# Intuitive explanation of Cauchy’s Integral Formula in Complex Analysis

There is a theorem that states that if $f$ is analytic in a domain $D$, and the closed disc {$z:|z-\alpha|\leq r$} contained in $D$, and $C$ denotes the disc’s boundary followed in the positive direction, then for every $z$ in the disc we can write:

My question is:
What is the intuitive explanation of this formula? (For example, but not necessary, geometrically.)

(Just to clarify – I know the proof of this theorem, I’m just trying to understand where does this exact formula come from.)

If you are looking for intuition then let us assume that we can expand $$f(ζ)f(\zeta)$$ into a power series around $$zz$$: $$f(ζ)=∑n≥0cn(ζ−z)nf(\zeta) = \sum_{n \geq 0} c_n(\zeta - z)^n$$. Note $$c0=f(z)c_0 = f(z)$$.
becomes $$∑n≥0∫cn(ζ−z)n−1dζ\sum_{n \geq 0} \int c_n(\zeta - z)^{n-1}d\zeta$$. 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 $$zz$$ (counterclockwise). Then you are basically reduced to
showing that $$∫(ζ−z)mdζ\int (\zeta - z)^{m}d\zeta$$ is 0 for $$m≥0m \geq 0$$ and is $$2πi2\pi i$$ for $$m=−1m = -1$$.
These can be done by direct calculations using polar coordinates with $$ζ=z+eit\zeta = z + e^{it}$$. Now divide by $$2πi2\pi i$$ 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.