# Math416 Lecture 10

## Fast reload on Power Series

Suppose $\sum_{n=0}^\infty a_n$ converges absolutely. ($\sum_{n=0}^\infty |a_n|<\infty$)

Then rearranging the terms of the series does not affect the sum of the series.

For any permutation $\sigma$ of the set of positive integers, $\sum_{n=0}^\infty a_{\sigma(n)}=\sum_{n=0}^\infty a_n$.

Proof:

Let $\epsilon>0$, then $\exists N\in\mathbb{N}$ such that $\forall n\geq N$,

$$
\sum_{n=N}^\infty |a_n|<\epsilon
$$

So there exists $N_0$ such that if $M\geq N_0$, then

$$
\sum_{n=N_0}^M |a_n|<\epsilon
$$

_for any first $M$ terms of $\sigma$, we choose $N_0$ such that all the terms (no overlapping with the first $M$ terms) on the tail is less than $\epsilon$_.

$$
\sum_{n=1}^{\infty} a_n=\sum_{n=1}^{M} a_n+\sum_{n=M+1}^\infty a_n
$$

Let $K>N$, $L>N_0$,

$$
\left|\sum_{n=1}^{K}a_n-\sum_{n=1}^{L}a_{\sigma(n)}\right|<2\epsilon
$$

QED

## Chapter 4 Complex Integration

### Complex Integral

#### Definition 6.1

If $\phi(t)$ is a complex function defined on $[a,b]$, then the integral of $\phi(t)$ over $[a,b]$ is defined as

$$
\int_a^b \phi(t) dt = \int_a^b \text{Re}\{\phi(t)\} dt + i\int_a^b \text{Im}\{\phi(t)\} dt
$$

#### Theorem 6.3 (Triangle Inequality)

If $\phi(t)$ is a complex function defined on $[a,b]$, then

$$
\left|\int_a^b \phi(t) dt\right| \leq \int_a^b |\phi(t)| dt
$$

Proof:

Let $\lambda(t)=\frac{\left|\int_a^t \phi(t) dt\right|}{\int_a^t |\phi(t)| dt}$, then $\left|\lambda(t)\right|=1$.

$$
\begin{aligned}
\left|\int_a^b \phi(t) dt\right|&=\lambda\int_a^b \phi(t) dt\\
&=\int_a^b \lambda(t)\phi(t) dt\\
&=\text{Re} \{\int_a^b \lambda(t)\phi(t) dt\}\\
&\leq\int_a^b |\lambda(t)\phi(t)| dt\\
&=\int_a^b |\phi(t)| dt
\end{aligned}
$$

Assume $\phi$ is continuous on $[a,b]$, the equality means $\lambda(t)\phi(t)$ is real and positive everywhere on $[a,b]$, which means $\arg \phi(t)$ is constant.

QED

#### Definition 6.4 Arc Length

Let $\gamma$ be a curve in the complex plane defined by $\gamma(t)=x(t)+iy(t)$, $t\in[a,b]$. The arc length of $\gamma$ is given by

$$
\Gamma=\int_a^b |\gamma'(t)| dt=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} dt
$$

N.B. If $\int_{\Gamma} f(z) dz$ depends on orientation of $\Gamma$, but not the parametrization.

We define

$$
\int_{\Gamma} f(z) dz=\int_{\Gamma} f(\gamma(t))\gamma'(t) dt
$$

Example:

Suppose $\Gamma$ is the circle centered at $z_0$ with radius $R$

$$
\int_{\Gamma} \frac{1}{z-z_0} dz
$$

Parameterize the unit circle:

$$
\gamma(t)=z_0+Re^{it}\quad
\gamma'(t)=iRe^{it}, t\in[0,2\pi]
$$

$$
f(z)=\frac{1}{z-z_0}
$$

$$
f(\gamma(t))=\frac{1}{(z_0+Re^{it})-z_0}
$$

$$
\int_{\Gamma} f(z) dz=\int_0^{2\pi} f(\gamma(t))\gamma'(t) dt=\int_0^{2\pi} \frac{1}{Re^{-it}}iRe^{it} dt=2\pi i
$$

#### Theorem 6.11 (Uniform Convergence)

If $f_n(z)$ converges uniformly to $f(z)$ on $\Gamma$, assume length of $\Gamma$ is finite, then

$$
\lim_{n\to\infty} \int_{\Gamma} f_n(z) dz = \int_{\Gamma} f(z) dz
$$

Proof:

Let $\epsilon>0$, since $f_n(z)$ converges uniformly to $f(z)$ on $\Gamma$, there exists $N\in\mathbb{N}$ such that for all $n\geq N$,

$$
\left|f_n(z)-f(z)\right|<\epsilon
$$

$$
\begin{aligned}
\left|\int_{\Gamma} f_n(z) dz - \int_{\Gamma} f(z) dz\right|&=\left|\int_{\Gamma} (f_n(\gamma(t))-f(\gamma(t)))\gamma'(t) dt\right|\\
&\leq \int_{\Gamma} |f_n(\gamma(t))-f(\gamma(t))||\gamma'(t)| dt\\
&\leq \int_{\Gamma} \epsilon|\gamma'(t)| dt\\
&=\epsilon\text{length}(\Gamma)
\end{aligned}
$$

QED

#### Theorem 6.6 (Integral of derivative)

Suppose $\Gamma$ is a closed curve, $\gamma:[a,b]\to\mathbb{C}$ and $\gamma(a)=\gamma(b)$.

$$
\begin{aligned}
\int_{\Gamma} f'(z) dz &= \int_a^b f'(\gamma(t))\gamma'(t) dt\\
&=\int_a^b \frac{d}{dt}f(\gamma(t)) dt\\
&=f(\gamma(b))-f(\gamma(a))\\
&=0
\end{aligned}
$$

QED

Example:

Let $R$ be a rectangle $\{-a,a,ai+b,ai-b\}$, $\Gamma$ is the boundary of $R$ with positive orientation.

Let $\int_{R} e^{-z^2}dz$.

Is $e^{-z^2}=\frac{d}{dz}f(z)$?

Yes, since

$$
e^{z^2}=1-\frac{z^2}{1!}+\frac{z^4}{2!}-\frac{z^6}{3!}+\cdots=\frac{d}{dz}\left(\frac{z}{1!}-\frac{1}{3}\frac{z^3}{2!}+\frac{1}{5}\frac{z^5}{3!}-\cdots\right)
$$

This is polynomial, therefore holomorphic.

So

$$
\int_{R} e^{z^2}dz = 0
$$

with some limit calculation, we can get

<!--TODO: Fill the parts-->

$$
\int_{R} e^{-z^2}dz = 2\pi i
$$