Expanding into Power Series (Complex)

  • Thread starter curtdbz
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  • #1
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Homework Statement


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


Homework Equations


Not sure, see below.


The Attempt at a Solution


I already know that

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

And the other expansion for the exponential (but we just replace the usual [tex] z \Rightarrow 1/z [/tex]. 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:

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

The reason for the minus in our "a" is because we want to expand to powers of [tex]z+1+i[/tex] 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!
 

Answers and Replies

  • #2
1,796
53
I would treat it as

[tex]f(z)=e^{1/z}\csc(z)[/tex]

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:

[tex]
\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*}
[/tex]

where:

[tex]c_n=\sum_{k=0}^n a_k b_{n-k}[/tex]

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.
 

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