Difference between revisions of "Cauchy's Integral Formula"

(added relation for nth derivatives)
(added discussion)
Line 53: Line 53:
 
complex-differentiable on some region, then it is ''infinitely
 
complex-differentiable on some region, then it is ''infinitely
 
differentiable'' on the interior of that region.
 
differentiable'' on the interior of that region.
 +
 +
Since the <math>(n+1)</math>th derivative exists in general, it follows that
 +
the <math>n</math>th derivative is continuous.  This is not true for functions
 +
of real variables!  For instance the real function
 +
<cmath> f(x) = \begin{cases} x \sin(1/x), & x \neq 0 \\ 0, & x=0 \end{cases} </cmath>
 +
is everywhere differentiable, but its derivative is mysteriously
 +
not continuous at <math>x=0</math>.  In complex analysis, the mystery disappears:
 +
the function <math>z\sin(1/z) = z\frac{e^{i/z} - e^{-i/z}}{2i}</math> has an
 +
[[essential singularity]] at <math>z=0</math>, so we can't establish a derivative
 +
there in any case.
 +
 +
The theorem is useful for estimating a function (or its <math>n</math>th derivative)
 +
at a point based on the behavior of the function around the point.
 +
For instance, the theorem yields an easy proof that [[holomorphic function]]s
 +
are in fact [[analytic function | analytic]].
  
 
== See also ==
 
== See also ==

Revision as of 22:42, 6 April 2009

Cauchy's Integral Formula is a fundamental result in complex analysis. It states that if $U$ is a subset of the complex plane containing a simple counterclockwise loop $C$ and the region bounded by $C$, and $f$ is a complex-differentiable function on $U$, then for any $z_0$ in the interior of the region bounded by $C$, \[\frac{1}{2\pi i} \int\limits_C \frac{f(z)}{z- z_0}dz = f(z_0) .\]

Proof

Let $D$ denote the interior of the region bounded by $C$. Let $C_r$ denote a simple counterclockwise loop about $z_0$ of radius $r$. Since the interior of the region bounded by $C$ is an open set, there is some $R$ such that $C_r \subset D$ for all $r \in (0, R)$. For such values of $r$, \[\int\limits_C \frac{f(z)}{z-z_0}dz = \int\limits_{C_r} \frac{f(z)}{z-z_0}dz ,\] by application of Cauchy's Integral Theorem.

Since $f$ is differentiable at $z_0$, for any $\epsilon$ we may pick an arbitarily small $r>0$ such that \[\left\lvert \frac{f(z)-f(z_0)}{z-z_0} - f'(z_0) \right\rvert < \epsilon\] whenever $\lvert z - z_0 \rvert \le r$. Let us parameterize $C_r$ as $h(t) = r e^{it}+ z_0$, for $t\in [0,2\pi]$. Since $\int\limits_{C_r} f'(z_0)dz = 0$ (again by Cauchy's Integral Theorem), it follows that \begin{align*} \biggl\lvert \int\limits_{C_r} \frac{f(z)}{z-z_0}dz - \int\limits_{C_r} \frac{f(z_0)}{z-z_0}dz \biggr\rvert &= \biggl\lvert \int\limits_{C_r} \left[ \frac{f(z) - f(z_0)}{z-z_0} - f'(z_0) \right] dz \biggr\rvert \\ &\le \int\limits_0^{2\pi} \left\lvert \frac{f(h(t)) - f(z_0)}{ h(t) - z_0} - f'(z_0) \right\vert \cdot r dt \\ &< \int\limits_0^{2\pi} \epsilon \cdot r dt = 2\pi \epsilon r . \end{align*} Since $\epsilon$ and $r$ can simultaneously become arbitrarily small, it follows that \begin{align*} \int\limits_C \frac{f(z)}{z-z_0}dz &= \int\limits_{C_r} \frac{f(z_0)} {z- z_0}dz \\ &= f(z_0) \int\limits_{0}^{2\pi} \frac{h'(t)}{h(t) - z_0} dt \\ &= f(z_0) \int\limits_{0}^{2\pi} \frac{ir e^{it}}{re^{it}} dt \\ &= f(z_0) \cdot 2\pi i , \end{align*} which is equivalent to the desired theorem. $\blacksquare$

Consequences

By induction, we see that the $n$th derivative of $f$ at $z_0$ is \[f^{(n)}(z_0) = (-1)^{n-1} (n-1)! \int\limits_C \frac{f(z)}{(z-z_0)^n}dz,\] for $n>0$. In particular, the $n$th derivative exists at $z_0$, for all $n>0$. In other words, if a function $f$ is complex-differentiable on some region, then it is infinitely differentiable on the interior of that region.

Since the $(n+1)$th derivative exists in general, it follows that the $n$th derivative is continuous. This is not true for functions of real variables! For instance the real function \[f(x) = \begin{cases} x \sin(1/x), & x \neq 0 \\ 0, & x=0 \end{cases}\] is everywhere differentiable, but its derivative is mysteriously not continuous at $x=0$. In complex analysis, the mystery disappears: the function $z\sin(1/z) = z\frac{e^{i/z} - e^{-i/z}}{2i}$ has an essential singularity at $z=0$, so we can't establish a derivative there in any case.

The theorem is useful for estimating a function (or its $n$th derivative) at a point based on the behavior of the function around the point. For instance, the theorem yields an easy proof that holomorphic functions are in fact analytic.

See also