Question

1 )) Find the Radius of convergent of the power series
n from 0 to ∞ ∑ [2 ^ n / i ^ n] . { (z+ pi ) } ^n
2)) Determine whether ∞ is a singularity of f(z)=[z^2] + [2/(z^3)] - [ 2 ]
if its a singularity ,classify it

Answer 1

nswer on Question #50234 Math, Complex Analysis

1) Given:

$$\sum_{n=0}^{\infty} \frac{2^n}{i^n} (z + \pi)^n$$

Find:

the Radius of convergent of the power series

Solution:

$$a_n = \frac{2^n}{i^n}$$ $$R = \lim_{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right| = \lim_{n \to \infty} \left| \frac{2^n}{i^n} \cdot \frac{i^{n+1}}{2^{n+1}} \right| = \lim_{n \to \infty} \left| \frac{2^n \cdot i \cdot i^n}{i^n \cdot 2 \cdot 2^n} \right| = \left| \frac{i}{2} \right| = \frac{1}{2}$$

Answer: $R = \frac{1}{2}$

2) Given:

$$f(z) = z^2 + \frac{2}{z^3} - 2$$

Determine:

whether $z_0 = \infty$ is a singularity of $f(z)$ if its singularity, classify it

Solution:

a) function $f(z)$ is not analytic in the point $z_0 = \infty$, so

$z_0 = \infty$ is a singularity of $f(z)$

b) $\lim_{z \to \infty} (z^2 + \frac{2}{z^3} - 2) = \lim_{z \to \infty} \frac{z^5 - 2z^3 + 2}{z^3} = \infty \Rightarrow z_0 = \infty$ is a pole of order 2,

because $f(z) = z^2 + \frac{2}{z^3} - 2 = z^2\left(1 + \frac{2}{z^5} - \frac{2}{z^2}\right) = z^m g(z)$, where $m$ is the order of pole, when $g(\infty) \neq 0$.

Answer: $z_0 = \infty$ is a singularity of $f(z)$, $z_0 = \infty$ is a pole of order 2.

www.assignmentexpert.com