Math416 Lecture 7

Review

Exponential function

e^z=e^{x+iy}=e^x(\cos y+i\sin y)

Logarithm Reviews

Definition 4.9 Logarithm

A logarithm of $a$ is any $b$ such that $e^b=a$ .

Branch of Logarithm

A branch of logarithm is a continuous function $f$ on a domain $D$ such that $e^{f(z)}=\exp(f(z))=z$ for all $z\in D$ .

Continue on Chapter 4 Elementary functions

Logarithm

Theorem 4.11

$\log(z)$ is holomorphic on $\mathbb{C}\setminus\{0\}$ .

Proof

We proved that $\frac{\partial}{\partial\overline{z}}e^{z}=0$ on $\mathbb{C}\setminus\{0\}$ .

Then $\frac{d}{dz}e^{z}=\frac{\partial}{\partial x}e^{z}=0$ if we know that $e^{z}$ is holomorphic.

Since $\frac{d}{dz}e^{z}=e^{z}$ , we know that $e^{z}$ is conformal, so any branch of logarithm is also conformal.

Since $\exp(\log(z))=z$ , we know that $\log(z)$ is the inverse of $\exp(z)$ , so $\frac{d}{dz}\log(z)=\frac{1}{e^{\log(z)}}=\frac{1}{z}$ .

We call $\frac{f'}{f}$ the logarithmic derivative of $f$ .

Definition 4.16

I don’t know if this material is covered or not, so I will add it here to prevent confusion for future readers

If $a$ and $c$ are complex numbers, with $a\neq 0$ , then by the values of $a^c$ one means the value of $e^{c\log a}$ .

For example, $1^i=e^{i (2\pi n i)}$

If you accidentally continue on this section and find it interesting, you will find Riemann zeta function

z(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}

And analytic continuation for such function for number less than or equal to $1$ .

And perhaps find trivial zeros for negative integers on real line. It is important to note that the Riemann zeta function has non-trivial zeros, which are located in the critical strip where the real part of $s$ is between 0 and 1. The famous Riemann Hypothesis conjectures that all non-trivial zeros lie on the critical line where the real part of $s$ is $\frac{1}{2}$ .

Chapter 5. Power series

Convergence

Necessary Condition for Convergence

If $\sum_{n=0}^{\infty}c_n$ converges, then $\lim_{n\to\infty}c_n=0$ exists.

Geometric series

Let $c$ be a complex number

\sum_{n=0}^{N}c^n=\frac{1-c^{N+1}}{1-c}

If $|c|<1$ , then $\lim_{N\to\infty}\sum_{n=0}^{N}c^n=\frac{1}{1-c}$ .

otherwise, the series diverges.

Proof

The geometric series converges if $\frac{c^{N+1}}{1-c}$ converges.

(1-c)(1+c+c^2+\cdots+c^N)=1-c^{N+1}

If $|c|<1$ , then $\lim_{N\to\infty}c^{N+1}=0$ , so $\lim_{N\to\infty}(1-c)(1+c+c^2+\cdots+c^N)=1$ .

If $|c|\geq 1$ , then $c^{N+1}$ does not converge to 0, so the series diverges.

Theorem 5.4 (Triangle Inequality for Series)

If the series $\sum_{n=0}^{\infty}c_n$ converges, then $\left|\sum_{n=0}^\infty c_n\right|\leq \sum_{n=0}^{\infty}|c_n|$ .

Definition 5.5

\sum_{n=0}^{\infty}c_n

converges absolutely if $\sum_{n=0}^{\infty}|c_n|$ converges.

Note: Some other properties of converging series covered in Math4111, bad, very bad.

Definition 5.6 Convergence of sequence of functions

A sequence of functions $f_n$ converges pointwise to $f$ on a set $G$ if for every $z\in G$ , $\forall\epsilon>0$ , $\exists N$ such that for all $n\geq N$ , $|f_n(z)-f(z)|<\epsilon$ .

