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

Note :
Please Radius is limit |an / an+1 | as n approach to infinity
and general for of power series is ∑ [ an . (z-center )^n]

Answer 1

Answer on Question #50248 – Math – Complex Analysis

1) Given:

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

Find:

the Radius of convergence 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 at the point $z_0 = \infty$, so

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

b) Function $g(z) = f\left(\frac{1}{z}\right) = \frac{1}{z^2} + 2z^3 - 2$ has a singularity $z = 0$, which is a pole of order 2, hence function $f(z)$ has a singularity $z_0 = \infty$, which is a pole of order 2.

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

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

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