .
pole $g = \frac{1}{f}$ $\Omega$ $z_0$ $z_0$ $g(z_0)=0$ .
The singularity is said to be essential if it is neither removable nor a pole.

Multiplicity of Poles

$f:Ω→\mathbb{C}$ $z_0 ∈Ω$ $U$ holomorphic function $h$ $n$ such that

f(z)=\left(z-z_{0}\right)^{-n} h(z) \quad \text { for all } z \in U

$n$ is called the multiplicityorder $f$ $n=1$ , we say that the pole is simple.

How do you judge singularities?

$a$ removable singularity $f(z)$ $\lim _{z \rightarrow a} f(z) = c_0$ $c_0 \neq \infin$ .
Question1
$f(z) = \frac{cosz - 1}{z^2}$ $z_0 = 0$ $z_0$ ?
Answer1

$z_0$ pole with order n $f(z)$ $z_0$ $g = \frac{1}{f}$ $n$ $c_0 \neq \infin$ .
Question2
$f(z) = \frac{1}{cosz - 1}$ $f$ . What kind of singularity are them?
Answer2

Representation Near Poles

$f:Ω→\mathbb{C}$ $n$ $z_0 ∈Ω$ exists a neighborhood $U ⊂ Ω$ $z_0$ $a_{−n}$ $a_{−1}$ $∈ \mathbb{C}$ holomorphic $G : U → \mathbb{C}$ such that

f(z)=\frac{a_{-n}}{\left(z-z_{0}\right)^{n}}+\frac{a_{-n+1}}{\left(z-z_{0}\right)^{n-1}}+\cdots+\frac{a_{-1}}{z-z_{0}}+G(z)

$z ∈ U$ .

Notice
neiborhood $z_0$ $f(z)(z-z_0)^n$ neiborhood $z_0$ .
But you cannot contain other poles inside such neiborhood, for G to be a analytic function.

The Principle Part

P(z)=\frac{a_{-n}}{\left(z-z_{0}\right)^{n}}+\frac{a_{-n+1}}{\left(z-z_{0}\right)^{n-1}}+\cdots+\frac{a_{-1}}{z-z_{0}}

$f$ $z_0$ .

The Residue

$a_{−1}$ $1$ $(z − z_0)$ $z_0$ , written

\operatorname{res}_{z_{0}} f=a_{-1}

Notice
One pole has exactly one residue.
$x$ $x$ residues accordingly.

How do you find the residue?

By direct expansion
By apply the theorem:
For a simple pole,
$\operatorname{res}_{z_{0}} f=\lim _{z \rightarrow z_{0}}\left(z-z_{0}\right) f(z)$
For a $n$ ,
$\operatorname{res}_{z_{0}} f=\frac{1}{(n-1) !} \lim _{z \rightarrow z_{0}} \frac{d^{n-1}}{d z^{n-1}}\left(\left(z-z_{0}\right)^{n} f(z)\right)$
$n = 2$ ,
$\operatorname{res}_{z_{0}} f=\lim _{z \rightarrow z_{0}}[\left(z-z_{0}\right)^2 f(z)] \prime$

Why does the theorem work?

Question
Find the residue of
$f(z)=\frac{1}{z\left(z^{2}+1\right)(z-2)^{2}}$

Answer

An Important Property of the Residue

$z^n$ $n = -1$ .

\oint_{S^{1}} \frac{d z}{z}=\int_{0}^{2 \pi} \frac{i e^{i t}}{e^{i t}} d t=2 \pi i \neq 0

\begin{aligned} \oint_{S^{1}} \frac{d z}{z^{n}} &=\int_{0}^{2 \pi} \frac{i e^{i t}}{e^{n i t}} d t=i \int_{0}^{2 \pi} e^{(1-n) i t} d t \\ &=i \int_{0}^{2 \pi}(\cos ((n-1) t)-i \sin ((n-1) t)) d t=0 \end{aligned}

A Tricky Question
Previously, we discuss about several theorems related to "whether a holomorphic function has primitives". And the most general one, Cauchy's Integral Theorem* says:
$U$ $\mathbb {C}$ simply connected $f:U\to \mathbb {C}$ holomorphic $\mathcal{C}$ $U$ , the integral vanishes to 0.
$z^n$ $n < 0$ ,... is there a contradiction to this theorem?

