# MATH3401

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 $$\C$$

Notes

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

Notes

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 $$0$$.