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}<br /> \sin z &amp; = \sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)!}z^{2n+1}<br /> \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:

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

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.