[proofplan] We prove all three formulas by extracting the coefficient $c_{-1}$ from the Laurent expansion. For a simple pole, multiplying by $(z - a)$ and taking the limit isolates $c_{-1}$. The quotient form follows from L'Hopital's rule applied to $(z-a)/h(z)$. For a pole of order $k$, multiplying by $(z-a)^k$ produces a holomorphic function whose $(k-1)$-th derivative at $a$ gives $(k-1)! \, c_{-1}$. [/proofplan] [step:Derive the simple pole formula by multiplying by $(z - a)$ and taking the limit] **(1)** If $a$ is a simple pole, the [Laurent series expansion](/theorems/350) of $f$ near $a$ is \begin{align*} f(z) = \frac{c_{-1}}{z - a} + c_0 + c_1(z - a) + \cdots \end{align*} Multiplying both sides by $(z - a)$: \begin{align*} (z - a) f(z) = c_{-1} + c_0(z - a) + c_1(z - a)^2 + \cdots \end{align*} Taking $z \to a$, all terms except $c_{-1}$ vanish, so $\operatorname{Res}(f, a) = c_{-1} = \lim_{z \to a} (z - a) f(z)$. [/step] [step:Derive the quotient form for $f = g/h$ with $h$ having a simple zero] **(2)** If $f(z) = g(z)/h(z)$ with $g, h$ holomorphic near $a$, $g(a) \neq 0$, and $h(a) = 0$ with $h'(a) \neq 0$ (a simple zero of $h$), then $f$ has a simple pole at $a$. Applying formula (1): \begin{align*} \operatorname{Res}(f, a) = \lim_{z \to a} (z - a) \frac{g(z)}{h(z)} = g(a) \cdot \lim_{z \to a} \frac{z - a}{h(z) - h(a)} = \frac{g(a)}{h'(a)}, \end{align*} where we used $h(a) = 0$ and the definition of $h'(a) = \lim_{z \to a} \frac{h(z) - h(a)}{z - a}$. [/step] [step:Derive the higher-order pole formula by differentiating $(z - a)^k f(z)$] **(3)** If $a$ is a pole of order $k$, the Laurent expansion is \begin{align*} f(z) = \frac{c_{-k}}{(z-a)^k} + \cdots + \frac{c_{-1}}{z - a} + c_0 + c_1(z-a) + \cdots \end{align*} Define \begin{align*} \psi: B(a, r) &\to \mathbb{C} \\ z &\mapsto (z - a)^k f(z) = c_{-k} + c_{-k+1}(z-a) + \cdots + c_{-1}(z-a)^{k-1} + c_0(z-a)^k + \cdots \end{align*} The function $\psi$ is holomorphic near $a$ (the pole has been removed by the factor $(z-a)^k$). The Taylor expansion of $\psi$ about $a$ shows that the coefficient of $(z-a)^{k-1}$ is $c_{-1}$. By the standard Taylor coefficient formula, this coefficient equals $\psi^{(k-1)}(a)/(k-1)!$. Therefore \begin{align*} \operatorname{Res}(f, a) = c_{-1} = \frac{1}{(k-1)!} \lim_{z \to a} \frac{d^{k-1}}{dz^{k-1}} \bigl[(z-a)^k f(z)\bigr]. \end{align*} [guided] The idea is to trade a pole of order $k$ for a holomorphic function by multiplying by $(z-a)^k$. The resulting function $\psi(z) = (z-a)^k f(z)$ is holomorphic near $a$ because the factor $(z-a)^k$ cancels the pole. Its Taylor expansion is $\psi(z) = \sum_{m=0}^\infty c_{m-k}(z-a)^m$. We want $c_{-1}$, which is the coefficient of $(z-a)^{k-1}$ (corresponding to $m = k-1$). The Taylor coefficient formula states that the $m$-th coefficient of a holomorphic function $\psi$ at $a$ equals $\psi^{(m)}(a)/m!$. Setting $m = k - 1$ gives $c_{-1} = \psi^{(k-1)}(a)/(k-1)!$. This formula reduces residue computation to differentiation, which is often more practical than computing the full Laurent expansion. For simple poles ($k = 1$), the formula recovers part (1): $\psi(z) = (z-a)f(z)$ and $\psi^{(0)}(a)/0! = \psi(a) = \lim_{z \to a}(z-a)f(z)$. [/guided] [/step]