Intuitive explanation of Cauchy's Integral Formula in Complex Analysis

If you are looking for intuition then let us assume that we can expand $f(\zeta)$ into a power series around $z$: $f(\zeta) = \sum_{n \geq 0} c_n(\zeta - z)^n$. Note $c_0 = f(z)$. If you plug this into the integral and interchange the order of integration and summation then that integral on the right side of the formula becomes $\sum_{n \geq 0} \int c_n(\zeta - z)^{n-1}d\zeta$. Let us also assume that an integral along a contour doesn't change if we deform the contour continuously through a region where the function is "nice". So let us take as our path of integration a circle going once around the point $z$ (counterclockwise). Then you are basically reduced to showing that $\int (\zeta - z)^{m}d\zeta$ is 0 for $m \geq 0$ and is $2\pi i$ for $m = -1$. These can be done by direct calculations using polar coordinates with $\zeta = z + e^{it}$. Now divide by $2\pi i$ and you have the formula. Of course this is a hand-wavy argument in places, but the question was not asking for a rigorous proof. Personally, this is how I first came to terms with understanding how Cauchy's integral formula could be guessed.

As Chandru1 pointed out, a beautiful particular case of Cauchy's integral formula gives us a very particular winding number. For the sake of simplicity, let us look only at closed paths on the boundary of the unit disc, $\gamma : [0,1] \longrightarrow S^1$. Then, the winding number of $\gamma$ is the number of times $\gamma$ goes round the circumference. That is, if we write the complex number $\gamma (t)$ in its exponential form

$$ \gamma (t) = e^{i\theta (t)} \qquad \qquad \qquad [1] $$

then you can prove that there is a continuous choice for the argument $\theta (t)$ of $\gamma (t)$; that is, a continuous function $\theta : [0, 1] \longrightarrow \mathbb{R} $ such that [1] holds for all $t$. In Algebraic Topology, $\theta$ is called a "lifting" of $\gamma$. Moreover, any two such continuous "liftings" of $\gamma$ differ necessarily by a constant integer multiple of $2\pi$: if $\widetilde{\theta}(t)$ is another (continuous) lifting of $ \gamma$, then there is a constant integer $k$, not depending on $t$, such that $\widetilde{\theta}(t) = \theta (t) + 2k\pi$ for all $t$. (This is called the "lifting lemma" in Algebraic Topology. For a proof in Complex Analysis, see theorem 7.1, in Stewart and Tall.) So the number of times $\gamma $ goes round the circumference (its winding number) is well-defined as

$$ w (\gamma , 0) = \frac{\theta (1) - \theta (0)}{2\pi} \ . $$

Then you can easily prove (see op.cit, section 7.5), that

$$ w (\gamma , 0 ) = \frac{1}{2\pi i}\int_\gamma \frac{1}{\zeta}d\zeta \ . $$

So, Cauchy's integral formula for the constant function $f \equiv 1$, $z = 0$ (and $\gamma (t) = e^{i2\pi t}$, hence $\theta (t) =2 \pi t$) tells us that

$$ \frac{1}{2\pi i}\int_\gamma \frac{1}{\zeta}d\zeta = 1 \ . $$

That is, if you go round the circumference just one time, well, you are going round the circumference indeed exactly once. Isn't math amazing? :-)

The trick is to remember that differentiable is really a way of saying 'locally linear', that is, close to $z$ we have $$f( \zeta )= f(z) + (\zeta -z)f'(z) +g( \zeta )$$

Where $g(\zeta)$ is an error term that vanishes as we get close to $z$ (in fact it is $o(\zeta -z)$).

Now, dividing through by $(\zeta -z)$ as in the integral formula gives expression under the integral sign as: $$\frac{f( \zeta )}{(\zeta -z)}= \frac{f(z)}{(\zeta -z)} + f'(z) +\frac{g( \zeta )}{(\zeta -z)}$$

We now integrate this termwise:

The last term still vanishes as $\zeta \to z$ (by the definition of little o), which allows us to ignore it by way of a homotopy argument. You see, since we can shrink the loop $\gamma$ around $z$ arbitrarily small without changing the value of the integral (an old theorem of Cauchy's says that the integral along a closed curve is homotopy invariant $(*)$) and, as the loop gets smaller, we see that the value of the last term on $\gamma$ gets closer to vanishing- and so, morally, its integral over $\gamma$ may be bounded above in modulus by successively smaller $\epsilon$. But, because of homotopy invariance, these $\epsilon$ bound the integral over all such loops- and $\epsilon$ is arbitrary- so the integral of the last term vanishes.

The second term is easy- this integrates to $[\zeta f'(z)]_{\gamma (0)}^{\gamma (1)}$, with $\gamma (0)=\gamma (1)$. Plug it in to see this vanishes.

The first term is the interesting one and others have covered it well in their posts, but I'll finish the argument anyways. Having eliminated the other terms (and observing that $f(z)$ is a constant) we have a pair of statements equivalent to the original formula:

$$\frac{1}{2\pi i}\int\frac{f(z)}{\zeta-z}d\zeta =f(z) \iff \int\frac{1}{\zeta-z}d\zeta =2\pi i$$

This last one we can use substitution $u=\zeta-z$ and homotopy to make about the integral over a circle $\gamma (t)= e^{2\pi it}$ around the origin giving:

$$\int_{\gamma} \frac{1}{u} du (**)= \int_0^1 \frac{\gamma '(t)}{\gamma (t)}dt= = \int_0^1 \frac{2 \pi i e^{2\pi it}}{e^{2\pi it}}dt$$

The last one is a total gift after cancellation and we all go home smiling. We could of course have morally calculated $(**)$ without resorting to technicalities by observing that the primitive of $\frac{1}{u}= log(u)$ and looking at the integrand $[log(u)]_{e^0}^{e^{(2\pi -\epsilon) i}}$ and letting $\epsilon \to 0$. But the 'right' intuition is certainly that of Augusti's answer.

As for $(*)$, the intuition is another story altogether, but I may add it later if it looks worth it...