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:
$$\textrm{Res}(g;z)= \lim_{x\to z} \frac{d}{dx}((x-z)^2 g(x))$$
 A: 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
$$g(x)=\frac{1}{(x-z)^2}(a_0+a_1(x-z)+a_2(x-z)^2+\ldots)\Longrightarrow$$
$$\frac{d}{dx}\left((x-z)^2g(z)\right)=\frac{d}{dx}\left(
a_0+a_1(x-z)+a_2(x-z)^2+\ldots\right)=$$
$$=a_1+2a_2(x-z)+\ldots \xrightarrow[x\to z]{}a_1=Res_{x=z}(g)$$
A: 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 $a$.