$\frac{1}{z - z_0}$ is a special case, and now we see:

contour $\mathcal{C}$ only the pole $z_0$ , by Cauthy's integral formula:

\frac{1}{2 \pi i} \int_{\mathcal{C}} f(z) d z=0+\frac{1}{2 \pi i} \int_{\mathcal{C}} \frac{a_{-1}}{z-z_{0}} d z=a_{-1}

A Tricky Reminder
only $z_0$ ? What if you want to integrate along a contour containning several poles?

Residue Calculus

The Residue Theorem

$f$ holomorphic $\mathcal{C}$ except for poles $z_1$ $z_N$ $\mathcal{C}$ . Then

\int_{\mathcal{C}} f(z) d z=2 \pi i \sum_{k=1}^{N} \operatorname{res}_{z_{k}} f

Question4
How do we generalize the residue theorem from the simple case where only one pole is contained?

Answer4

How can we apply this theorem?
Calculate certain integral $\mathbb{R}$ .

A Tricky Question
How do you compare the residue theorem with Cauchy Integral Formulas, which we can also use to find integrals?
$f(z)=\frac{1}{2 \pi i} \oint_{C} \frac{f(\zeta)}{\zeta-z} d \zeta \quad \text { for all } z \in D$

Question5
Redo the integration of the below function using the Residue Theorem.
Where C is the circle with radius 3 and centered at the origin.
$\int_{c} \frac{1}{(z-2)(z-5)} d z$

Answer5

The Complex Logarithm, Power, and Root

Complex Logarithm

The idea to define the complex logarithm can be traced back to the Cauthy's Integral Theorem...

The principal branch of the logarithm:

\Omega=\mathbb{C} \backslash\{x \in \mathbb{R}: x \leq 0\}

in which case

\ln \left(r e^{i \varphi}\right)=\ln r+\varphi i \quad(r>0,-\pi<\varphi<\pi)

For brevity, we set:

\mathbb{R}_{-}^{0}:=\{x \in \mathbb{R}: x \leq 0\}, \quad \mathbb{R}_{+}^{0}:=\{x \in \mathbb{R}: x \geq 0\}

We can also define other branches of the logarithm. Often, a convenient choice (especially for residue calculus) is

\ln : \mathbb{C} \backslash \mathbb{R}_{+}^{0} \rightarrow \mathbb{C}, \quad \quad \ln z=\ln r+\varphi i

$r > 0$ $0 < φ < 2π$ .

$\ln \left(z_{1} z_{2}\right) \neq \ln z_{1}+\ln z_{2}$ .

Complex Powers

z^{\alpha}:=e^{\alpha \ln z}, \quad \alpha \in \mathbb{C}, z \in \mathbb{C} \backslash \mathbb{R}_{-}^{0}

Complex Roots

\sqrt[n]{z}:=z^{1 / n}

Residue Calculus

Overview: Evaluation of Real Integrals

Extend the real domain to complex domain
- If only containing x, always directly extend to z
- $e^{iz}$
$f(z)$
Decide the countour and the branch if needed
Calculate the residue for poles in the countour
Apply residue theorem (or Cauchy’s theorem)
Except for the integral part we need, solve or vanish other parts one by one
- You may need to use Jordan’s Lemma or do similar operation

Review: Jordan’s Lemma

$R_0 > 0$ $g: \mathbb{C} \backslash \overline{B_{R_{0}}(0)}→\mathbb{C}$ isholomorphic. Let

f(z)=e^{i a z} g(z), \quad \text { for some } a>0

Let

C_{R}=\left\{z \in \mathbb{C}: z=R \cdot e^{i \theta}, 0 \leq \theta \leq \pi\right\}

be a semi-circle segment in the upper half-plane and assume that

\sup _{0 \leq \theta \leq \pi}\left|g\left(R e^{i \theta}\right)\right| \stackrel{R \rightarrow \infty}{\longrightarrow} 0

Then

\lim _{R \rightarrow \infty} \int_{C_{R}} f(z) d z=0

Semicircle and Indented Semicircle

Screen Shot 2020-10-26 at 14.30.03

Can be used to solve (hw6):

