11 - Cauchy Integrals

Cauchy Integral Formula for the Winding Number lemma

Let $\gamma$ be a piecewise $C^1$ closed path
Let $z_0\in \mathbb{C}$ be a point not in the image of $\gamma$ then

Winding number $I(\gamma ,z_0)$ of $\gamma$ around $z_0$ is given by

$$I(\gamma ,z_0)=\frac{1}{2\pi i}{\int }_{\gamma }\frac{1}{z-z_0}dz$$

Proof

If $\gamma :[0,1]\to \mathbb{C}$ then write

$$\gamma (t)=z_0+r(t)e^{2\pi ia(t)}$$

where $r(t)=|\gamma (t)-z_0|>0$ is continuous and existence of $a(t)$ is guaranteed by Existence of a Continuous Argument

So

$$\begin{array}{rl}{\int }_{\gamma }\frac{1}{z-z_0}dz& ={\int }_0^1\frac{1}{r(t)e^{2\pi ia(t)}}(r^{\prime }(t)+r(t)\cdot 2\pi i{a}^{\prime }(t))e^{2\pi ia(t)}dt\\ & ={\int }_0^1\frac{r^{\prime }(t)}{r(t)}+2\pi i{a}^{\prime }(t)dt\\ & =[\log (r(t))+2\pi ia(t){]}_0^1\\ & =2\pi i(a(1)-a(0))\end{array}$$

as $r(1)=r(0)=|\gamma (0)-z_0|$

General Cauchy Integral Formula for the Winding Number

Let $U$ be an open set in $\mathbb{C}$
Let $\gamma :[0,1]\to U$ be a closed path

If $f(z)$ is a continuous function on ${\gamma }^{\ast }$ then

$$I_f(\gamma ,w)=\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(z)}{z-w}dz$$

is analytic in $w$

Particularly if $f(z)=1$ then $w\mapsto I(\gamma ,w)$ is a continuous function on $\mathbb{C}\setminus {\gamma }^{\ast }$
Since it is integer-valued then it is constant on connected components of $\mathbb{C}\setminus {\gamma }^{\ast }$

Proof

By translation, assume $z_0=0$

Since $\mathbb{C}\setminus {\gamma }^{\ast }$ is open, there exists some $r>0$ such that

$$B(0,2r)\cap {\gamma }^{\ast }=\varnothing$$

For $w\in B(0,r)$ and $z\in {\gamma }^{\ast }$ then

$$\left|\frac{w}{z}\right|<\frac{1}{2}$$

Since ${\gamma }^{\ast }$ is compact then

$$M=\sup \{|f(z)|:z\in {\gamma }^{\ast }\}\text{ is finite}$$

Hence

$$\left|f(z)\frac{w^n}{z^{n+1}}\right|=|f(z)||z{|}^{-1}{\left|\frac{w}{z}\right|}^n<\frac{M}{2r}{\left(\frac{1}{2}\right)}^n\forall z\in {\gamma }^{\ast }$$

By Weierstrass M-test then series

$$\sum _{n=0}^{\infty }\frac{f(z)w^n}{z^{n+1}}=\sum _{n=0}^{\infty }\frac{f(z)}{z}{\left(\frac{w}{z}\right)}^n=\frac{f(z)}{z}{\left(1-\frac{w}{z}\right)}^{-1}=\frac{f(z)}{z-w}$$

is viewed as a function of $z$ converges uniformly on ${\gamma }^{\ast }$ to $\frac{f(z)}{z-w}$

Hence for all $w\in B(0,r)$ then

$$I_f(\gamma ,w)=\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(z)}{z-w}dz=\sum _{n=0}^{\infty }\left(\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(z)}{z^{n+1}}dz\right)w^n$$

Thus $I_f\gamma ,w$ is given by a power series in $B(0,r)$ and hence analytic and holomorphic

Finally, if $f=1$ then as $I_1(\gamma ,z)=I(\gamma ,z)$ is integer-valued then
It must be constant on any connected component of $\mathbb{C}\setminus {\gamma }^{\ast }$ as required

Derivatives of the General Cauchy Integral Formula

Let

$$g(w)=I_f(\gamma ,w)$$

at $z_0$ for some continuous function $f$ then

$$g^{(n)}(z_0)=\frac{n!}{2\pi i}{\int }_{\gamma }\frac{f(z)}{(z-z_0{)}^{n+1}}dz$$

Cauchy Formula for Multiple Curves corollary

Let $U$ be a bounded domain with piecewise $C^1$ boundary with finitely many components
Let $f$ be a function holomorphic in the closure of $U$

Parametrise each boundary component of $U$ by contour ${\gamma }_i$ such that

$$i{\gamma }_i^{\prime }(t)\text{ is an inward normal}$$

Hence the outer boundary is positively oriented (counter clockwise)
So the inner components are negatively oriented (clockwise)

Denoting

$${\int }_{\partial U}=\sum {\int }_{{\gamma }_i}$$

then

$${\int }_{\partial U}f(z)dz=0$$

and

$$\frac{1}{2\pi i}{\int }_{\partial U}\frac{f(z)}{z-w}dz=f(w),w\in U$$

Proof

Idea of proof is to add a few cuts that avoid $w$ and connect inner boundary components to each other and to the outer boundary in such a way that $U$ without those extra curves is simply connected

Let $\gamma$ be the boundary of the domain
Apply Cauchy’s Integral Formula to $\gamma$ so

$$\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(z)}{z-w}=f(w)$$

As curve $\gamma$ is made up of boundary components ${\gamma }_i$ in the right orientation and cuts ${\eta }_i$
where ${\eta }_i$ appear twice with different orientations so the integral cancels out

Taylor Series for Holomorphic Functions corollary

Let $f:U\to \mathbb{C}$ be a function

If $f$ is holomorphic on open set $U$ then for any $z_0\in U$ then

$$f(z)\text{ is equal to its Taylor Series at }{z}_0$$

and the Taylor Series converges on any open disk centred at $z_0$ lying in $U$

Derivatives of $f$ at $z_0$ are given by

$$f^{(n)}(z_0)=\frac{n!}{2\pi i}{\int }_{\gamma (z_0,r)}\frac{f(z)}{(z-z_0{)}^{n+1}}dz$$

for any $r<R$ where $R$ is such that $B(z_0,R)\subset U$

Note that the integral formulae for the derivatives of $f$ are also referred to as Cauchy’s Integral Formulae

Proof Integral Formula, General cauchy integral formula for the winding number and Derivatives of the general cauchy integral formula

Follows from