Complex Analysis

Lecture 31

Taylor series in \(\C\)

Example 3: Find Maclaurin series for \(f(z) = \dfrac{1}{1-z}\).

Note \(f\) is entire for \(|z|\lt 1\), indeed, on \(\C\setminus \{1\}\): and \[ f^{(n)}(z) = \dfrac{n!}{(1-z)^{n+1}}, z\neq 1. \]

\(\implies f^{(n)}(0) = n!\) \(\implies \) Taylor series of \(f\) at \(0\),

Taylor series in \(\C\)

Example 3 (cont): \(T_{f,0}\) is given by \[ T_{f,0}(z) = \sum_{n=0}^{\infty} z^n, \] where this converges. \(\Lambda = \ds \lim_{n\ra \infty } \frac{1}{1} = 1 \implies R=1\)

Theorem (Lect 30) \(\implies T_{f,0}\) converges to \(f\) for \(|z|\lt 1\).

Taylor series in \(\C\)


  1. Geometric series formula \(\implies\) \[ \frac{1}{1-z} =\sum_{n=0}^{\infty}z^n \text{ for } |z|\lt 1. \]
  2. Series converges "out to the first singularity",
    which here is \(1\).

Taylor series in \(\C\)

Example 4: Find Maclaurin series for \(f(z) = \dfrac{1}{2+ 4z}\) on \(\C\setminus \{1/2\}\).

Note that

\(\ds \frac{1}{2+4z} = \frac{1/2}{1+2z}\) \(\ds =\frac{1/2}{1-(-2z)}\qquad \qquad \)

\(\quad = \ds\frac{1}{2} \sum_{n=0}^{\infty} \left(-2z\right)^{n}\) for \(|-2z|\lt 1\)

\(\qquad \; = \ds\frac{1}{2} \sum_{n=0}^{\infty} (-1)^{n}2^{n-1}z^n\,\) for \(\,|z|\lt 1/2\).

Taylor series in \(\C\)

Example 5:   \(f(z) = \dfrac{1+2z^2}{z^3+z^5}\) \( = \ds \frac{1}{z^3} \left( \frac{2+2z^2}{1+z^2} - \frac{1}{1+z^2} \right)\)

\(\qquad \qquad\;\; \quad = \ds \frac{1}{z^3} \left( 2 - \frac{1}{1+z^2} \right)\), which is analytic on \(\C\setminus\{0, \pm i\}\).

For \(|z|\lt 1\), \(\ds \frac{1}{1+z^2} = \sum_{n=0}^{\infty} (-1)^nz^{2n}\). So, for \(0\lt |z|\lt 1\),

\(f(z) = \ds \frac{1}{z^3} \left( 2- \left( 1-z^2+z^4-\cdots \right) \right)\)

\( = \ds \frac{1}{z^3} + \frac{1}{z} -z + z^3 - \cdots\)

Laurent Series

Theorem: Let \(f\) be analytic on the open annulus \[A= \{z: r_1 \lt |z-z_0|\lt r_2\}.\]

Let \(C\) be a positively oriented, simple closed curve in \(A\), \(z_0\in \text{Int}\, C\).

Then \(f\) has a series representation (Laurent series)

\(f(z)=\ds\sum_{n=0}^{\infty} a_n(z-z_0)^n+\ds\sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n}\)   on  \(A\),

where:  \( a_n=\ds\frac{1}{2\pi i} \int_C \frac{f(\zeta)}{(\zeta-z_0)^{n+1}}d\zeta\)   and   \(b_n=\ds \frac{1}{2\pi i}\int_C \frac{f(\zeta)}{(\zeta-z_0)^{-n+1}}d\zeta\).

Laurent Series

We can rewrite the Laurent series as \[f(z) = \sum_{n=-\infty}^{\infty} c_n(z-z_0)^n\] where \[c_n = \frac{1}{2 \pi i} \int_C \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}d\zeta.\]

Laurent Series

In particular:

\(b_1 = \ds \frac{1}{2 \pi i} \int_C \frac{f(\zeta)}{(\zeta - z_0)^{-1+1}}d\zeta \)

\(= \ds \frac{1}{2 \pi i} \int_Cf(\zeta)d\zeta. \;\)

This is called the residue of \(\,f\) at \(z_0\): \( \;\underset{z=z_0}{\res} f(z)\).

Laurent Series


  1. If \(f\) is analytic on and in \(C\), then all the \(b_n\)'s are zero.
  2. \(r_1 \searrow 0 \), \( r_2 \nearrow \infty\) are ok.
  3. Taylor and Laurent series are unique.

Laurent Series

Example 6: Find the Laurent series for \(e^{1/z}\):

\(e^{1/z} = \ds \sum_{n=0}^{\infty} = \frac{1}{n!z^n} \) \(= 1 + \ds \frac{1}{z} + \frac{1}{2!z^2} + \cdots \)

for all \(|z|\gt 0\), and note \[\ds \frac{1}{2\pi i } \int_C e^{1/\zeta}d\zeta = b_1 = 1.\]

Laurent Series

Example 7: Compute \(I = \ds \int_C \frac{5z-2}{z(z-1)} dz\), where \( C = 2 e^{i \theta } \), \( \theta \in [0,2\pi]\).

Note \(f(z) = \ds \frac{5z-2}{z(z-1)} \) is analytic on \(\C \setminus\{0,1\}\).

Solution coming soon!

Residues and Poles

\(f: \Omega \ra \C\) has a singularity at \( z_0\) if:

  1. \(f\) is not analytic at \(z_0\);
  2. Given any \( \varepsilon \gt 0 \) \( \exists\, z_1\in B_{\varepsilon}(z_0)\) such that \(f\) is analytic at \(z_1\).

Residues and Poles

\(f\) has an isolated singularity at \( z_0\) if \(f\) is analytic on \[B_{\varepsilon}(z_0)\setminus\{z_0\} \text{ for some } \varepsilon \gt 0.\]

E. g.  \(f\) from Example 7 has isolated singularities at \(0, 1\);

\(\Log z\) has non-isolated singularities on the negative real axis \(union\) \(\{0\}\);

Residues and Poles

\(\ds \frac{\sin z}{z}\) has one isolated singularity at \(0\);

\(\ds \frac{1}{\sin\left(\frac{\pi}{z}\right)}\) has isolated singularities at \(z=\dfrac{1}{k}, k\in \Z\)

\(\qquad \quad \)and a non-isolated singularity at \(0\).