Cauchy's Integral Formula

From Vanishing Integrals to an Integral Formula

Cauchy’s theorem tells us that the integral of a holomorphic function around a suitable closed curve is zero. Cauchy’s Integral Formula turns that fact into something more precise: the values of a holomorphic function along a circle determine its value at every point inside the circle. The integrand in the formula has a factor that becomes singular at the point where the function is evaluated. The proof uses Cauchy’s theorem away from that point, then shrinks a circle around the singularity.

We use the convention from the previous tutorial that a circle traversed counterclockwise has positive orientation. We also use the Evaluation by a Primitive theorem and Cauchy’s theorem on a convex domain. In the proof below, the latter is applied to small convex neighborhoods that avoid the singularity; the function being integrated need not be holomorphic at the point inside the small circle.

Definition (Positively Oriented Circle): For a center $a\in\mathbb{C}$ and radius $R>0$, the positively oriented circle $C(a,R)$ is parametrized by $\gamma(t)=a+Re^{it}$, $0\leq t\leq2\pi$. The closed disk it bounds is $\overline{D(a,R)}=\{z\in\mathbb{C}:|z-a|\leq R\}$.

Cauchy’s Integral Formula

Theorem (Cauchy’s Integral Formula): Let $U\subseteq\mathbb{C}$ be open, let $f:U\to\mathbb{C}$ be holomorphic, and suppose $\overline{D(a,R)}\subseteq U$ for some $a\in\mathbb{C}$ and $R>0$. For every $w$ with $|w-a|<R$, $$ f(w)=\frac{1}{2\pi i}\int_{C(a,R)}\frac{f(z)}{z-w}\,dz. $$

The condition on the closed disk is important: $f$ must be holomorphic on an open set containing the entire disk, not just defined on the circle. The point $w$ may be anywhere strictly inside the circle. The denominator $z-w$ is nonzero along the path of integration, so the line integral in the formula is well-defined.

Proof. Fix $w$ with $|w-a|<R$, and define $h(z)=f(z)/(z-w)$ for $z\neq w$. This function is holomorphic on the disk with $w$ removed. Choose $\rho>0$ small enough that the closed disk $\overline{D(w,\rho)}$ lies inside $D(a,R)$. We first show that

\int_{C(a,R)}h(z)\,dz = \int_{C(w,\rho)}h(z)\,dz,

where both circles in this equation are positively oriented. The function $h$ is holomorphic on the region between these circles, but it is not defined at $w$. Thus we cannot apply Cauchy’s theorem on a convex domain to a region that includes $w$. Instead, approximate both circles by inscribed regular polygons, with enough sides that the inner polygon stays a positive distance from $w$ and lies strictly inside the outer polygon. The polygonal region between them can be divided into finitely many triangles. Each triangle lies in a small open convex neighborhood contained in the region where $h$ is holomorphic. Cauchy’s theorem on a convex domain makes the integral around each triangle zero.

When we add these triangle integrals, every internal edge occurs twice in opposite directions and cancels by the Additivity and Reversal proposition. The remaining edges are the outer polygon with positive orientation and the inner polygon with negative orientation. Their integrals therefore agree when both polygons are given positive orientation. As the numbers of sides tend to infinity, the polygonal paths and their parametrized line integrals converge to those on the corresponding circles: the polygonal paths converge uniformly to the circles, and their constant velocities on each side converge uniformly, piece by piece, to the circle’s velocity. Since $h$ is continuous on a neighborhood of both circles, the polygonal integrals converge to the circle integrals. This proves the displayed equality.

We now let $\rho$ tend to zero. Parametrize the inner circle by $z=w+\rho e^{it}$, so $dz=i\rho e^{it}\,dt$. Then

\int_{C(w,\rho)}\frac{f(z)}{z-w}\,dz = i\int_0^{2\pi}f(w+\rho e^{it})\,dt.

Continuity of $f$ at $w$ implies that $f(w+\rho e^{it})$ tends uniformly to $f(w)$ for $0\leq t\leq2\pi$. Hence the right-hand side tends to $2\pi i f(w)$. The outer-circle integral does not depend on $\rho$, by the equality just proved, so it equals this limit. Dividing by $2\pi i$ gives the stated formula. $\square$

What the Formula Tells Us

The formula expresses an interior value through the values of $f$ on the whole surrounding circle, weighted by $1/(z-w)$. It is not an assertion that the integral of $f$ itself is $2\pi i f(w)$. The factor $1/(z-w)$ is essential: it creates a singularity at $w$, while remaining holomorphic everywhere on the integration path.

When $w=a$, the formula has a particularly simple form. Parametrizing the circle directly gives

\int_{C(a,R)}\frac{f(z)}{z-a}\,dz = i\int_0^{2\pi}f(a+Re^{it})\,dt = 2\pi i f(a).

This special case is useful for evaluating integrals quickly. More generally, the formula is a strong constraint: changing the values of a holomorphic function in the interior while keeping all of its values on the circle fixed is impossible, because the integral recovers each interior value.

