How to compute residues at simple and higher-order poles?

**Example 1: Simple pole at $z = i$** $$f(z) = \frac{e^z}{z^2 + 1} = \frac{e^z}{(z-i)(z+i)}$$ \begin{align*} \text{Res}(f, i) &= \lim_{z \to i} (z - i) \cdot \frac{e^z}{(z-i)(z+i)} \\ &= \lim_{z \to i} \frac{e^z}{z+i} = \frac{e^i}{2i} \end{align*} **Example 2: Pole of order 2 at $z = 1$** $$f(z) = \frac{1}{(z-1)^2(z-2)}$$ Using the formula with $m = 2$: \begin{align*} \text{Res}(f, 1) &= \frac{1}{(2-1)!} \lim_{z \to 1} \frac{d}{dz}\left[(z-1)^2 \cdot \frac{1}{(z-1)^2(z-2)}\ ight] \\ &= \lim_{z \to 1} \frac{d}{dz}\left[\frac{1}{z-2}\ ight] = \lim_{z \to 1} \left(-\frac{1}{(z-2)^2}\ ight) = -1 \end{align*} **Shortcut for rational functions $f(z) = \frac{p(z)}{q(z)}$ at simple poles:** If $q(z)$ has a simple zero at $z_0$, then: $$\text{Res}\left(\frac{p}{q}, z_0\ ight) = \frac{p(z_0)}{q'(z_0)}$$ For $f(z) = \frac{e^z}{z^2+1}$ at $z = i$: $q'(z) = 2z$, so $\text{Res} = \frac{e^i}{2i}$, matching our earlier result. **Geometric meaning of the residue:** The residue is $\frac{1}{2\pi i}$ times the integral of $f$ around a small circle enclosing $z_0$. It measures the "strength" of the singularity — how much the function fails to have an antiderivative locally. The sum of all residues inside a contour determines the contour integral.