Expand $f(z)=\frac{1}{z^2(z-i)}$ in 2 different Laurent expansions around $z=i$ and tell where each converges.

Question

My attempt:

$$f(z)=\frac{1}{z^2(z-i)}$$ $$\frac{1}{z^2(z-i)}=\frac{Az+B}{z^2}+\frac{C}{(z-i)}$$

Solving for the unknown constants yields

$$A=1$$ $$B=i$$ $$C=-1$$

Thus,

$$f(z)=\frac{z+i}{z^2} - \frac{1}{(z-i)}$$

There are singularities at $z=0$ and $z=i$, but this is where I get stuck. How do the singularities relate to the part of the question that says

Expand...around $z=i$

?

the rest of my answer is:

for $|z|<i$: $$\frac{z+i}{z^2}-i\sum_{n=0}^\infty (-1)^n(zi)^n$$

$$=\frac{z+i}{z^2} -i[1-zi-z^2+z^3i+...]$$

---

The given answer to this problem is

for $0<|z-i|<1$: $$f(z)=-\frac{1}{z-i}-2i+3(z-i)-4i(z-i)^2+...$$ for $|z-i|>1$:

$$...\frac{4i}{(z-i)^6}-\frac{3}{(z-i)^5}-\frac{2i}{(z-i)^4}+\frac{1}{(z-i)^3}$$

Answer 1

When you expand a given holomorphic function in Laurent series on a set of the form $\Delta=\{z:~r< |z-z_0|< R\}$, where $0\leq r<R\leq +\infty$, by definition you should have a series of the form $\displaystyle \sum_{n=-\infty}^{+\infty}a_n(z-z_0)^n$, where $z_0$ is the center of $\Delta$. Since you are asked to expand around $i$, there will be two regions, $\Delta_1=\{z:~0<|z-i|<1\}$ and $\Delta_2=\{z:~1<|z-i|<+\infty\}$ (your function has singularities on $i,~0$), so these regions can not contain either. Note that your series is expanded as having center $0$, not $i,$ therefore you have answered a different problem.

Answer 2

In your attempt, you gave series in $z$, but they must be series in $z-i$. Here’s my way of doing the job, purely formally.

First, to simplify notation, let $z-i=w$ and $z=w+i$, when your function becomes $$ \frac1{(w+i)^2w}=\frac{-1}w\frac1{(1-iw)^2}\,. $$ Now, using the expansion $1/(1-t)^2=1+2t+3t^2+4t^3+\cdots$, you get your expansion to be $(-1/w)(1 + 2iw - 3w^2 - 4iw^3 + 5w^4 +\cdots)=-w^{-1}-2i+3w+4iw^2-5w^3-\cdots$. Notice that the common ratio is $iw$ in the limit, and you need $|iw|<1$ for convergence. So when you replace $w$ by $z-i$, this expansion is good for $0<|z-i|<1$.

But we also want a Laurent series convergent for $|z-i|=|w|>1$. This must be a series in $w^{-1}$. Again, for convenience, let $w=1/u$, $u=1/w$, and expand our function $$ \frac1{(w+i)^2w}=\frac1{(\frac1u+i)^2\frac1u}=\frac{u^3}{(1+iu)^2}\,, $$ which becomes $u^3(1 - 2iu - 3u^2 + 4iu^3 + 5u^4 -\cdots)=w^{-3}-2iw^{-4}-3w^{-5}+4iw^{-6}+\cdots\,$. Again you’ll make the resubstitution $w=z-i$, to get a series that’s convergent for $|1/z-i|<1$, that is, $|z-i|>1$.

Notice that partial fraction decomposition has no part to play in this method.