Expanding into Power Series (Complex)

[curtdbz]

Homework Statement

Expand $$e^{1/z}/\sin z$$ in powers of $$z+1+i$$.

Homework Equations

Not sure, see below.

The Attempt at a Solution

I already know that

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

And the other expansion for the exponential (but we just replace the usual $$z \Rightarrow 1/z$$. Now when I do that I get two infinite sums, one on top the other. I also know that the power series is defined as:

$$\sum_{n=0} ^ {\infin } \frac {f^{(n)}(a)}{n!} \, (z-a)^{n}, a = -1 -i$$

The reason for the minus in our "a" is because we want to expand to powers of $$z+1+i$$ and so we need the negative. Anyway, when I differentiate I get no pattern that I can see and it just becomes a HUGE mess. Can someone help me clean it up, or just guide me? Thanks!

 

[jackmell]

I would treat it as

$$f(z)=e^{1/z}\csc(z)$$

and since you're expanding around an analytic center of both of these (z_0=-1-i), then the series for each of those functions have no singular terms so I could use the Cauchy product formula and write:

$$\begin{align*} f(z)=e^{1/z}\csc(z)&=\sum_{n=0}^{\infty}a_n(z-z_0)^n \sum_{n=0}^{\infty}b_n(z-z_0)^n\\ &=\sum_{n=0}^{\infty}\sum_{k=0}^n a_k b_{n-k}(z-z_0)^n \\ &=\sum_{n=0}^{\infty} c_n (z-z_0)^n \end{align*}$$

where:

$$c_n=\sum_{k=0}^n a_k b_{n-k}$$

We've now reduced it to a somewhat simpler form of finding the power series for each of e^{1/z} and csc(z) around the point z_0=-1-i.