# Cauchy Integral Formula

**Cauchy integral formula**states that the values of a holomorphic function inside a disk are determined by the values of that function on the boundary of the disk. More precisely, suppose \(f: U \to \mathbb{C}\) is holomorphic and \(\gamma\) is a circle contained in \(U\). Then for any \(a\) in the disk bounded by \(\gamma\),

\[f(a) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{z-a} \, dz.\]

More generally, \(\gamma\) is the boundary of any region whose interior contains \(a\).

Cauchy is useful for solving integrations of complex numbers.

## Cauchy differentiation formula

A direct corollary of the Cauchy integral formula is the following (using the above definitions of \(f\) and \(\gamma\).

\[f^{(n)}(a) = \frac{n!}{2\pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}} \, dz\]

The content of this formula is that if one knows the values of \(f(z)\) on some closed curve \(\gamma\), then one can compute the derivatives of \(f\) inside the region bounded by \(\gamma\), via an integral. The formula can be proved by induction on \(n\):

The case \(n=0\) is simply the Cauchy Integral Formula. Suppose the differentiation formula holds for \(n=k\). Using differentiation under the integral, we have

\[\begin{align} f^{(k+1)}(a) &= \frac{d}{da} f^{(k)}(a) \\ &= \frac{k!}{2\pi i} \int_{\gamma} \frac{d}{da} \frac{f(z)}{(z-a)^{k+1}} \, dz \\ &= \frac{k!}{2\pi i} \int_{\gamma} \frac{(k+1)f(z)}{(z-a)^{k+2}} \, dz \\ &= \frac{(k+1)!}{2\pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{k+2}} \, dz. \end{align}\]

Thus, the formula is proven.

## Examples

One can use the Cauchy integral formula to compute contour integrals which take the form given in the integrand of the formula.

Compute \(\displaystyle \int_{C} \frac{(z-2)^2}{z+i} \, dz\) where \(C\) is the circle of radius \(2\) centered at the origin.

Let \(f(z) = (z-2)^2\); \(f\) is holomorphic everywhere in the interior of \(C\). Hence, by the Cauchy integral formula,

\[\int_{C} \frac{(z-2)^2}{z+i} \, dz = 2\pi i f(-i) = -8\pi + 6\pi i.\]

Similarly, one can use the Cauchy differentiation formula to evaluate less straightforward integrals:

Compute \(\displaystyle \int_{C} \frac{z+1}{z^4 + 2z^3} \, dz \) where \(C\) is the circle of radius \(1\) centered at the origin.

Let \(g(z) = (z+1)/(z+2)\); \(g\) is holomorphic everywhere inside \(C\). Hence,

\[\int_{C} \frac{z+1}{z^4 + 2z^3} \, dz = \frac{2\pi i}{2!} g''(0) = -\frac{\pi i}{4}.\]

## Liouville's Theorem

As an application of the Cauchy Integral Formula, one can prove **Liouville's Theorem**, an important theorem in complex analysis. This theorem states that if a function is holomorphic everywhere in \(\mathbb{C}\) and is bounded, then the function must be constant.

If \(f: \mathbb{C} \to \mathbb{C}\) is holomorphic and there exists \(M > 0\) such that \(|f(z)| \le M\) for all \(z\in \mathbb{C}\), then \(f\) is constant.

Suppose \(f: \mathbb{C} \to \mathbb{C}\) satisfies the conditions of the theorem. Given \(a\in \mathbb{C}\), let \(C_r\) denote the circle of radius \(r\) centered at \(a\). By the Cauchy Differentiation Formula and the triangle inequality, we have \[|f^{(n)}(a)| = \frac{n!}{2\pi} \left\vert \int_{C_r} \frac{f(z)}{(z-a)^{n+1}} \, dz \right\vert \le \frac{n!}{2\pi} \int_{C_r} \frac{|f(z)|}{|z-a|^{n+1}}\, dz \le \frac{n! M}{2\pi} \frac{1}{r^{n+1}}\] Taking \(r\to\infty\) shows that \(f^{(n)}(a) = 0\). Thus, all derivatives of \(f\) are 0 everywhere, and it follows \(f\) is constant.

**Cite as:**Cauchy Integral Formula.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/cauchy-integral-formula/