Cauchy Integral Formula

The 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's formula is useful for evaluating integrals of complex functions.

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

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

Thus, the formula is proven. $_\square$

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.

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

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

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.

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Let $g(z) =\dfrac{z+1}{z+2}$; $g$ is holomorphic everywhere inside $C$. Hence,

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

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 that $f$ is constant. $_\square$