Q5 TUT 5301

[Victor Ivrii]

Find the power-series expansion about the given point for the given function; find the largest disc in which the series is valid:
 $$\frac{z+2}{z+3}\qquad\text{about}\; z_0 = -1.$$

[Tianfangtong Zhang]

\begin{align*}
 \frac{z+2}{z+3} &= \frac{z+3-1}{z+3} \\ &= 1-\frac{1}{z+3}\\
 &= 1- \frac{1}{z+1+2} \\ &= 1 - \frac{1}{2} \frac{1}{1+\frac{z+1}{2}}\\
 &= 1 - \frac{1}{2} \frac{1}{1-\frac{-(z+1)}{2}}\\
 &= 1 - \sum_{n=0}^{\infty}(\frac{-(z+1)}{2})^n \\
 &= 1 - \sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}(z-(-1))^n
\end{align*}

[Ende Jin]

For the former one, the largest disc is $\{z: |z + 1| < 2\}$

For the latter one, the largest disc is the whole complex plane.