# Math416 Lecture 13

## Review on Cauchy's Theorem

Cauchy's Theorem states that if a function is holomorphic (complex differentiable) on a simply connected domain, then the integral of that function over any closed contour within that domain is zero.

Last lecture we proved the case for convex regions.

### Cauchy's Formula for a Circle

Let $C$ be a counterclockwise oriented circle, and let $f$ be a holomorphic

function defined in an open set containing $C$ and its interior. Then,

$$
f(z_0)=\frac{1}{2\pi i}\int_C\frac{f(z)}{z-z_0}dz
$$

for all points $z$ in the interior of $C$.

## New materials

### Mean value property

#### Theorem 7.6: Mean value property

Special case: Suppose $f$ is holomorphic on some $\mathbb{D}(z_0,R)\subset \mathbb{C}$, by cauchy's formula,

$$
f(z_0)=\frac{1}{2\pi i}\int_{C_r}\frac{f(z)}{z-z_0}dz
$$

Parameterizing $C_r$, we get $\gamma(t)=z_0+re^{it}$, $0\leq t\leq 2\pi$

$$
\int f(z)dz=\int f(\gamma) \gamma'(t) d t
$$

So,

$$
f(z_0)=\frac{1}{2\pi i}\int_0^{2\pi}\frac{f(z_0+re^{it})}{re^{it}} ire^{it} dt=\frac{1}{2\pi}\int_{0}^{2\pi} f(z_0+re^{it}) dt
$$

This concludes the mean value property for the holomorphic function

If $f$ is holomorphic, $f(z_0)$ is the mean value of $f$ on any circle centered at $z_0$

#### Area representation of mean value property

Area of $f$ on $\mathbb{D}(z_0,r)$

$$
\frac{1}{\pi r^2}\int_{0}^{2\pi}\int_0^r f(z_0+re^{it})
$$

/*Track lost*/

### Cauchy Integral

#### Definition 7.7

Let $\gamma:[a,b]\to \mathbb{C}$ be piecewise $\mathbb{C}^1$, let $\phi$ be condition on $\gamma$. Then the Cauchy interval of $\phi$ along $\gamma$ is 

$$
F(z)=\int_{\gamma}\frac{\phi(\zeta)}{\zeta-z}d \zeta
$$

#### Theorem

Suppose $F(z)=\int_{\gamma}\frac{\phi(z)}{\zeta-z}d z$. Then $F$ has a local power series representation at all points $z_0$ not in $\gamma$.

Proof:

Let $R=B(z_0,\gamma)>0$, let $z\in \mathbb{D}(z_0,R)$

So

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

Since $|z-z_0|<R$ and $|\zeta-z_0|>R$, $|\frac{z-z_0}{\zeta-z_0}|<1$.

Converting it to geometric series

$$
\frac{1}{1-z_0}\frac{1}{1-\frac{z-z_0}{\zeta-z_0}}=\sum_{n=0}^\infty \left(\frac{z-z_0}{\zeta-z_0}\right)^n
$$

So,

$$
\begin{aligned}
F(z)&=\int_\gamma \frac{\phi(\zeta)}{\zeta - z} d\zeta\\
&=\int_\gamma \frac{\phi(\zeta)}{z-z_0} \sum_{n=0}^\infty \left(\frac{z-z_0}{\zeta-z_0}\right)^n dz\\
&=\sum_{n=0}^\infty (z-z_0)^n \int_\gamma \frac{\phi(\zeta)}{(\zeta-z_0)^{n+1}} 
\end{aligned}
$$

Which gives us an power series representation if we set $a_n=\int_\gamma \frac{\phi(\zeta)}{(\zeta-z_0)^{n+1}}$

QED

#### Corollary 7.7

Suppose $F(z)=\int_\gamma \frac{\phi(\zeta)}{\zeta-z_0} dz$,

Then,

$$
f^{(n)}(z)=n!\int_\gamma \frac{\phi(z)}{(\zeta-z_0)^{n+1}} d\zeta
$$

where $z\in \mathbb{C}\setminus \gamma$.

Combine with Cauchy integral formula:

If $f$ is in $O(\Omega)$, then $\forall z\in \mathbb{D}(z_0,r)$.

$$
f(z)=\frac{1}{2\pi i}\int_{C_r}\frac{f(\zeta)}{\zeta-z} d\zeta
$$

We have proved that If $f\in O(\Omega)$, then $f$ is locally given  by a convergent power series

power series has radius of convergence at $z_0$ that is $\geq$ dist($z_0$,boundary $\Omega$)

### Liouville's Theorem

#### Definition 7.11

A function that is holomorphic in all of $\mathbb{C}$ is called an entire function.

#### Theorem 7.11 Liouville's Theorem

Any bounded entire function is constant.

> Basic Estimate of integral
>
> $$\left|\int_\gamma f(z) dz\right|\leq L(\gamma) \max\left|f(z)\right|$$

Since,

$$
f'(z)=\frac{1}{2\pi i} \int_{C_r} \frac{f(z)}{(\zeta-z)^2} dz
$$

So the modulus of the integral is bounded by

$$
\frac{1}{2\pi} |M|\cdot \frac{1}{R^2}=2\pi R\cdot M \frac{1}{R^2}=\frac{M}{R}
$$

### Fundamental Theorem of Algebra

#### Theorem 7.12 Fundamental Theorem of ALgebra

Every nonconstant polynomial with complex coefficients can be factored over
$\mathbb{C}$ into linear factors.

#### Corollary

For every polynomial with complex coefficients.

$$
p(z)=c\prod_{j=i}^n(z-z_0)^{t_j}
$$

where the degree of polynomial is $\sum_{j=0}^n t_j$

Proof:

Let $p(z)=a_0+a_1z+\cdots+a_nz^n$, where $a_n\neq 0$ and $n\geq 1$.

So

$$
|p(z)|=|a_nz_n|\left[\left|1+\frac{a_{n-1}}{a_nz}+\cdots+\frac{a_0}{a_nz^n}\right|\right]
$$

If $|z|\geq R$, $\left|1+\frac{a_{n-1}}{a_nz}+\cdots+\frac{a_0}{a_nz^n}\right|<\frac{1}{2}$