I need to expand the function $f(z)=\frac{1}{(z-a)^{k}}$ where $a \in \mathbb{C}$, $a \neq 0$, $k \in \mathbb{Z}$, $k>0$ in a Laurent series in the annuli
(a) $0< |z|<|a|$
(b) $|a|<|z|$
Note that earlier, I found the Laurent series for $\displaystyle g(z) = \frac{1}{(z-a)}$ to be $\displaystyle – \sum_{n=0}^{\infty}\frac{z^{n}}{a^{n+1}}$ for the annulus in part (a) and the laurent series for $g$ to be $\displaystyle \sum_{n=0}^{\infty}\frac{a^{n}}{z^{n+1}}$ for the annulus in part (b).
Since I have these results, I've been given the hint to use Weierstrass Theorem to take the derivative of $\displaystyle \frac{1}{z-a}$. The particular Weierstrass Theorem of which I speak is Weierstrass’ theorem on uniformly convergent series of analytic functions:
"If the series $\sum_{n=0}^{\infty} f_{n}(z) = f(z)$ is uniformly
convergent on every compact subset of a domain $G$, and if every term
$f_{n}(z)$ is analytic on $G$, then $f(z)$ is analytic on $G$.
Moreover, the series can be differentiated term by term any number of
times – i.e.,$ \displaystyle \sum_{n=0}^{\infty}f_{n}^{(k)}(z)= f^{(k)}(z)$, $\forall
> z \in G$, $\forall k \geq 0$,and each differentiated series is uniformly convergent on every
compact subset of $G$."
Do I need to show anything special about $\displaystyle g(z) = \frac{1}{z-a}$ before I can do that here? It's pretty obvious that it's analytic on both of the annuli, so the only problem we might run into is the uniformly convergent part. But it's a geometric series, and those are known to be uniformly convergent, are they not?
Once I have made sure all the conditions are met to apply the theorem, then, since the $k-1$st derivative gives us the denominator we want; i.e., $ \displaystyle f^{(k-1)}(z) = \frac{(-1)^{k-1}(k-1)!}{(z-a)^{k}}$, we have, by the theorem that $\displaystyle \frac{1}{(z-a)^{k}}= \frac{-1}{(-1)^{k-1}(k-1)!}\sum_{n=0}^{\infty}\frac{(z^{n})^{(k-1)}}{a^{n+1}}$ (where ^${(k-1)}$ means the $k-1$st derivative) for the annulus $0 < |z|<|a|$.
Is this correct?
It just looks awfully ugly…
So, (1) let me know if I need to prove any conditions in order to use this Weierstrass test, and (2), let me know if I have used it correctly for $0<|z|<|a|$ (I do realize it needs to be simplified a bit and have derivatives taken, but is the gist of what I'm supposed to do correct?)
Best Answer
Hint:
For $|z|\gt|a|$, use $$ \begin{align} \frac1{(z-a)^k} &=\frac1{z^k}\left(1-\frac az\right)^{-k}\\ &=\sum_{j=0}^\infty\binom{-k}{j}\frac{(-a)^{\,j}}{z^{\,j+k}}\\ &=\sum_{j=0}^\infty\binom{j+k-1}{k-1}\frac{a^{\,j}}{z^{\,j+k}}\\ \end{align} $$ and for $|z|\lt|a|$, use $$ \begin{align} \frac1{(z-a)^k} &=\frac1{(-a)^k}\left(1-\frac za\right)^{-k}\\ &=\sum_{j=0}^\infty\binom{-k}{j}\frac{z^{\,j}}{(-a)^{\,j+k}}\\ &=(-1)^k\sum_{j=0}^\infty\binom{j+k-1}{k-1}\frac{z^{\,j}}{a^{\,j+k}}\\ \end{align} $$