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I'm studying for a midterm and my teacher warned this would be a good question to understand for the test. The problem is, I do not know how to go about explaining it.

Suppose g(x) has a pole of order 2 at the point z. Explain why:

$$\textrm{Res}(g;z)= \lim_{x\to z} \frac{d}{dx}((x-z)^2 g(x))$$

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Do you know (the general form of) Cauchy's integral formula? – mrf Feb 25 '13 at 16:13
@Tom If you're going to post so many questions, please pause for a moment and learn how to use LaTeX markup:… – Emily Feb 25 '13 at 16:15
Have you tried following the proof for the more general theorem? – Git Gud Feb 25 '13 at 16:56
This exact question was asked by another use not 1 hour ago, and subsequently deleted. – Emily Feb 25 '13 at 16:57
Sock puppet? Another new user named "Tom" just asked this exact same question for which I left a comment. That question has now been deleted. – mrf Feb 25 '13 at 16:58
up vote 2 down vote accepted

Definition: $\,g(x)\,$ has a pole of order $\,2\,$ at $\,x=z\,$ iff when we develop $\,g\,$ in a Laurent series around $\,x=z\,$ , we get


with $\,h(z)\,$ analytic and non-zero in some neighborhood of $\,z\,$ , and then


$$\frac{d}{dx}\left((x-z)^2g(z)\right)=\frac{d}{dx}\left( a_0+a_1(x-z)+a_2(x-z)^2+\ldots\right)=$$

$$=a_1+2a_2(x-z)+\ldots \xrightarrow[x\to z]{}a_1=Res_{x=z}(g)$$

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If $g$ has a pole of order $2$ at $a$, then its Laurent expansion at $a$ looks like $$ g(z)=\frac{c_{-2}}{(z-a)^2}+\frac{c_{-1}}{z-a}+c_0+c_1(z-a)+\dots\tag{1} $$ where the residue, the coefficient of $\frac1{z-a}$, is $c_{-1}$.

Multiplying by $(z-a)^2$, we have $$ (z-a)^2g(z)=c_{-2}+c_{-1}(z-a)+c_0(z-a)^2+c_1(z-a)^3+\dots\tag{2} $$ Taking the derivative yields $$ \frac{\mathrm{d}}{\mathrm{d}z}\left((z-a)^2g(z)\right)=c_{-1}+2c_0(z-a)+3c_1(z-a)^2+\dots\tag{3} $$ Taking the limit gives $$ \lim_{z\to a}\frac{\mathrm{d}}{\mathrm{d}z}\left((z-a)^2g(z)\right)=c_{-1}\tag{4} $$ which is again the residue of $g$ at $a$.

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