# Math 416 Lecture 12

## Continue on last class

### Cauchy's Theorem on triangles

Let $T$ be a triangle in $\mathbb{C}$ and $f$ be holomorphic on $T$. Then

$$
\int_T f(\zeta) d\zeta = 0
$$

### Cauchy's Theorem for Convex Sets

Let's start with a simple case: $f(\zeta)=1$.

For any closed curve $\gamma$ in $U$, we have

$$
\int_\gamma f(\zeta) d\zeta = \int_a^b f(\gamma(t)) \gamma'(t) dt \approx \sum_{i=1}^n f(\gamma(t_i)) \gamma'(t_i) \Delta t_i
$$

#### Definition of a convex set

A set $U$ is convex if for any two points $\zeta_1, \zeta_2 \in U$, the line segment $[\zeta_1, \zeta_2] \subset U$.

Let $O(U)$ be the set of all holomorphic functions on $U$.

#### Definition of primitive

Say $f$ has a primitive on $U$. If there exists a holomorphic function $g$ on $U$ such that $g'(\zeta)=f(\zeta)$ for all $\zeta \in U$, then $g$ is called a primitive of $f$ on $U$.

#### Cauchy's Theorem for a Convex region

Cauchy's Theorem holds if $f$ has a primitive on a convex region $U$.

$$
\int_\gamma f(\zeta) d\zeta = \int_\gamma \left[\frac{d}{d\zeta}g(\zeta)\right] d\zeta = g(\zeta_1)-g(\zeta_2)
$$

Since the curve is closed, $\zeta_1=\zeta_2$, so $\int_\gamma f(\zeta) d\zeta = 0$.

Proof:

It is sufficient to prove that if $U$ is convex, $f$ is holomorphic on $U$, then $f=g'$ for some $g$ holomorphic on $U$.

We pick a point $z_0\in U$ and define $g(\zeta)=\int_{[\zeta_0,\zeta]}f(\xi)d\xi$.

We claim $g\in O(U)$ and $g'=f$.

Let $\zeta_1$ close to $\zeta$, since $f$ is holomorphic on $U$, using the Goursat's theorem, we can find a triangle $T$ with $\xi\in T$ and $\zeta\in T$ and $T\subset U$.

$$
\begin{aligned}
0&=\int_{\zeta_0}^{\zeta}f(\xi)d\xi+\int_{\zeta}^{\zeta_1}f(\xi)d\xi\\
&=g(\zeta)-g(\zeta_1)+\int_{\zeta}^{\zeta_1}f(\xi)d\xi+\int_{\zeta_1}^{\zeta_0}f(\xi)d\xi\\
\frac{g(\zeta)-g(\zeta_1)}{\zeta-\zeta_1}&=-\frac{1}{\zeta-\zeta_1}\left(\int_{\zeta}^{\zeta_1}f(\xi)d\xi\right)\\
\frac{g(\zeta_1)-g(\zeta_0)}{\zeta_1-\zeta_0}-f(\zeta_1)&=-\frac{1}{\zeta_1-\zeta_0}\left(\int_{\zeta}^{\zeta_1}f(\xi)d\xi\right)-f(\zeta_1)\\
&=-\frac{1}{\zeta_1-\zeta_0}\left(\int_{\zeta}^{\zeta_1}f(\xi)-f(\zeta_1)d\xi\right)\\
&=I
\end{aligned}
$$

Use the fact that $f$ is holomorphic on $U$, then $f$ is continuous on $U$, so $\lim_{\zeta\to\zeta_1}f(\zeta)=f(\zeta_1)$.

There exists a $\delta>0$ such that $|\zeta-\zeta_1|<\delta$ implies $|f(\zeta)-f(\zeta_1)|<\epsilon$.

So

$$
|I|\leq\frac{1}{\zeta_1-\zeta_0}\int_{\zeta}^{\zeta_1}|f(\xi)-f(\zeta_1)|d\xi<\frac{\epsilon}{\zeta_1-\zeta_0}\int_{\zeta}^{\zeta_1}d\xi=\epsilon
$$

So $I\to 0$ as $\zeta_1\to\zeta$.

Therefore, $g'(\zeta_1)=f(\zeta_1)$ for all $\zeta_1\in U$.

EOP