Remark: f(z) = 1/(z-i)^2 is already a Laurent series about i with b_2 = 1 and other coefficients zero.
§73 (8 Ed §67)
Theorem. Suppose f(z) = \sum_{n=0}^\infty a_n z^n and g(z) = \sum_{n=0}^\infty b_nz^n converge. Then, (fg)(z) = \sum_{n=0}^\infty c_nz^n where c_n = \sum_{k=0}^n a_k b_{n-k}.
Example: Consider for |z|<1, \begin{aligned} \frac {e^z}{1+z} = e^z \frac{1}{1-(-z)} &= \left(1 + z + \frac {z^2}{2!} + \cdots\right)\left(1-z+z^2-\cdots\right) \\ &= 1 + (-1+1)z + (1+1/2-1)z^2 + \cdots \\ &= 1 + \frac{z^2}2 + \cdots \end{aligned} Remark: We can take term-by-term derivatives and integrals of series (see §71 or 8 Ed §65).
Theorem. Suppose C is a positively oriented simple closed curve and that f is analytic in and on C except at finitely many isolated points \{z_1, z_2, \ldots, z_k\}. Then, \int_C f(z)\,dz = 2\pi i \sum_{j=1}^k \underset{z=z_j}{\operatorname{res}}f(z). Proof. Take disjoint positively oriented circles C_1, \ldots, C_k around each z_1, \ldots, z_k with disjoint interiors, all lying in the interior of C. Then, C, C_1, \ldots, C_k form the boundary of a multiply-connected domain \Omega. Then, f is analytic on \Omega and its boundary, so the Cauchy-Goursat extension implies that \int_C f(z)\,dz = \sum_{j=1}^k \int_{C_j}f(z)\,dz = 2\pi i \sum_{j=1}^k \operatorname{res}_{z=z_j}f(z). \quad \square Example: Apply this to example 7 from last lecture. Note that f is analytic on \mathbb C \setminus \{0,1\} and C is the circle of radius 2 around the origin. I = \int_C \frac{5z-2}{z(z-1)}\,dz. There are two methods we can try. First, look close to zero with 0<|z|<1 to see that \begin{aligned} f(z) = -\frac{1}{1-z}\cdot\frac{5z-2}{z} &= (-1)(1+z+z^2+\cdots)(5-2/z) \\ \implies \underset{z=0}{\operatorname{res}}f(z) &= 2 \end{aligned} Now, we also need a Laurent series around 1 (which is even less fun). We can write \begin{aligned} f(z) &= \frac{5(z-1)+3}{z-1}\cdot\frac{1}{1-(-(z-1))} \\ &= \left(5+\frac3{z-1}\right)(1+(-(z-1))+(-(z-1))^2+\cdots) \\ &\qquad\implies \underset{z=1}{\operatorname{res}}f(z) = 3 \end{aligned} Because the only coefficient of (z-1)^{-1} comes from the 3/(z-1).
Alternatively, we can use partial fractions. Note that f can be decomposed into parts which are analytic on \mathbb C \setminus \{1\} and \mathbb C \setminus \{0\}, respectively. f(z) = \frac{3}{z-1}+\frac 2 z Therefore, this is its own Laurent series around the points 0 and 1. The other fraction is analytic near the other singularity, so does not affect the Laurent series. Therefore, \operatorname{res_{z=0}}f(z) = 2 \qquad \operatorname{res}_{z=1}=3. Note that the expression 3/(z-1) describes, in some sense, the prototypical singularity of a whole family of functions.
If z_0 is an isolated singularity of f (i.e. f analytic on a deleted neighbourhood), then there exists R such that f has a Laurent series expansion on B_R(z_0) \setminus \{z_0\} given by f(z) = \sum_{n=0}^\infty a_n(z-z_0)^n + \sum_{n=1}^\infty b_n(z-z_0)^{-n}. There are three cases to consider.