# Explain why the residue is equal to the limit?

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:

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

-
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: meta.math.stackexchange.com/questions/5020/… –  Arkamis 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. –  Arkamis 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

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

$$g(x)=\frac{h(x)}{(x-z)^2}$$

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

$$g(x)=\frac{1}{(x-z)^2}(a_0+a_1(x-z)+a_2(x-z)^2+\ldots)\Longrightarrow$$

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

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

-

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 $a$.

-