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I'm trying to find the residue of $$z \cos\left(\frac{1}{z}\right)$$ at $z=0$.

This is how I did it:

$\cos(z)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}z^{2n}$. $\\$ Then $\cos(\frac{1}{z})=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}z^{2n-1}$. $\\$ Finally, $z\cos(z)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}z^{2n}$, which is where I started. I thought the residue would by $0$ because there are no negative powers, but it's supposed to be $-\frac{1}{2}$. Did I do something incorrectly?

DonAntonio
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Alti
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    If you plug in $\frac{1}{z}$ into the definition of $\cos(z)$ you don't get the sum that you meantion. It would be: \begin{align} \cos(\frac{1}{z}) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} (\frac{1}{z})^{2n} =\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} (z)^{-2n} \end{align} – Thomas Apr 04 '13 at 19:00
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    $\cos(\frac{1}{z})=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}z^{-2n}$. –  Apr 04 '13 at 19:01
  • @macydanim Oh yeah, wow I totally zoned out on that. Thanks! – Alti Apr 04 '13 at 19:01

1 Answers1

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The residue is the coefficient of the $z^{-1}$ term in the Laurent Series.

As you probably already know:

$$\cos z \equiv 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} \pm \cdots $$

It follows that

$$\cos\left(z^{-1}\right) \equiv 1 - \frac{z^{-2}}{2!} + \frac{z^{-4}}{4!} - \frac{z^{-6}}{6!} \pm \cdots $$

Finally, multiplying through by $z$ gives:

$$z\cos\left(z^{-1}\right) \equiv z - \frac{z^{-1}}{2!} + \frac{z^{-3}}{4!} - \frac{z^{-5}}{6!} \pm \cdots $$

The coefficient of $z^{-1}$ is $-\frac{1}{2!} = -\frac{1}{2}$.

Fly by Night
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