Cauchy integral formula

This article gives the statement, and possibly proof, of a basic fact in complex analysis.
View a complete list of basic facts in complex analysis

Statement

For a circle with respect to any point in the interior

Suppose $U$ is a domain in $\mathbb {C}$ and $f:U\to \mathbb {C}$ is a holomorphic function. The Cauchy integral formula states that if $z_{0}\in U$ is such that the disk $D$ of radius $r$ about $z_{0}$ lies completely inside $U$, and if $z$ is such that $|z-z_{0}|<r$, we have:

$f(z)={\frac {1}{2\pi i}}\oint _{\partial D}{\frac {f(\xi )}{\xi -z}}\,d\xi$

In other words, we can determine the value of $f$ at any point in the interior, by knowing its value at all points on the boundary.

For any union of piecewise smooth curves and any point

Further information: Winding number version of Cauchy integral formula

Related facts

• Cauchy integral formula for derivatives
• Winding number version of Cauchy integral formula
• Homotopy-invariance formulation of Cauchy's theorem

Facts used

1. Cauchy integral formula for constant functions: In other words, the fact that for a constant function, the Cauchy integral formula holds.
2. A version of Fubini's theorem, that allows the exchange of a real and a complex integral when the total quantities are absolutely integrable.
3. fundamental theorem of complex calculus.

Proof

Given: $U$ is a domain in $\mathbb {C}$ and $f:U\to \mathbb {C}$ is a holomorphic function. $z_{0}\in U$ is such that the disk $D$ of radius $r$ about $z_{0}$ lies completely inside $U$, and $z$ is such that $|z-z_{0}|<r$.

To prove:

$f(z)={\frac {1}{2\pi i}}\oint _{\partial D}{\frac {f(\xi )}{\xi -z}}\,d\xi$

Proof: The proof proceeds in two steps. We first use the Cauchy integral formula for constant functions (fact (1)) to show that for the constant function $w\mapsto f(z)$, we have:

$f(z)=\oint _{\partial D}{\frac {f(z)}{\xi -z}}\,d\xi$

We then show that:

$\oint _{\partial D}{\frac {f(\xi )}{\xi -z}}\,d\xi =\oint _{\partial D}{\frac {f(z)}{\xi -z}}\,d\xi$.

Let's do the second step now. We do this by a homotopy method. Namely, consider the function $[0,1]\times {\overline {D}}\setminus \{z\}\to \mathbb {C}$:

$F(t,w)={\frac {f((1-t)z+tw)}{w-z}}$.

Clearly, $F$ is holomorphic in $w$, and we have:

${\frac {\partial F}{\partial t}}=f'((1-t)z+tw)$.

This yields, by fact (3), that for any $t\in [0,1]$:

$\oint _{\partial D}{\frac {\partial F(t,\xi )}{\partial t}}\,d\xi =0$

Integrating over $t\in [0,1]$ yields:

$\int _{0}^{1}\oint _{\partial D}{\frac {\partial F(t,\xi )}{\partial t}}\,d\xi \,dt=0$

We now exchange the two integrals by fact (2). We get:

$\oint _{\partial D}\int _{0}^{1}{\frac {\partial F(t,\xi )}{\partial t}}\,dt\,d\xi =0$.

Next, we apply the fundamental theorem of real calculus to evaluate the inner integral, and obtain:

$\oint _{\partial D}F(1,\xi )-F(0,\xi )\,d\xi =0$.

Plugging in values yields:

$\oint _{\partial D}{\frac {f(z)-f(\xi )}{z-\xi }}=0$

which, on rearrangement, gives the desired equality:

$\oint _{\partial D}{\frac {f(\xi )}{\xi -z}}\,d\xi =\oint _{\partial D}{\frac {f(z)}{\xi -z}}\,d\xi$.