\int_{0}^{\infty} \frac{\sin x}{x} d x, \int_{-\infty}^{\infty} \frac{x \sin x}{x^{2}+a^{2}} d x, \int_{0}^{\infty} \frac{x \sin x}{\left(x^{2}+4\right)^{2}} d x,

\int_{0}^{\infty} \frac{1-\cos x}{x^{2}} d x, \int_{-\infty}^{\infty} \frac{\cos x}{x^{2}+a^{2}} d x,

\int_{-\infty}^{\infty} \frac{d x}{1+x^{4}}, \int_{-\infty}^{\infty} \frac{d x}{\left(1+x^{2}\right)^{n+1}} d x, \dots

Question6
Show that
$\int_{-\infty}^{\infty} \frac{d x}{1+x^{2}}=\pi$

Answer6

Circle with Keyholes

Screen Shot 2020-10-26 at 14.58.21

$\sqrt{x}$ $lnx$ $\ln : \mathbb{C} \backslash \mathbb{R}_{+}^{0} \rightarrow \mathbb{C}, \quad \quad \ln z=\ln r+\varphi i$

Such as (hw6):

\int_{0}^{\infty} \frac{\sqrt{x}}{x^{2}+a^{2}} d x, \int_{0}^{\infty} \frac{\ln x}{x^{2}+a^{2}} d x

Multiple Kehole

Screen Shot 2020-10-26 at 14.46.46

Question7
Compute
$\int_{1}^{\infty} \frac{d x}{x \sqrt{x^{2}-1}}$

Answer7

Square

Screen Shot 2020-10-26 at 15.14.29

Question8
Show that
$\int_{-\infty}^{\infty} \frac{e^{a x}}{1+e^{x}} d x=\frac{\pi}{\sin \pi a}, \quad 0<a<1$

Answer8

Residue Calculus for Functions with Branch points

$P$ $Q$ $m$ $n$ $n ≥ m + 2$ $Q(x) \neq 0$ $x > 0$ $Q$ has a zero of order at most 1 at the origin and if

f(z)=\frac{z^{\alpha} P(z)}{Q(z)}, \quad 0<\alpha<1

then

\text { p.v. } \int_{0}^{\infty} \frac{x^{\alpha} P(x)}{Q(x)} d x=\frac{2 \pi i}{1-e^{2 \pi \alpha i}} \sum_{j=1}^{k} \operatorname{res}_{z_{j}} f

$z_1, \dots , z_k$ $P/Q$ .

The Heaviside Operator Method

The Heaviside Function

H: \mathbb{R} \rightarrow \mathbb{R}, \quad H(t)=\left\{\begin{array}{ll} 1 & t>0 \\ 0 & t \leq 0 \end{array}\right.

The Heaviside Operator Method(Basic)

This method first comes up without solid proof of its correctness, but just with an idea that maybe you can perform "differentiation" by "multiplying a number".

The Heaviside Operator

$d/dt$ $D$ .

D f(t)=f^{\prime}(t), \quad \frac{1}{D} f(t)=\int_{0}^{t} f(s) d s+f(0)

$t < 0$ $H(t)$ . In particular,

\frac{1}{D} H(t)=t

Apply to Solve the Initial Value Problem

But this basic method might fail in some cases. For example the final solution you get does not converge. So we want to consider the idea behind the Heaviside Operator Method more intuitively.

$D$ $\frac{1}{D}$ $H(t)$ .
$y(t) = g(D)\cdot p(t)$ .
$g(D)$ $D^a$ $\frac{1}{D^b}$
$y(t)$ $p(t)$ and suming up.

