Cauchy integral formula

Let $f$ be a complex-valued holomorphic function on an open connected set $U \subseteq C$ , and let $γ$ be a contour contained entirely in $U$ and homotopic to a point. If $a \notin γ ([0, 1])$ , then for any non-negative integer $n$ , we have:

f^{(n)} (a) \cdot n (a, γ) = \frac{n!}{2 π i} \oint_{γ} \frac{f (z)}{(z - a)^{n + 1}} d z,

where $n (a, γ)$ denotes the winding number.

We have the important particular cases $n = 0$ :

f (a) \cdot n (a, γ) = \frac{1}{2 π i} \oint_{γ} \frac{f (z)}{z - a} d z

and $n = 1$ :

f^{'} (a) \cdot n (a, γ) = \frac{1}{2 π i} \oint_{γ} \frac{f (z)}{(z - a)^{2}} d z .

This formula expresses the $n$ -th derivative of a holomorphic function at a point $a$ as a contour integral around $a$ . It highlights the remarkable fact that all derivatives of $f$ inside the region bounded by $γ$ are completely determined by the values of $f$ on the contour itself.

Consequences:

Holomorphic functions are infinitely differentiable and analytic.
Knowledge of a holomorphic function on a closed curve determines its behavior inside.
The formula underpins many core results in complex analysis, including the development of power series and the residue theorem.

Proof

Here is the classic derivation showing that the Cauchy Integral Formula

f (a) = \frac{1}{2 π i} \oint_{C} \frac{f (z)}{z - a} d z

is a corollary of the Cauchy--Goursat theorem.
Fix a point $a$ in the interior of $C$ . Define

g (z) = \frac{f (z) - f (a)}{z - a} .

Since $f$ is holomorphic on and inside $C$ , $g$ is holomorphic there as well (the apparent pole at $z = a$ is removable, because $lim_{z \to a} g (z) = f^{'} (a)$ ).
By Cauchy–Goursat, $\oint_{C} g (z) d z = 0.$

Write

0 = \oint_{C} \frac{f (z) - f (a)}{z - a} d z = \oint_{C} \frac{f (z)}{z - a} d z - f (a) \oint_{C} \frac{1}{z - a} d z .

Rearrange:

\oint_{C} \frac{f (z)}{z - a} d z = f (a) \oint_{C} \frac{1}{z - a} d z .

One shows by parameterizing $z = a + R e^{i θ}$ , $0 \leq θ \leq 2 π$ , that

\oint_{C} \frac{1}{z - a} d z = \int_{0}^{2 π} \frac{1}{R e^{i θ}} i R e^{i θ} d θ = 2 π i .

So therefore

f (a) = \frac{1}{2 π i} \oint_{C} \frac{f (z)}{z - a} d z .

Theorem Let $γ : [a, b] \to C$ be a piecewise $C^{1}$ closed curve and let $f$ be a function continuous on $γ ([a, b])$ . Then the function

G (z) = \frac{1}{2 π i} \int_{γ} \frac{f (w)}{w - z} d w

is holomorphic in $C ∖ {γ ([a, b])}$ and

G^{'} (z) = \frac{1}{2 π i} \int_{γ} \frac{f (w)}{(w - z)^{2}} d w

Remark. Only continuity of $f$ on $γ$ is required!

Personal approach

My visualization: see this video of mine.

The Winding Field `1/z`

Let's recall first the special case of the function $f (z) = 1 / z$ . We established that integrating this function along a curve $γ$ that winds once around the origin yields a non-zero result:

\oint_{γ} \frac{1}{z} d z = 2 π i

This is because $1 / z$ is the derivative of the multi-valued logarithm function, so we don't need to reconstruct the curve by multiplying each tangent vector by $1 / z$ , but it is enough to evaluate the primitive and subtract. See interpretation of complex integration.

Reconstructing the Function's Value

Now, let's consider the integral of a new function, $\frac{f (z)}{z}$ , where $f (z)$ is holomorphic inside and on the curve $γ$ , and $γ$ encloses the point $z = 0$ .

\oint_{γ} \frac{f (z)}{z} d z

Using the Deformation Theorem, we can shrink the curve $γ$ to an infinitesimally small circle around the point $0$ without changing the value of the integral.
Since $f$ is holomorphic around $z = 0$ , we use the local approximation $f (z) \approx f (0) + f^{'} (0) z$ .

By dividing by $z$ , the integrand $f (z) / z$ (i.e., the amplitwist we provide to each tangent vector) is approximated by the sum of two actions:

\frac{f (z)}{z} \approx f^{'} (0) + f (0) \cdot \frac{1}{z}

The overall transformation applied to each vector $v_{i}$ of the curve $γ$ is approximately:

{\tilde{v}}_{i} \approx f^{'} (0) v_{i} + f (0) \cdot \frac{1}{z_{i}} v_{i}

The integral $\int_{γ} f (z) / z d z$ is approximated by the sum of these transformed vectors:

\sum {\tilde{v}}_{i} = \sum f^{'} (0) v_{i} + \sum f (0) \cdot \frac{1}{z_{i}} v_{i}

We sum the two terms:

First Term (Constant Factor): The sum of the terms $f^{'} (0) v_{i}$ is $f^{'} (0) \sum v_{i}$ . Since $γ$ is closed, $\sum v_{i} = 0$ , so this part yields zero. This geometrically means the curve defined by constantly rotating and scaling the $v_{i}$ 's remains closed.
Second Column ( $1 / z$ Factor): The sum is $f (0) \sum (\frac{1}{z_{i}}) v_{i}$ .
The sum $\sum (1 / z_{i}) v_{i}$ is precisely the approximation for the integral $\int_{γ} 1 / z d z$ , which has already been established to equal $2 π i$ .

Conclusion:The total sum (the integral $\int_{γ} f (z) / z d z$ ) is approximated by:

\int_{γ} \frac{f (z)}{z} d z \approx f (0) \cdot 2 π i

This construction, performed around the origin, can be generalized to any point $A$ by shifting the coordinate system.

Cauchy integral formula

Proof

Related technical (and surprising) result

Personal approach

The Winding Field 1/z

Reconstructing the Function's Value

The Winding Field `1/z`