Worked Example: Recovering a Value at the Center

Take $f(z)=z^2+3z+2$, the circle $C(0,2)$, and the interior point $w=0$. Since $f$ is holomorphic on the whole plane, Cauchy’s Integral Formula applies. The value at the center is $f(0)=2$, so

\int_{C(0,2)}\frac{z^2+3z+2}{z}\,dz = 2\pi i f(0) = 4\pi i.

The value can also be checked term by term: $(z^2+3z+2)/z=z+3+2/z$. The integrals of $z$ and $3$ around the circle are zero because they have primitives $z^2/2$ and $3z$. The remaining term gives $2\int_{C(0,2)} dz/z=4\pi i$, in agreement with the formula.

Worked Example: Evaluating at a Point Away from the Center

Let $f(z)=z^2+1$, and use the circle $C(0,3)$ to evaluate an integral with $w=1$. The point lies strictly inside the circle because $|1-0|=1<3$, and $f$ is holomorphic on a neighborhood of the closed disk. The formula gives

\int_{C(0,3)}\frac{z^2+1}{z-1}\,dz = 2\pi i f(1) = 2\pi i(1^2+1) = 4\pi i.

In particular, the value is not obtained by substituting the center of the circle into $f$. The formula evaluates $f$ at the point $w$ appearing in the denominator, here $w=1$.

A Derivative Formula

Cauchy’s Integral Formula also gives an integral representation for the derivative. Keep $a$ and $R$ fixed, and allow $w$ to vary inside the circle. For a small nonzero complex number $s$ such that $w+s$ is also inside the circle, subtract the formulas at $w+s$ and $w$, then divide by $s$. The algebraic identity

\frac{1}{s}\left(\frac{1}{z-w-s}-\frac{1}{z-w}\right) = \frac{1}{(z-w-s)(z-w)}

shows that the resulting difference quotient is

\frac{f(w+s)-f(w)}{s} = \frac{1}{2\pi i}\int_{C(a,R)} \frac{f(z)}{(z-w-s)(z-w)}\,dz.

Corollary (Cauchy’s Integral Formula for the First Derivative): Under the hypotheses of Cauchy’s Integral Formula, for every $|w-a|<R$, $$ f'(w)=\frac{1}{2\pi i}\int_{C(a,R)}\frac{f(z)}{(z-w)^2}\,dz. $$

Proof. Fix $w$ strictly inside $C(a,R)$. The distance from $w$ to the circle is $R-|w-a|>0$. For sufficiently small $s$, both $w$ and $w+s$ remain inside the circle, and the denominators $z-w-s$ stay uniformly bounded away from zero as $z$ ranges over $C(a,R)$. As $s\to0$, the integrand in the difference-quotient identity therefore converges uniformly on the circle to $f(z)/(z-w)^2$. Uniform convergence on a piecewise $C^1$ curve permits passage to the limit in the line integral. The left-hand side tends to $f'(w)$, which proves the formula. $\square$

Worked Example: Evaluating a Derivative by Integration

Let $f(z)=z^3$, and evaluate the derivative formula at $w=1$ using $C(0,2)$. The hypotheses hold because $z^3$ is holomorphic everywhere and $|1|<2$. Since $f'(1)=3$, the formula gives

\int_{C(0,2)}\frac{z^3}{(z-1)^2}\,dz = 2\pi i f'(1) = 6\pi i.

This example illustrates that a higher power of $z-w$ in the denominator changes which quantity the contour integral returns. Here the squared denominator recovers the first derivative, not the function value.

Hypotheses and a Common Pitfall

A contour integral can be evaluated with Cauchy’s Integral Formula only after its geometric and analytic conditions have been checked. The point $w$ must be strictly inside the circle, and $f$ must be holomorphic on an open set containing the entire closed disk. Being defined or holomorphic only along the circle does not suffice. As in the $1/z$ example in the previous tutorial, an interior singularity can prevent use of Cauchy’s theorem on the region needed for the proof.

The formula is also sensitive to orientation. Reversing the direction of travel changes the sign of a line integral by the Additivity and Reversal proposition. Thus a clockwise circle gives the negative of the displayed formula. The standard statement uses counterclockwise orientation so that the factor is $1/(2\pi i)$.

The derivative formula makes clear why holomorphicity has strong consequences: an integral over a fixed circle controls both the value and the rate of change at every interior point. In the next tutorial, this kind of control will be used to study holomorphic functions that are bounded on the whole complex plane.

Check Your Understanding

Use the statements and arguments in this tutorial to answer the following questions.

Why must the closed disk bounded by the contour lie in an open set on which $f$ is holomorphic?
In the proof, why is $h(z)=f(z)/(z-w)$ not eligible for Cauchy’s theorem on a region containing $w$?
What limit does the integral over the small circle $C(w,\rho)$ approach as $\rho\to0$, and why?
How does the denominator in the derivative formula differ from the denominator in Cauchy’s Integral Formula?
What changes in the formula if the circle is traversed clockwise rather than counterclockwise?

What You'll Learn