The Heaviside Operator Method(More General)

The General Idea

Screen Shot 2020-10-26 at 11.15.35

Understand the graph...
A Tricky Question
$D$ $D = \frac{d}{dt}$ which is defined previously?
A Tricky Question
What transformations, or operations do we still need to find explicitly in this graph?

transformation $f \mapsto\{f\}$ operation $D$ $f$ } to satisfy:

$\{\alpha f+\beta g\}=\alpha\{f\}+\beta\{g\}$
$D\{f\}(p)=\left\{f^{\prime}\right\}(p)+f(0)$

Then actually automatically the following properpies hold:

$D\{\alpha f+\beta g\}=\alpha D\{f\}+\beta D\{g\}$
$\{0\}=0$
$D\{1\}=\{0\}+1=1$
$D^{n}\left\{\frac{t^{n}}{n !}\right\}=\{1\}$

The Explicit Transformation

$f$ be an analytic function, so

\{f\}=\left\{\sum_{n=0}^{\infty} f^{(n)}(0) \frac{t^{n}}{n !}\right\}=\sum_{n=0}^{\infty} f^{(n)}(0)\left\{\frac{t^{n}}{n !}\right\}

$p^{n+1}\left\{\frac{t^{n}}{n !}\right\}=1$ (why?), further we have

\{f\}=\frac{1}{p} \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{p^{n}}

Recall the definition of the Euler gamma function

\Gamma(t)=\int_{0}^{\infty} z^{t-1} e^{-z} d z

$\Gamma(n+1)=n !$ . Then

\begin{aligned} \{f\}(p) &=\frac{1}{p} \sum_{n=0}^{\infty} \int_{0}^{\infty} z^{n} e^{-z} d z \frac{f^{(n)}(0)}{n ! p^{n}} \\ &=\frac{1}{p} \int_{0}^{\infty} \sum_{n=0}^{\infty}\left(\frac{z}{p}\right)^{n} \frac{f^{(n)}(0)}{n !} e^{-z} d z \\ &=\frac{1}{p} \int_{0}^{\infty} f(z / p) e^{-z} d z \\ &=\int_{0}^{\infty} f(z) e^{-p z} d z \end{aligned}

$D$ . From

\begin{aligned} \left\{f^{\prime}\right\}(p) &=\int_{0}^{\infty} f^{\prime}(z) e^{-p z} d z=\left.f(z) e^{-p z}\right|_{0} ^{\infty}+p \int_{0}^{\infty} f(z) e^{-p z} d z \\ &=-f(0)+p \cdot\{f\}(p) \end{aligned}

$D\{f\}(p)=\left\{f^{\prime}\right\}(p)+f(0)$ , we obtain

D\{f\}(p)=p \cdot\{f\}(p)

Here, a new possible method for you to solve differential equations has come up.
$f(x)$ $\{f\}(p)$ $\{f\}(p)$ ,
$\{f\}(p)$ ,
$\{f\}(p)$ $f(x)$ .
You will see more... （Laplace Transformation & Fourier Transformation）

Note5 Complex Analysis II

More About Holomorphic Functions

Holomorphic Functions are Analytic

Uniqueness of Holomorphic Functions

Analytic Continuation

Singularities and Poles

Singularities

Multiplicity of Poles

How do you judge singularities?

Question1

Let f(z) = \frac{cosz - 1}{z^2}, z_0 = 0. What kind of singularity is z_0?

Answer1

Question2

Let f(z) = \frac{1}{cosz - 1}, find all singularities of f. What kind of singularity are them?

Answer2

Representation Near Poles

The Principle Part

The Residue

How do you find the residue?

Question

Find the residue of

Answer

An Important Property of the Residue

Residue Calculus

The Residue Theorem

Question5

Redo the integration of the below function using the Residue Theorem.

Answer5

The Complex Logarithm, Power, and Root

Complex Logarithm

Complex Powers

Complex Roots

Residue Calculus

Overview: Evaluation of Real Integrals

Review: Jordan’s Lemma

Semicircle and Indented Semicircle

Question6

Answer6

Circle with Keyholes

Multiple Kehole

Question7

Answer7

Square

Question8

Answer8

Residue Calculus for Functions with Branch points

The Heaviside Operator Method

The Heaviside Function

The Heaviside Operator Method(Basic)

The Heaviside Operator

Apply to Solve the Initial Value Problem

The Heaviside Operator Method(More General)

The General Idea

The Explicit Transformation

$f(z) = \frac{cosz - 1}{z^2}$ $z_0 = 0$ $z_0$ ?

$f(z) = \frac{1}{cosz - 1}$ $f$ . What kind of singularity are them?