7.2 Basic applications of the cauchy integral formula

As a major application of the Cauchy Integral Formula, let us show the much alluded to remarkable fact that a function that is a differentiable function of a complex variable on an open set $U$ is actually expandable in a Taylor series around every point in $U,$ i.e., is an analytic function on $U.$

Suppose $f$ is a differentiable function of a complex variable on an open set $U\subseteq C,$ and let $c$ be an element of $U.$ Then $f$ is expandable in a Taylor series around $c.$ In fact, for any $r>0$ for which ${\overline{B}}_r\left(c\right)\subseteq U,$ we have

for all $z\in B_r\left(c\right).$

Choose an $r>0$ such that the closed disk ${\overline{B}}_r\left(c\right)\subseteq U,$ and write $C_r$ for the boundary of this disk. Note that, for all points $\zeta$ on the curve $C_r,$ and any fixed point $z$ in the open disk $B_r\left(c\right),$ we have that $|z-c|<r=|\zeta -c|,$ whence $|z-c|/|\zeta -c|=|z-c|/r<1.$ Therefore the geometric series

Moreover, by the Weierstrass $M$ -Test, as functions of the variable $\zeta ,$ this infinite series converges uniformly on the curve $C_r.$ We will use this in the calculation below. Now, according to [link] , we have that

where we are able to bring the summation sign outside the integral by part (3) of [link] , and where

This proves that $f$ is expandable in a Taylor series around the point $c,$ as desired.

Using what we know about the relationship between the coefficients of a Taylor series and the derivatives of the function, together with the Cauchy Integral Theorem, we obtain the followingformulas for the derivatives of a differentiable function $f$ of a complex variable. These are sometimes also called the Cauchy Integral Formulas.

Suppose $f$ is a differentiable function of a complex variable on an open set $U,$ and let $c$ be an element of $U.$ Then $f$ is infinitely differentiable at $c,$ and

for any piecewise smooth geometric set $S\subseteq U$ whose boundary $C_S$ has finite length, and for which $c$ belongs to the interior $S^0$ of $S.$

[link] and [link] constitute what we called the “identity theorem” for functions that are expandable in a Taylor series around a point $c.$ An even stronger result than that is actually true for functions of a complex variable.

Identity theorem

Let $f$ be a continuous complex-valued function on a piecewise smooth geometric set $S,$ and assume that $f$ is differentiable on the interior $S^0$ of $S.$ Suppose $\left\{z_k\right\}$ is a sequence of distinct points in $S^0$ that converges to a point $c$ in $S^0.$ If $f\left(z_k\right)=0$ for every $K,$ then $f\left(z\right)=0$ for every $z\in S.$

It follows from [link] that there exists an $r>0$ such that $f\left(z\right)=0$ for all $z\in B_r\left(c\right).$ Now let $w$ be another point in $S^0,$ and let us show that $f\left(w\right)$ must equal $0.$ Using part (f) of [link] , let $\phi :[\widehat{a},\widehat{b}]\to C$ be a piecewise smooth curve, joining $c$ to $w,$ that lies entirely in $S^0.$ Let $A$ be the set of all $t\in [\widehat{a},\widehat{b}]$ such that $f\left(\phi \right(s\left)\right)=0$ for all $s\in [\widehat{a},t).$ We claim first that $A$ is nonempty. Indeed, because $\phi$ is continuous, there exists an $ϵ>0$ such that $|\phi \left(s\right)-c|=|\phi \left(s\right)-\phi \left(\widehat{a}\right)|<r$ if $|s-\widehat{a}|<ϵ.$ Therefore $f\left(\phi \right(s\left)\right)=0$ for all $s\in [\widehat{a},\widehat{a}+ϵ),$ whence, $\widehat{a}+ϵ\in A.$ Obviously, $A$ is bounded above by $\widehat{b},$ and we write $t_0$ for the supremum of $A.$ We wish to show that $t_0=\widehat{b},$ whence, since $\phi$ is continuous at $\widehat{B},$ $f\left(w\right)=f\left(\phi \left(\widehat{b}\right)\right)=f\left(\phi \left(t_0\right)\right)=0.$ Suppose, by way of contradiction, that $t_0<\widehat{b},$ and write $z_0=\phi \left(t_0\right).$ Now $z_0\in S^0,$ and $z_0=lim\phi (t_0-1/k)$ because $\phi$ is continuous at $t_0.$ But $f\left(\phi (t_0-1/k)\right)=0$ for all $k.$ So, again using [link] , we know that there exists an $r^{\text{'}}>0$ such that $f\left(z\right)=0$ for all $z\in B_{r^{\text{'}}}\left(z_0\right).$ As before, because $\phi$ is continuous at $t_0,$ there exists a $\delta >0$ such that $t_0+\delta <\widehat{b}$ and $|\phi \left(s\right)-\phi \left(t_0\right)|<r^{\text{'}}$ if $|s-t_0|<\delta .$ Hence, $f\left(\phi \right(s\left)\right)=0$ for all $s\in (t_0-\delta ,t_0+\delta ),$ which implies that $t_0+\delta$ belongs to $A.$ But then $t_0$ could not be the supremum of $A,$ and therefore we have arrived at a contradiction. Consequently, $t_0=\widehat{b},$ and therefore $f\left(w\right)=0$ for all $w\in S^0.$ Of course, since every point in $S$ is a limit of points from $S^0,$ and since $f$ is continuous on $S,$ we see that $f\left(z\right)=0$ for all $z\in S,$ and the theorem is proved.

The next exercise gives some consequences of the Identity Theorem. Part (b) may appear to be a contrived example,but it will be useful later on.