Series

## Question 1

a)Find the Taylor series and the radius of convergence for the Taylor series.

$sin{z}$ at z=0

Using:
$= sin{z} = z - \frac{z^3}{3!}+\frac{z^5}{5!}-\frac{z^7}{7!}+\cdots$

$= sin{z^2} = z^2 - \frac{z^(6)}{3!}+\frac{z^(10)}{5!}-\frac{z^(14)}{7!}+\cdots$

Radius of Convergence: R < infinity

b) Find the Taylor series and the radius of convergence for the Taylor series.

$\frac{1}{4-2z}$ at z=0

Using:
$\frac{1}{1-z}= 1+z+z^2+z^3+\cdots$

$\frac{1}{4-2z}= \frac{1}{4}(1+\frac{z}{2}+\frac{z^2}{4}+\frac{z^3}{8}+\cdots)$

$=(\frac{1}{4}+\frac{z}{8}+\frac{z^2}{16}+\frac{z^3}{32}+\cdots)$

Radius of Convergence: R=2 (from the center at $z=0$ out to the nearest singularity at $z=2$)

## Question 2

##### find the Laurent Series

++++2. b)
$\frac{1}{z(z-3)}$ on the annulus $0<\left | z \right |<3$

(1)
\begin{align} \left ( \frac{1}{z} \right )\left ( \frac{1}{z-3} \right )= \left ( \frac{-1}{3z} \right )\left ( \frac{1}{1-\frac{z}{3}} \right ) \end{align}
(2)
\begin{align} \left ( \frac{-1}{3z} \right )\left ( \frac{1}{1-\frac{z}{3}} \right )= \left ( \frac{-1}{3z} \right )\left ( 1+\frac{z}{3}+\frac{z^2}{9}+\frac{z^3}{27}+\cdots \right ) \end{align}
(3)
\begin{align} =\left ( \frac{-1}{3z}-\frac{1}{9}-\frac{z}{27}+\frac{z^2}{81}+\cdots \right ) \end{align}

#### c. $\frac{1}{z(z-3)}$ $\;\;\;$ on annulus $0<|z-3|<3$ $\;\;\;\;\;$ expand in powers of $z-3$.

(4)
\begin{align} \frac{1}{z(z-3)} = \frac{1}{z-3} \left [\frac{1}{z} \right] \end{align}
(5)
\begin{align} \frac{1}{z-3} \left [ \frac{1}{3+(z-3)} \right ] = \frac{1}{z-3} \left [\frac{1}{3}\left ( \frac{1}{1+\frac{z-3}{3}} \right ) \right] \end{align}
(6)
\begin{align} \frac{1}{3(z-3)} \left [ 1-\left(\frac{z-3}{3}\right)+\left(\frac{z-3}{3}\right)^{2} -\left(\frac{z-3}{3}\right)^{3}+\left(\frac{z-3}{3}\right)^{4}+\cdots \right ] \end{align}
(7)
\begin{align} \frac{1}{\, 3(z-3)}- \frac{1}{9}+\frac{(z-3)}{27}-\frac{(z-3)^{2}}{81} +\frac{(z-3)^{3}}{243}+\cdots \end{align}

## Question 3c

Classify Singularity
$z^{3}sin(\frac{1}{2})$
$z^{3}[ \frac{1}{z} - \frac{1}{3!z^{3}} +\frac{1}{5!z^{5}} -\frac{1}{7!z^{7}} +…]$
$z^{2} - \frac{1}{3!} + \frac{1}{5!z^{2}} - \frac{1}{7!z^{4}} + …$
Essential Singularity because infinite number of negative powers