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A nice way to remember a formula is to connect it to something you already know. For example, to remember the Cauchy integral formula, I remember that $f(z_0)=\frac{1}{2\pi i}\int_\Gamma\frac{f(z)}{z-z_0}\,dz$ because if you write out the Laurent series for $\frac{f(z)}{z-z_0}$ you're likely to get something like:

$$ \frac{f(z)}{z-z_0}=\frac{a_{-1}}{z-z_0}+a_0 +a_1(z-z_0)+\dots \,\,\,\,(1)$$

If you take the integral of the right hand side of (1), you're going to be left with $a_{-1} 2\pi i$, but we were really looking for the behavior of $f(z)$ at $z_0$, and since we used an integral to find the behavior we got $2\pi i$ as an artifact, so we're just going to divide our answer by $2\pi i$ and be left with $f(z_0)$ a.k.a $Res(f;z_o).$

My questions:

  • is there a similar way to think about $f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_\Gamma\frac{f(z)}{(z-z_0)^{n+1}}\,dz$? I've tried writing out a Laurent series for $\frac{f(z)}{(z-z_0)^{n+1}}\,dz$, but it didn't get me anywhere.

  • if $f(z)=\frac{g(z)}{(z-z_0)^m}$, is $g(z_0)=a_{-1}$ ? (where $a_{-1}$ is from its Laurent series)

Obviously there is a strong connection between the singularities of a function, its Laurent series, residue theory and the Cauchy integral formulas. After a semester of introductory complex analysis, I'm struggling to place everything in a coherent narrative. Any help would be greatly appreciated.

Edit: Based on the answers I've formed a new narrative to remember the formulas. I thought I'd share it in case someone finds it helpful.

Let's say we have a function:

$$ \frac{ f(z) }{ (z-z_0)^{n+1} } $$

We want to study the behavior of $f^{(n)}(z)$ at $z=z_0$. We know that a Taylor coefficient is on the form $a_n=\frac{f^{(n)}(z_0)}{n!}$, so we can find $f^{(n)}(z_0)$ by manipulating the Taylor series.

So expand to Taylor form:

$$ \frac{ f(z) }{ (z-z_0)^{n+1} }=\frac{c_0}{(z-z_0)^{n+1}}+\frac{c_{1}}{(z-z_0)^{n}} +\dots+\frac{c_{k}}{(z-z_0)^{1}}+c_{k+1}+c_{k+2}(z-z_0)+\dots $$

We see that we can recover $c_k=\frac{f^{(k)}(z_0)}{n!}$ by integrating the equation above. But since we used an integral to isolate $c_k$, we get $2\pi i$ as an artifact, and since we used the Taylor series we also got $\frac{1}{n!}$ as an artifact.

We can think of the forumla $f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_\Gamma\frac{f(z)}{(z-z_0)^{n+1}}$ a formula for recovering $f^{(n)}(z_0)$ from a Taylor series: Use $\int_\Gamma\frac{f(z)}{(z-z_0)^{(n+1)}}$ to isolate the Taylor coefficient and remove the extraneous information by multiplying by $\frac{n!}{2\pi i}$.

When $n=0$ it's just the regular Cauchy integral formula, of course; $\frac{0!}{2\pi i}=\frac{1}{2\pi i}$.

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up vote 2 down vote accepted

Expand $f$ into its Taylor series,

$$f(z) = \sum_{k=0}^\infty c_k(z-z_0)^k.$$

Then you have

$$\frac{f(z)}{(z-z_0)^{n+1}} = \sum_{k=0}^\infty c_k (z-z_0)^{k-n-1}.$$

When you integrate termwise, the only term with a nonzero integral is the one with the exponent $k-n-1 = -1$, that is, $k = n$, so the integral is $2\pi i c_n$. But $$c_k = \frac{f^{(k)}(z_0)}{k!}$$ for all $k$.

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Ahh, it's much easier to remember this argument and just let the Cauchy integral formula be when $n=0$. Thanks! – john.abraham Nov 30 '13 at 14:14

The Laurent series argument goes like this. If $f(z) = a_0 + a_1 (z - z_0) + a_2 (z - z_0)^2 + \ldots$, then $$\dfrac{1}{2\pi i} \oint_\Gamma \dfrac{f(z)\ dz}{(z - z_0)^n} = \dfrac{1}{2 \pi i} \left( \oint_\Gamma \dfrac{a_0 dz}{(z - z_0)^n} + \oint_\Gamma \dfrac{a_1 dz}{(z - z_0)^{n-1}} + \ldots \right)$$ All terms are $0$ except the one with denominator $(z - z_0)^1$ ...

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If you believe in Cauchy's formula $$f(z)={1\over2\pi i}\int_{\partial\Omega}{f(\zeta)\over \zeta-z}\,d\zeta\qquad(z\in\Omega)$$ then the formula for the $n$th derivative $f^{(n)}(z)$ follows by simple differentiation under the integral sign. Note that on $\partial\Omega$ the function $f$ is supposed to be at least continuous.

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Differentiating $(\zeta-z)^{-1}$, doesn't that give me a minus sign in the integral? How is that dealt with? – john.abraham Nov 30 '13 at 14:17
@john.abraham: You get two minus signs, one for the negative exponent and one for the "inner derivative". – Christian Blatter Nov 30 '13 at 15:28

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