TT2 Problem 5

[Victor Ivrii]

Consider $$f(z)= \frac{5}{(z-2)(z+3)}$$ and decompose it into Laurent's series converging

(a) As $|z|<2$;

(b)  As $2<|z|<3$;

(c) As $|z|>3$.

[Jihang Yu]

We have
\begin{equation}
f(z) = \frac{5}{(z-2)(z+3)} \\
Res(f;2)=\frac{5}{z+3}\bigg\rvert_{z=2} = 1 \\
Res(f;-3)=\frac{5}{z-2}\bigg\rvert_{z=-3} = -1
\end{equation}

Question a
As $\mid z \mid < 2$, $r=0, R=2$, so we have
\begin{equation}
a_k = -(2^{-k-1} \times 1+(-3)^{-k-1} \times -1) = -2^{-k-1} + (-3)^{-k-1}, k \geq 0 \\
f(z) =  \sum_{k=0}^{\infty} a_k z^k = \sum_{k=0}^{\infty} (-2^{-k-1}+(-3)^{-k-1}) z^k
\end{equation}

Question b
As $2 < \mid z \mid < 3$, $r=2, R=3$, so we have $$a_k =
\begin{cases}
2^{k-1} \times 1 & \text{for } k \leq -1 \\
-(-3)^{-k-1} \times (-1) = -3^{-k-1} & \text{for } k \neq 0
\end{cases}
$$
$$a_k =
\begin{cases}
2^{k-1} & \text{for } k \leq -1 \\
-3^{-k-1} & \text{for } k \neq 0
\end{cases}
$$
Therefore
\begin{equation}
f(z) = \sum_{k=-\infty}^{\infty} a_k z^k = \sum_{k=-\infty}^{-1} 2^{k-1} z^k + \sum_{0}^{\infty} -3^{-k-1}z^k
\end{equation}

Question c
As $\mid z \mid >3$, $r=3, R=\infty$. So we have
\begin{align*}
a_k &=2^{k-1} \times 1 + (-3)^{k-1} \times (-1) \qquad k \leq -1 \\
& = 2^{k-1} - (-3)^{k-1} \qquad k \leq -1
\end{align*}
Therefore Line above is correct, line below is wrong (substitution error). V.I.
\begin{equation}
f(z) = \sum_{k=-\infty}^{-1} a_k z^k = \sum_{k=-\infty}^{-1} (2^{k-1} -(-1)^{k-1}) z^k
\end{equation}

[Zechen Wang]

Here is my solution by partial fraction.

[Jihang Yu]

There is a typo in (b) where it should be k>=0 rather than k not equal to zero. Sorry about it.

You could correct in the text