Lecture 32
\((1)\) \(f(z) = \dfrac{1}{(z-i)^2}\) is already
a Laurent series
\(\quad\;\) about \(\,i\),
with
\(b_2 = 1,\qquad \quad \;\;\)
\(b_k = 0 \text{ for } k\neq 2\),
\(a_k = 0 \text{ } \forall k. \quad \quad \; \)
\((2)\) 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 \(c_n =\ds \sum_{n=0}^{\infty} a_k b_{n-k} \).
Example:
\(\ds \frac{e^z}{1+z} \) \(= e^z\cdot \dfrac{1}{1-(-z)}\)
\( \qquad \;= \left( 1+ z + \ds \frac{z^2}{2!} +\cdots \right) \left( 1 - z + z^2 - \cdots \right)\)
\( \qquad \;= 1+(-1+1)z + \left(1+ \frac{1}{2}-1\right)z^2 + \cdots \)
\( \qquad \;= 1+ \frac{1}{2}z^2 + \cdots \)
\((3)\) We can take term-by-term derivatives and
\(\quad \;\) integrals of series.
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 \[ \int_C f(x)dz = 2\pi i \sum_{j=1}^{k}\underset{z=z_j}{\res} f(z) . \]
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
\(\ds \int_Cf(z)dz = \sum_{j=1}^k \int_{C_j} f(z)dz\) \(= 2 \pi i \ds \sum_{j= 1}^k\underset{z=z_j}{\res} f(z) \).
Recall Example 7 from last lecture: \[I = \ds \int_C \frac{5z-2}{z(z-1)} dz\]
Note \(f(z) \) is analytic on \(\C \setminus\{0,1\}\).
Method 1: Look at \(0\lt |z|\lt 1\). Then
\(f(z) = \ds \frac{1}{1-z} (-1) \frac{5z-2}{2}\qquad \qquad\qquad \;\;\)
\(= \left(1+z+z^2+\cdots \right) (-1) \left(5 - \frac{2}{z}\right) \)
\(\implies \underset{z=0}{\res} f(z) = \text{coefficient of } \frac{1}{z}\) \(= 2\).
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 \(\;I = 2 \pi i (2+ 3)\) \(= 10 \pi i\).
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., \(\; I= 10 \pi i\).
If \(z_0\) is an isolated singularity of \(f\), then there exists \(R\) such that \(f\) has a Laurent series expansion on \(B_R(z_0)\setminus\{z_0\}\): \[f(z) = \sum_{n=0}^{\infty}a_n(z-z_0)^n + \sum_{n=1}^{\infty}b_n(z-z_0)^{-n}.\]
Case I: \(b_n=0 \;\forall n\).
Singularity is removable. By setting \(f(z_0) = a_0\) makes \(f\) analytic on \(B_R(z_0)\).
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.
Case III: Infinitely many of the \(b_n\)'s \( \neq 0\).
This is called an essential singularity.