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Problem:find the residue Res$(f(z),\infty)$ for $f(z)$ as follows

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

I tried two methods to solve the problem,but obtained two different results.

The first method was like this:

enter image description here

Because Res$(f(z),\infty)$+Res$(f(z),1)$+Res$(f(z),-1)$=0,then I get the answer:

Res$(f(z),\infty)=-\sinh1$

The second method was like this:

Res$(f(z),\infty)$=-Res$(\frac{1}{z^2}f(1/z),0)$ $$ \frac{1}{z^2}f(1/z)=\frac{e^{1/z}}{1-z^2}=(1+z^2+z^4+z^6+...)(1+1/z+1/(2!z^2)+1/(3!z^3)+...) $$

Then $b_1=1$

so Res$(f(z),\infty)=-1$

I don't know where went wrong? I would appreciate it if you could help me find the reason.Thank you!

wit
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1 Answers1

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Let's look again at the product you got at the end of your second approach:

$$\frac{e^{1/z}}{1-z^2} = (1 + z^2 + z^4 + \ldots)\left(1 + \frac{1}{z} + \frac{1}{2z^2} + \frac{1}{6z^3} + \ldots\right)$$

Now notice when we multiply out the last two given terms in each polynomial, we get another $\frac{1}{z}$ term, with a coefficient $\frac{1}{3!}.$ This is where your answer goes wrong, it fails to account for other terms which give multiples of $\frac{1}{z}.$

In general, any term of the form $z^{2k}$ from the first sum will multiply with a term of the form $\frac{1}{z^{2k+1}(2k+1)!}$ in the second sum to give a term $\frac{1}{(2k+1)! z},$ for any whole number $k.$ These should be the only such terms to multiply to a multiple of $\frac1{z}$ because given any term in the first sequence there will be no other term with the correct power.

So, the correct residue at infinity is $-\sum_{k=0}^{\infty} \frac{1}{(2k+1)!} = -\sinh(1).$

Stephen Donovan
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