Question

f(z) = sin ( 1/(z-pi)) / 2z
1- Find the singularities of f , and all possible annulus centered at each singularity
2- Find the laurent series of f in each annulus
3- classify each singularity
4- compute the residue of at each singular point

Answer 1

50047

Given:

$$f(z) = \frac{\sin\left(\frac{1}{z - \pi}\right)}{2z}$$

1) Find the singularities of $f$, and all possible annulus centered at each singularity

2) Find the laurent series of $f$ in each annulus

3) classify each singularity

4) compute the residue of at each singular point

Solution:

1) singularities is those points wherever a function is analytical, so we obtain

$z = 0$ isolated singularity

$z = \pi$ isolated singularity

$z = \infty$ singularity

2) the Laurent series in each annulus is

$$f(z) = \frac{\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n + 1)!} z^{2n+1}}{2z} = \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n + 1)!} z^{2n}$$

3) classifying of each singularity

a) $z = 0$

$$\lim_{z \to 0} \frac{\sin\left(\frac{1}{z - \pi}\right)}{2z} = \infty \quad \Rightarrow \quad z = 0 \quad \text{simple pole}$$

b) $z = \pi$

$$\lim_{z \to \pi} \frac{\sin\left(\frac{1}{z - \pi}\right)}{2z} = \frac{1}{2\pi} \sin\left(\frac{1}{0}\right) \quad \exists \quad \Rightarrow \quad z = \pi \quad \text{substantially singularity}$$

c) $z = \infty$

$$\lim_{z \to \infty} \frac{\sin\left(\frac{1}{z - \pi}\right)}{2z} = 0 \quad \Rightarrow \quad z = \infty \quad \text{removable singularity}$$

4) the residue at each singular point

a) $z = 0$

$$\operatorname{Res}_{z=0} f(z) = \lim_{z \to 0} (z - 0)^1 \cdot f(z) = \lim_{z \to 0} z \cdot \frac{\sin\left(\frac{1}{z - \pi}\right)}{2z} = \frac{1}{2} \sin\left(-\frac{1}{\pi}\right)$$

b) $z = \pi$

$$\underset {z = \pi} {\operatorname {R e}} s f (z) = c _ {- 1} = 0 \quad \text {because} \quad z = \pi \quad \text {substantially singularity}$$

c) $z = \infty$

$$\underset {z = \infty} {\operatorname {R e}} s f (z) = - \left(\underset {z = 0} {\operatorname {R e}} s f (z) + \underset {z = \pi} {\operatorname {R e}} s f (z)\right) = - \frac {1}{2} \sin \left(- \frac {1}{\pi}\right)$$

www.assignmentexpert.com