7.1 Cauchy's theorem  (Page 3/3)

Perhaps the main application of [link] is what's called the Cauchy Integral Formula. It may not appear to be useful at first glance, but we will be able to use it over and over throughout this chapter.In addition to its theoretical uses, it is the basis for a technique for actually evaluating contour integrals, line integrals, as well as ordinary integrals.

Cauchy integral formula

Let $S$ be a piecewise smooth geometric set whose boundary $C_S$ has finite length, and let $f$ be a continuous function on $S$ that is differentiable on the interior $S^0$ of $S.$ Then, for any point $z\in S^0,$ we have

REMARK This theorem is an initial glimpse at how differentiable functions of a complex variable are remarkablydifferent from differentiable functions of a real variable. Indeed, Cauchy's Integral Formula shows that the values of a differentiable function $f$ at all points in the interior of a geometric set $S$ are completely determined by the values of that function on the boundary of the set. The analogous thing for a function of a real variable would be to say thatall the values of a differentiable function $f$ at points in the open interval $(a,b)$ are completely determined by its values at the endpoints $a$ and $b.$ This is patently absurd for functions of a real variable, so there surely is something marvelous going on for differentiable functions of a complex variable.

Let $r$ be any positive number such that ${\overline{B}}_r\left(z\right)$ is contained in the interior $S^0$ of $S,$ and note that the close disk ${\overline{B}}_r\left(z\right)$ is a piecewise smooth geometric set $\widehat{S}$ contained in $S^0.$ We will write $C_r$ instead of $C_{\widehat{S}}$ for the boundary of this disk, and we will use as a parameterization of the curve $C_r$ the function $\phi :[0,2\pi ]\to C_r$ given by $\phi \left(t\right)=z+r{e}^{it}.$ Now the function $g\left(\zeta \right)=f\left(\zeta \right)/(\zeta -z)$ is continuous on $S\cap \widetilde{{\widehat{S}}^0}$ and differentiable on $S^0\cap \tilde{\widehat{S}},$ so that [link] applies to the function $g.$ Hence

Since the equality established above is valid, independent of $r,$ we may take the limit as $r$ goes to 0, and the equality will persist. We can evaluate such a limit by replacing the $r$ by $1/n,$ in which case we would be evaluating

where $f_n\left(t\right)=f(z+frac1n{e}^{it}).$ Finally, because the function $f$ is continuous at the point $z,$ it follows that the sequence $\left\{f_n\right\}$ converges uniformly to the constant function $f\left(z\right)$ on the interval $[0,2\pi ].$ So, by Theorem 5.6, we have that

Therefore,

and the theorem is proved.

The next exercise gives two simple but strong consequences of the Cauchy Integral Formula, and it would be wise to spend a few minutesderiving other similar results.

The preceding exercise shows that two differentiable functions of a complex variable are equal everywhere on a piecewise smooth geometric set $S$ if they agree on the boundary of the set. More is true. We will see below in the Identity Theorem that they are equal everywhere on a piecewise smooth geometric set $S$ if they agree just along a single convergent sequence in the interior of $S.$

Combining part (b) of [link] , [link] , and [link] , we obtain the following corollary:

Let $S_1,...,S_n$ be pairwise disjoint, piecewise smooth geometric sets whose boundaries have finite length, all contained in the interior of a piecewise smooth geometric set $S$ whose boundary has finite length. Suppose $f$ is continuous at each point of $S$ that is not in the interior of any of the $S_k$ 's, and that $f$ is differentiable at each point of $S^0$ that is not an element of any of the $S_k$ 's. Then, for any $z\in S^0$ that is not an element of any of the $S_k$ 's, we have

Let $r>0$ be such that ${\overline{B}}_r\left(z\right)$ is disjoint from all the $S_k$ 's. By part (b) of [link] , let $T_1,...,T_m$ be a partition of $S$ such that $T_k=S_k$ for $1\le k\le n,$ and $T_{n+1}={\overline{B}}_r\left(z\right).$ By [link] , we know that

From the Cauchy Integral Formula, we know that

Also, since $f\left(\zeta \right)/(\zeta -z)$ is differentiable at each point of the interior of the sets $T_k$ for $k>n+1,$ we have from [link] that for all $k>n+1$

Therefore,

which completes the proof.