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...
Cauchy's Formula has a remarkable interpretation in terms of hyperbolic geometry.
To understand it, you need to know very little about hyperbolic geometry.
In fact, you only need to know that if you define the "plane" as the open unit disk $D$ in $\mathbb C$, and the "lines" as those arcs of circle inside $D$ which are orthogonal to the unit circle $\partial D$, you get a model for the hyperbolic plane.
[There is a first miracle: the orientation preserving isometries of $D$ into itself are precisely the holomorphic automorphisms of $D$.]
If $f$ is holomorphic in a neighborhood of the closure of $D$, and $z$ is in $D$, then Cauchy's Formula says that
$f(z)$ is the average of $f$ on $\partial D$ "as you see it from $z$".
By this I mean that $f(z)$ is the integral of $f$ on $\partial D$ with respect to the measure which assigns to an arc in $\partial D$ the number $\theta/2\pi$ where $\theta\in[0,2\pi]$ is the angle between the (hyperbolic) half-lines going from $z$ to the end points of the arc. (That's the measure you think the arc has, as a part of your horizon, if you look at it from $z$.)
EDIT 1 OF NOV 1, 2010.
This is to clarify the relationship between the Cauchy's and Poisson's Formulas.
For $|z|<1$ and $\theta$ real define the Cauchy kernel by $$C(\theta,z):=\frac{1}{2\pi}\ \frac{e^{i\theta}}{e^{i\theta}-z}\quad,$$ and define the Cauchy transform of a continuous function $g$ on the unit circle $\partial D$ by $$(Cg)(z):=\int_0^{2\pi}g(e^{i\theta})\ C(\theta,z)\ d\theta$$ for $|z|<1$. In particular Cauchy's Formula says that if $g$ is the boundary value of a holomorphic function $f$ on $D$, then $Cg=f$.
For $|z|<1$ and $\theta$ real define the Poisson kernel by $$P(\theta,z):=C(\theta,z)+\overline{C(\theta,z)}-\frac{1}{2\pi}\quad,$$ and define the Poisson transform of a continuous function $g$ on the unit circle $\partial D$ by $$(Pg)(z):=\int_0^{2\pi}g(e^{i\theta})\ P(\theta,z)\ d\theta$$ for $|z|<1$. In particular Poisson's Formula says that if $g$ is the boundary value of a harmonic function $f$ on $D$, then $Pg=f$.
EDIT 2 OF NOV 1, 2010. I stole this from Bill Thurston: go to page 180 of (or search for "visual" in) http://www.math.unl.edu/~mbrittenham2/classwk/990s08/public/thurston.notes.pdf/8a.pdf .
EDIT OF NOV 2, 2010.
Stricto sensu, there is no geometric interpretation of the Cauchy kernel, because it is not invariant.
Indeed, if $G$ denotes the group of biholomorphic transforms of the open unit disk $D$, then the Cauchy kernel $C(z,\theta)d\theta$, viewed as a 1-form on $D\times\partial D$, is not $G$-invariant.
However, its restriction to $\{0\}\times\partial D$ is invariant under the stabilizer of $0\in D$ in $G$ (which is the circle group).
As a result, this restriction extends in a unique way to a $G$-invariant 1-form on $D\times\partial D$, and this 1-form is the Poisson kernel.
More precisely, the (complex) vector space of $G$-invariant $(0,1)$-forms on $D\times\partial D$ is one dimensional, generated by the Poisson kernel. [This is because the action of $G$ on $D\times\partial D$ is simply transitive, and the $(0,1)$-forms on $D\times\partial D$ are the sections of a homogeneous line bundle over $D\times\partial D$.]
In short, the intuitive reason the Cauchy integral formula is true is that (a) it is true in the limit for small circle paths $C_r$ about $z$, when it reduces to averaging, and (b) the integral does not change as the path deforms nicely from $C_r$ to $C$:
Integrals of complex differentiable functions $\zeta\mapsto F(\zeta)$ over closed paths don't change when the paths are deformed nicely in the domain of differentiability. For a super proof of this that does not use continuous differentiability, see H. Hanche-Olsen, ``On Goursat's proof of the Cauchy integral theorem,'' Amer. Math. Monthly 115 (2008) 648-652.
Integration on circle paths $C_r: t\mapsto \zeta=z+re^{it}$ ($t\in[0,2\pi]$) corresponds to averaging: $$ \frac{d\zeta}{\zeta-z} = \frac{i r e^{it} dt}{re^{it}} =i\,dt, \quad\mbox{hence}\ \frac{1}{2\pi i}\int_{C_r} \frac{f(\zeta)}{\zeta-z}d\zeta = \frac1{2\pi} \int_0^{2\pi}f(z+re^{it})\,dt. $$
Given $z\in D$, the function $$\zeta\mapsto F(\zeta)=\frac{f(\zeta)}{\zeta-z}$$ is differentiable in $D\setminus\{z\}$, so for any path, like $C$, that can be deformed nicely in $D\setminus\{z\}$ to a circle path $C_r$ as above, we can further deform by taking $r\to0$ and conclude that for all small positive $r$, $$ f(z) =\frac{1}{2\pi i} \int_{C_r} \frac{f(\zeta)}{\zeta-z}d\zeta =\frac{1}{2\pi i} \int_C \frac{f(\zeta)}{\zeta-z}d\zeta. $$