# MATH3401

Lecture 32

### Remarks on series

$$(1)$$   $$f(z) = \dfrac{1}{(z-i)^2}$$ is already a Laurent series
$$\quad\;$$ about $$\,i$$, with

$$b_2 = 1,\qquad \quad \;\;$$

$$b_k = 0 \text{ for } k\neq 2$$,

$$a_k = 0 \text{ } \forall k. \quad \quad \;$$

### Remarks on series

$$(2)$$   Cauchy product of series:
Suppose   $$f(z) = \ds \sum_{n=0}^{\infty} a_n z^n$$, $$g(z) = \ds \sum_{n=0}^{\infty} b_n z^n$$. Then

$$\left(\, f g \right)(z) = \ds\sum_{n=0}^{\infty} c_n z^n$$,   where   $$c_n =\ds \sum_{n=0}^{\infty} a_k b_{n-k}$$.

### Remarks on series

Example:

$$\ds \frac{e^z}{1+z}$$ $$= e^z\cdot \dfrac{1}{1-(-z)}$$

$$\qquad \;= \left( 1+ z + \ds \frac{z^2}{2!} +\cdots \right) \left( 1 - z + z^2 - \cdots \right)$$

$$\qquad \;= 1+(-1+1)z + \left(1+ \frac{1}{2}-1\right)z^2 + \cdots$$

$$\qquad \;= 1+ \frac{1}{2}z^2 + \cdots$$

### Remarks on series

$$(3)$$   We can take term-by-term derivatives and
$$\quad \;$$ integrals of series.

### Cauchy residue Theorem

Suppose $$C$$ is a positively oriented simple closed curve and that $$f$$ is analytic on $$C\cup \{\Int(C)\setminus \{z_1,\ldots, z_k\}\}$$. Then $\int_C f(x)dz = 2\pi i \sum_{j=1}^{k}\underset{z=z_j}{\res} f(z) .$

### Cauchy residue Theorem

Proof: Take disjoint positively oriented circles $$C_1, \ldots , C_k$$ around each $$z_1, \ldots , z_k$$ with disjoint interiors, all lying in $$\Int(C)$$. The contours $$C, C_1,\ldots , C_k$$ form the boundary of a multiply-connected domain $$\Omega$$. Then $$f$$ is analytic on $$\Omega$$ and $$\partial \Omega$$, so the Cauchy-Goursat extension implies

$$\ds \int_Cf(z)dz = \sum_{j=1}^k \int_{C_j} f(z)dz$$ $$= 2 \pi i \ds \sum_{j= 1}^k\underset{z=z_j}{\res} f(z)$$.

### Cauchy residue Theorem

Recall Example 7 from last lecture: $I = \ds \int_C \frac{5z-2}{z(z-1)} dz$

Note $$f(z)$$ is analytic on $$\C \setminus\{0,1\}$$.

### Cauchy residue Theorem

Method 1: Look at $$0\lt |z|\lt 1$$. Then

$$f(z) = \ds \frac{1}{1-z} (-1) \frac{5z-2}{2}\qquad \qquad\qquad \;\;$$

$$= \left(1+z+z^2+\cdots \right) (-1) \left(5 - \frac{2}{z}\right)$$

$$\implies \underset{z=0}{\res} f(z) = \text{coefficient of } \frac{1}{z}$$ $$= 2$$.

### Cauchy residue Theorem

Method 1 (cont): Now look at $$z$$ such that $$0\lt |z-1|\lt 1$$. Then

$$f(z) = \ds \frac{5(z-1)+3}{z-1} \cdot \frac{1}{(z-1)+1} \qquad \quad$$

$$= \left( 5+\frac{3}{z-1} \right) \dfrac{1}{1-(-(z-1))} \qquad$$

$$= \left( 5+\frac{3}{z-1} \right) \left( 1+(-(z-1))+\cdots \right)$$

$$\implies \underset{z=1}{\res} f(z) = \text{coefficient of } \frac{1}{z-1}$$ $$= 3$$.

Cauchy residue theorem implies $$\;I = 2 \pi i (2+ 3)$$ $$= 10 \pi i$$.

### Cauchy residue Theorem

Method 2: Note $$\;f(z) = \ds \frac{5z-2}{z(z-1)} =\frac{3}{z-1} + \frac{2}{z}$$.

$$\ds \frac{2}{z}$$ is analytic on $$\C\setminus \{0\}$$. $$\;\ds \frac{3}{z-1}$$ is analytic on $$\C\setminus \{0\}$$.

Therefore $$\;\underset{z=0}{\res} f(z) = 2\;$$ and $$\;\underset{z=1}{\res} f(z) = 3$$.

We get the same result, i.e., $$\; I= 10 \pi i$$.

### Classifying isolated singularities

If $$z_0$$ is an isolated singularity of $$f$$, then there exists $$R$$ such that $$f$$ has a Laurent series expansion on $$B_R(z_0)\setminus\{z_0\}$$: $f(z) = \sum_{n=0}^{\infty}a_n(z-z_0)^n + \sum_{n=1}^{\infty}b_n(z-z_0)^{-n}.$

### Classifying isolated singularities

Case I: $$b_n=0 \;\forall n$$.

Singularity is removable. By setting $$f(z_0) = a_0$$ makes $$f$$ analytic on $$B_R(z_0)$$.

### Classifying isolated singularities

Case II: At least one, but only finitely many of the $$b_n$$'s are nonzero.

Such a point is called a pole. Define $m = \max \{ n: b_n\neq 0\}$

$$m$$ is the order of the pole. $$\;m=1 \iff$$ simple pole.

### Classifying isolated singularities

Case III: Infinitely many of the $$b_n$$'s $$\neq 0$$.

This is called an essential singularity.