(choose $N$ based on $z$ )

A sequence of functions $f_n$ converges uniformly to $f$ on a set $G$ if for every $\epsilon>0$ , there exists a positive integer $N$ such that for all $n\geq N$ and all $z\in G$ , $|f_n(z)-f(z)|<\epsilon$ .

(choose $N$ based on $\epsilon$ )

A sequence of functions $f_n$ converges locally uniformly to $f$ on a set $G$ if for every $z\in G$ , $\forall\epsilon>0$ , $\exists r>0$ such that for all $z\in B(z,r)$ , $\forall n\geq N$ , $|f_n(z)-f(z)|<\epsilon$ .

(choose $N$ based on $z$ and $\epsilon$ )

A sequence of functions $f_n$ converges uniformly on compacta to $f$ on a set $G$ if it converges uniformly on every compact subset of $G$ .

Theorem 5.7

If the subsequence (or partial sum) of a converging sequence of functions converges (a), then the original sequence converges (a).

The N-th partial sum of the series $\sum_{n=0}^\infty f_n$ is $\sum_{n=0}^{N}f_n$

You can replace (a) with locally uniform convergence, uniform convergence, pointwise convergence, etc.

Corollary from definition of $a^b$ in complex plane

We defined $a^b=\{e^{b\log a}\}$ if $b$ is real, then $a^b$ is unique, if $b$ is complex, then $a^b=e^{b\log a}\{e^{2k\pi ik b}\},k\in\mathbb{Z}$ .

Power series

Definition 5.8

A power series is a series of the form $\sum_{n=0}^{\infty}c_n(z-z_0)^n$ .

Definition 5.9 Region of Convergence

For every power series, there exists a radius of convergence $r$ such that the series converges absolutely and locally uniformly on $B_r(z_0)$ .

And it diverges pointwise outside $B_r(z_0)$ .

Proof

Without loss of generality, we can assume that $z_0=0$ .

Suppose that the power series is $\sum_{n=0}^{\infty}c_n (z)^n$ converges at $z=re^{i\theta}$ .

We want to show that the series converges absolutely and uniformly on $\overline{B_r(0)}$ (closed disk, I prefer to use this notation, although they use $\mathbb{D}$ for the disk (open disk)).

We know $c_n r^ne^{in\theta}\to 0$ as $n\to\infty$ .

So there exists $M\geq|c_n r^ne^{in\theta}|$ for all $n\in\mathbb{N}$ .

So $\forall z\in\overline{B_r(0)}$ , $|c_nz^n|\leq |c_n| |z|^n \leq M \left(\frac{|z|}{r}\right)^n$ .

So $\sum_{n=0}^{\infty}|c_nz^n|$ converges absolutely.

So the series converges absolutely and uniformly on $\overline{B_r(0)}$ .

If $|z| > r$ , then $|c_n z^n|$ does not tend to zero, and the series diverges.

We denote this $r$ captialized by te radius of convergence

Possible Cases for the Convergence of Power Series

Convergence Only at $z = 0$ :
- Proof: If the power series $\sum_{n=0}^{\infty} c_n (z - z_0)^n$ converges only at $z = 0$ , it means that the radius of convergence $R = 0$ . This occurs when the terms $c_n (z - z_0)^n$ do not tend to zero for any $z \neq 0$ . The series diverges for all $z \neq 0$ because the terms grow without bound.
Convergence Everywhere:
- Proof: If the power series converges for all $z \in \mathbb{C}$ , the radius of convergence $R = \infty$ . This implies that the terms $c_n (z - z_0)^n$ tend to zero for all $z$ . This can happen if the coefficients $c_n$ decrease rapidly enough, such as in the exponential series.
Convergence Within a Finite Radius:
- Proof: For a power series with a finite radius of convergence $R$ , the series converges absolutely and uniformly for $|z - z_0| < R$ and diverges for $|z - z_0| > R$ . On the boundary $|z - z_0| = R$ , the series may converge or diverge depending on the specific series. This is determined by the behavior of the terms on the boundary.