Solution: We recall that if $f$ is a holomorphic function in the punctured disc $\mathbb{D} (a,r) - \{ a \} $, then the residue of $f$ at $a$ is the coefficient of $(z-a)^{-1} $ in the Laurent series of $f$ around $a$. We have the following important result to find out the residue.

• $f(z) = \frac{\sin z}{(z - \iota )(z + 2)^2}$. The singularities of $f$ are $z = \iota $ (simple pole) and $z = -2$ (pole of order $2$). Therefore, \begin{align*} \mathrm{Res} (f(z); \iota ) & = \lim_{z \to \iota } (z - \iota ) \frac{\sin z }{(z - \iota )(z + 2)^2} \\ & = \frac{\sin \iota }{(\iota +2)^2} \\ & = \frac{\sin \iota }{1 + 4\iota }. \\[2ex] \mathrm{Res} (f(z); -2) & = \frac{1}{1!} \lim_{z \to -2} \frac{\mathrm{d}}{\mathrm{d}z} \left[ (z+2)^2 \frac{\sin z}{(z - \iota )(z + 2)^2} \right] \\ & = \lim_{z \to -2} \frac{\mathrm{d}}{\mathrm{d}z} \left( \frac{\sin z}{z - \iota } \right) \\ & = \lim_{z \to -2} \frac{(z - \iota ) \cos z - \sin z}{(z - \iota )^2} \\ & = \frac{(-2 - \iota )\cos (-2) - \sin (-2)}{(-2 - \iota )^2} \\ & = \frac{\sin 2 - (2 + \iota )\cos 2}{3 + 4\iota }. \end{align*}

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• $f(z) = \frac{\sin z- z}{z^4}$. It is clear that $z = 0$ is a pole of the given function. But the order of the pole is not immediate as $\sin z - z$ is also vanishing at $z = 0$. We will consider the expansion of $\sin z$. \begin{align*} \sin z - z & = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots - z\\ & = - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots. \end{align*} Thus, \begin{align*} \frac{\sin z - z}{z^4} & = -\frac{1}{3!z} + \frac{z}{5!} - \dots. \end{align*} Therefore, \[ \mathrm{Res} (f(z); 0) = -\frac{1}{3!} = -\frac{1}{6}. \]